MI IV — Equation Bank

All 295 equations from the five units, in teaching order. Designed for whiteboard recall + phone reference. Tap any card's Show more for derivation, worked example, and the visual cue.

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Trigonometry

52 equations

Right-Triangle Trig (SOH-CAH-TOA)

Trig Fundamentals
$$ \sin\theta = \frac{\text{opp}}{\text{hyp}}, \quad \cos\theta = \frac{\text{adj}}{\text{hyp}}, \quad \tan\theta = \frac{\text{opp}}{\text{adj}} $$

In words. In a right triangle, sine, cosine, and tangent of an acute angle equal opposite-over-hypotenuse, adjacent-over-hypotenuse, and opposite-over-adjacent respectively.

Domain / range. Defined for acute angles in a right triangle: 0 < θ < π/2.

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Derivation
Definitional. The three primary trig ratios are read off the legs and hypotenuse of a right triangle relative to the chosen acute angle. SOH-CAH-TOA is the mnemonic that locks the ratios in place.
Example
Triangle with legs 3, 4 and hypotenuse 5. For the angle opposite the leg of length 3: sin θ = 3/5, cos θ = 4/5, tan θ = 3/4.
Picture
Right triangle with angle θ marked; opposite leg, adjacent leg, and hypotenuse labeled.

Circle Trig (x, y, r definitions)

Trig Fundamentals
$$ \sin\theta = \frac{y}{r}, \quad \cos\theta = \frac{x}{r}, \quad \tan\theta = \frac{y}{x} $$

In words. For a point (x, y) on the terminal side of angle θ at distance r from the origin, sine is y/r, cosine is x/r, and tangent is y/x.

Domain / range. θ is any real number (extends right-triangle trig to all angles, including obtuse and negative).

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Derivation
Extend SOH-CAH-TOA by placing the angle in standard position with vertex at the origin and initial side on the positive x-axis. The terminal side intersects a circle of radius r at point (x, y). Signs of x and y track the quadrant of θ.
Example
Point (4, -3) on terminal side. r = √(16+9) = 5. sin θ = -3/5, cos θ = 4/5, tan θ = -3/4. (Q4, so sin negative, cos positive.)
Picture
Coordinate plane with terminal ray from origin through (x, y); r is the length of the ray, theta is the angle from positive x-axis.

Unit Circle Point Coordinates

Trig Fundamentals / WS8
$$ P(\theta) = (\cos\theta,\; \sin\theta) $$

In words. On the unit circle, the point at angle θ from the positive x-axis has coordinates (cos θ, sin θ).

Domain / range. θ is any real number; coordinates always satisfy x² + y² = 1.

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Derivation
Special case of circle trig with r = 1. Sine and cosine are literally the y and x coordinates of the point obtained by rotating (1, 0) counter-clockwise by θ.
Example
At θ = π/3 (60°), the unit-circle point is (cos 60°, sin 60°) = (1/2, √3/2).
Picture
Unit circle centered at origin; ray from origin at angle θ meets circle at point (cos θ, sin θ).

Reciprocal Identities

Trig Fundamentals / WS8
$$ \csc\theta = \frac{1}{\sin\theta}, \quad \sec\theta = \frac{1}{\cos\theta}, \quad \cot\theta = \frac{1}{\tan\theta} = \frac{\cos\theta}{\sin\theta} $$

In words. Cosecant, secant, and cotangent are the reciprocals of sine, cosine, and tangent respectively.

Domain / range. csc undefined where sin = 0; sec undefined where cos = 0; cot undefined where sin = 0.

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Derivation
Definitional. The three secondary trig functions are introduced as reciprocals of the primary three. The identity cot θ = cos θ / sin θ follows by dividing both sides of tan θ = sin θ / cos θ by the original ratio.
Example
If sin θ = 3/5, then csc θ = 5/3. If cos θ = 4/5, then sec θ = 5/4. If tan θ = 3/4, then cot θ = 4/3.
Picture
Three pairs of reciprocal-bonded labels: sin/csc, cos/sec, tan/cot.

Quotient Identities

Trig Fundamentals / WS8
$$ \tan\theta = \frac{\sin\theta}{\cos\theta}, \quad \cot\theta = \frac{\cos\theta}{\sin\theta} $$

In words. Tangent equals sine over cosine; cotangent equals cosine over sine.

Domain / range. tan undefined where cos = 0 (i.e., θ = π/2 + πk). cot undefined where sin = 0 (i.e., θ = πk).

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Derivation
From the unit-circle definitions sin θ = y, cos θ = x: tangent is rise/run = y/x = sin θ / cos θ. Cotangent is the reciprocal.
Example
θ = π/3: sin = √3/2, cos = 1/2, so tan = (√3/2)/(1/2) = √3. cot = 1/√3 = √3/3.
Picture
Triangle of identities: sin and cos on top, with tan and cot as ratios connecting them.

Special-Angle Exact Values

Trig Fundamentals
$$ \sin(\pi/6)=\tfrac{1}{2},\ \sin(\pi/4)=\tfrac{\sqrt2}{2},\ \sin(\pi/3)=\tfrac{\sqrt3}{2};\ \cos(\pi/6)=\tfrac{\sqrt3}{2},\ \cos(\pi/4)=\tfrac{\sqrt2}{2},\ \cos(\pi/3)=\tfrac{1}{2} $$

In words. Sine and cosine of 30°, 45°, and 60° in their exact radical form. Tangent is the corresponding ratio.

Domain / range. Reference angles 30°, 45°, 60° in the first quadrant; extend to other quadrants using reference-angle rules and quadrant sign chart.

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Derivation
From a 30-60-90 right triangle with sides 1, √3, 2: sin 30° = 1/2, cos 30° = √3/2. From a 45-45-90 isoceles right triangle with legs 1, 1, hypotenuse √2: sin 45° = cos 45° = √2/2.
Example
tan(π/3) = sin(π/3)/cos(π/3) = (√3/2)/(1/2) = √3. tan(π/6) = (1/2)/(√3/2) = 1/√3 = √3/3.
Picture
Two reference right triangles: 30-60-90 with sides 1-√3-2, and 45-45-90 with sides 1-1-√2.

Reference-Angle Rule

Trig Fundamentals
$$ |\text{trig}(\theta)| = |\text{trig}(\theta_{\text{ref}})|, \text{ where } \theta_{\text{ref}} \text{ is the acute angle between the terminal side and the x-axis} $$

In words. The absolute value of any trig function at angle θ equals the same trig function evaluated at its reference angle; the sign comes from the quadrant.

Domain / range. Reference angle θ_ref is always in [0, π/2] (or [0°, 90°]). Quadrant sign chart: Q1 all positive; Q2 only sin/csc positive; Q3 only tan/cot positive; Q4 only cos/sec positive ("All Students Take Calculus").

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Derivation
Trig functions are determined by the coordinates of the terminal-side point on the unit circle. Reflections across the axes preserve the absolute value of the coordinates but flip signs. So |sin θ| = |sin θ_ref|; the sign of sin θ matches the sign of y in the appropriate quadrant.
Example
θ = 210° is in Q3 with reference angle 30°. sin(210°) = -sin(30°) = -1/2 (negative because Q3 has negative y).
Picture
Unit circle with terminal side in Q3; reference triangle dropped to the x-axis showing the acute reference angle.

SAS Area Formula

WS1 SAS Area
$$ K = \tfrac{1}{2}ab\sin C = \tfrac{1}{2}bc\sin A = \tfrac{1}{2}ac\sin B $$

In words. The area of any triangle equals one-half the product of two sides times the sine of the included angle, and this can be written three equivalent ways depending on which pair of sides plus included angle you pick.

Domain / range. C must be an interior angle of a triangle: 0 < C < π.

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Derivation
In triangle ABC with sides a, b and included angle C, drop an altitude from the third vertex. The altitude has length h = b sin C (right-triangle trig on the small right triangle). Then area = (1/2)(base)(height) = (1/2)(a)(b sin C). By choosing different vertex for the altitude, the formula re-labels to the other two forms.
Example
Triangle with sides a = 70 m, b = 122 m, included angle C = 102°: K = (1/2)(70)(122)sin(102°) ≈ 4177 m².
Picture
Triangle ABC with two sides a, b meeting at angle C; altitude h drawn from vertex A perpendicular to side a.

Parallelogram Area (SAS)

WS2 (Problem 5)
$$ K_{\text{parallelogram}} = ab\sin\theta $$

In words. The area of a parallelogram with adjacent sides a and b and included angle θ is the product of the sides times the sine of the included angle.

Domain / range. θ ∈ (0, π) is the interior angle of the parallelogram.

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Derivation
A parallelogram is two congruent SAS triangles glued along a diagonal. Each triangle has area (1/2)ab sin θ. Doubling gives K = ab sin θ.
Example
Parallelogram with sides 116 and 143 and included angle 110.6°: K = (116)(143) sin(110.6°) ≈ 15,527.
Picture
Parallelogram with two adjacent sides a, b meeting at angle θ; diagonal drawn to show two SAS triangles.

Law of Sines

WS2 Law of Sines
$$ \frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} \quad \text{equivalently} \quad \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} $$

In words. In any triangle, the ratio of the sine of an angle to the length of the opposite side is constant. Equivalently, the ratio of a side to the sine of its opposite angle is constant.

Domain / range. Each angle is in (0, π); each side is positive. The three angles must sum to π.

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Derivation
Start with the three equivalent expressions for the SAS area: K = (1/2)ab sin C = (1/2)bc sin A = (1/2)ac sin B. Multiply every term by 2/(abc) and cancel matching letters. The result is sin A / a = sin B / b = sin C / c.
Example
Triangle PQR with ∠P = 18°, ∠Q = 29°, side r = 20 opposite ∠R. Third angle ∠R = 180° - 18° - 29° = 133°. Then p = 20 sin(18°)/sin(133°) ≈ 8.45 and q = 20 sin(29°)/sin(133°) ≈ 13.25.
Picture
Triangle with each side labeled lowercase and each opposite angle labeled uppercase; three matching side-angle pairs.

Ambiguous SSA Case Conditions

WS3 Ambiguous Triangles
$$ h = c\sin A. \text{ Compare side } a \text{ to } h \text{ and to } c: $$

In words. Given angle A and sides a (opposite A) and c, the critical height is h = c sin A. If a < h, no triangle exists. If a = h, one right triangle. If h < a < c, two triangles. If a ≥ c, exactly one triangle.

Domain / range. A is an acute interior angle (the ambiguity table assumes A is acute; if A is right or obtuse, at most one triangle exists).

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Derivation
Place angle A at the origin with side c along one ray. The third vertex B lies at distance c from A along this ray. From B, swing an arc of radius a; the number of intersections with the other ray determines the number of valid triangles. The critical height h = c sin A is the altitude from B perpendicular to the other ray — it's the shortest possible value of a that still reaches the line.
Example
∠B = 24°, d = 12 (side opposite ∠D). Critical height: h = 12 sin(24°) ≈ 4.88. So b < 4.88: no triangle; b = 4.88 or b ≥ 12: one triangle; 4.88 < b < 12: two triangles.
Picture
Swing-arc diagram: angle A at origin, side c on one ray, circular arc of radius a swept from the far end of c showing 0, 1, or 2 intersections with the other ray.

Triangle Inequality

WS3 (Problem 4)
$$ a + b > c, \quad a + c > b, \quad b + c > a $$

In words. Each side of a triangle must be strictly less than the sum of the other two.

Domain / range. All three sides are positive real numbers. Violation of any one inequality means no triangle exists.

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Derivation
Geometrically, a triangle's two shorter sides must, when stretched in a straight line, exceed the longest side; otherwise they can't reach across to meet. Equivalent to the statement that the shortest path between two points is a straight line.
Example
Sides 15, 9, 25: check 15 + 9 = 24 < 25. Fails. No triangle exists.
Picture
Three line segments of lengths a, b, c laid out; if a + b just barely exceeds c, the triangle is very flat; if a + b = c, degenerate; if a + b < c, the segments can't close.

Law of Cosines

WS4 Law of Cosines
$$ c^2 = a^2 + b^2 - 2ab\cos C $$

In words. The square of any side of a triangle equals the sum of squares of the other two sides minus twice their product times the cosine of the included angle. Pattern repeats: the angle in the cosine is opposite the side on the left.

Domain / range. C ∈ (0, π); when C = π/2, cos C = 0 and the formula reduces to the Pythagorean theorem.

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Derivation
Place vertex C at origin, side a along the positive x-axis so B = (a, 0). Then A = (b cos C, b sin C). Drop perpendicular from A to the x-axis; foot is P = (b cos C, 0). Right triangle APB has legs b sin C and (a - b cos C), hypotenuse c. Pythagoras: c² = (b sin C)² + (a - b cos C)² = b²sin²C + a² - 2ab cos C + b²cos²C. Apply sin² + cos² = 1: c² = a² + b² - 2ab cos C.
Example
Triangle with a = 7, b = 8, ∠C = 47°: c² = 49 + 64 - 2(7)(8)cos(47°) ≈ 113 - 76.4 ≈ 36.6, so c ≈ 6.05.
Picture
Triangle with vertex C at origin, side a along x-axis; altitude from opposite vertex perpendicular to extended x-axis showing the rectangular components.

Law of Cosines (Solved for Angle)

WS4 Law of Cosines
$$ \cos C = \frac{a^2 + b^2 - c^2}{2ab} $$

In words. Rearranged Law of Cosines: the cosine of any angle equals the sum of squares of the two adjacent sides minus the square of the opposite side, all divided by twice the product of the adjacent sides.

Domain / range. Output cos C ∈ [-1, 1]; if computed value exceeds 1 in absolute value, the three given sides cannot form a triangle.

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Derivation
Algebraic rearrangement of c² = a² + b² - 2ab cos C by isolating the cosine term: 2ab cos C = a² + b² - c², then divide by 2ab.
Example
Sides a = 12, b = 7, c = 16: cos B = (144 + 256 - 49)/(2·12·16) wait — adjusting to find ∠B opposite b: cos B = (a² + c² - b²)/(2ac) = (144 + 256 - 49)/(2·12·16) = 351/384 ≈ 0.914, so B ≈ 24°. Or for ∠C opposite c = 16: cos C = (12² + 7² - 16²)/(2·12·7) = (144+49-256)/168 = -63/168 ≈ -0.375, so C ≈ 112°.
Picture
Same triangle as Law of Cosines; arccos of the formula gives the unique angle in [0°, 180°] — no ambiguity (unlike Law of Sines).

Angle Bisector Theorem

WS4 (Problem 8)
$$ \frac{x}{y} = \frac{a}{b} $$

In words. An angle bisector from a vertex divides the opposite side into two segments whose ratio equals the ratio of the two adjacent sides.

Domain / range. Applies to any triangle with a bisector drawn from one vertex to the opposite side.

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Derivation
Apply the Law of Sines (or SAS area) to the two sub-triangles formed by the bisector. Both share the bisector as a common side and have congruent bisected angles, so the ratio of their opposite sides (the segments x, y on the original opposite side) equals the ratio of the corresponding adjacent sides of the original triangle.
Example
Triangle with adjacent sides 140 and 173, opposite side total length 124.57 (split by bisector into x and y). x = (140/313)(124.57) ≈ 55.72, y ≈ 68.85.
Picture
Triangle with bisector from apex; opposite side split into segments x and y; sides labeled a and b on the bisected angle's two arms.

Inverse Sine (arcsin) Domain & Range

WS6 Inverse Trig Functions
$$ y = \sin^{-1}(x) \iff \sin y = x \text{ and } y \in [-\pi/2, \pi/2] $$

In words. The inverse sine function maps a value in [-1, 1] to the unique angle in [-π/2, π/2] whose sine equals that value.

Domain / range. Domain: [-1, 1]. Range: [-π/2, π/2]. Equivalent in degrees: range [-90°, 90°].

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Derivation
Sine is not one-to-one over all real numbers (it's periodic). To define an inverse function, restrict sine to its largest monotonic-increasing interval [-π/2, π/2] where sin is one-to-one taking values [-1, 1]. The inverse of this restricted sine is arcsin, with range equal to the restricted domain.
Example
arcsin(sin(3π/2)): inside, sin(3π/2) = -1. Then arcsin(-1) = -π/2 (the unique angle in [-π/2, π/2] with sine -1). Note this is NOT 3π/2 — arcsin snaps outputs to its restricted range.
Picture
Sine graph with the restricted piece (one increasing branch from -π/2 to π/2) highlighted; reflection across y = x gives the arcsin graph.

Inverse Cosine (arccos) Domain & Range

WS6 Inverse Trig Functions
$$ y = \cos^{-1}(x) \iff \cos y = x \text{ and } y \in [0, \pi] $$

In words. The inverse cosine function maps a value in [-1, 1] to the unique angle in [0, π] whose cosine equals that value.

Domain / range. Domain: [-1, 1]. Range: [0, π]. Equivalent in degrees: range [0°, 180°].

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Derivation
Cosine is not one-to-one; restrict to [0, π] where cosine is monotonically decreasing from 1 to -1 (one-to-one). The inverse on this restricted domain is arccos, with range [0, π].
Example
arccos(1/2) = π/3 (60°), NOT π/6. arccos(√3/2) = π/6 (30°). arccos(-1/2) = 2π/3 (120°). arccos(-1) = π.
Picture
Cosine graph with the restricted piece (decreasing branch from 0 to π) highlighted; reflection across y = x gives the arccos graph, a decreasing curve from (-1, π) to (1, 0).

Inverse Tangent (arctan) Domain & Range

WS6 Inverse Trig Functions
$$ y = \tan^{-1}(x) \iff \tan y = x \text{ and } y \in (-\pi/2, \pi/2) $$

In words. The inverse tangent function maps any real number to the unique angle in the open interval (-π/2, π/2) whose tangent equals that value.

Domain / range. Domain: all real numbers (-∞, ∞). Range: open interval (-π/2, π/2). Equivalent in degrees: open range (-90°, 90°).

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Derivation
Tangent has period π and is monotonically increasing from -∞ to ∞ on (-π/2, π/2) (one-to-one). The inverse on this restricted domain is arctan, with range as the open interval (the endpoints are vertical asymptotes of tan, so they're excluded).
Example
arctan(1) = π/4. arctan(-1) = -π/4. arctan(√3) = π/3. arctan(0) = 0. arctan(very large) → π/2 (asymptote).
Picture
Tangent graph with the restricted piece (one branch between vertical asymptotes at ±π/2) highlighted; reflection across y = x gives arctan as an S-curve with horizontal asymptotes at ±π/2.

2D Rotation Matrix

WS7 Rotation Matrix
$$ R_\theta = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix} $$

In words. The matrix that rotates any point in the plane counter-clockwise by angle θ about the origin has cosine on the diagonal, -sine in the top-right entry, and +sine in the bottom-left entry.

Domain / range. θ is any real number (in radians, conventionally). Determinant = cos²θ + sin²θ = 1, so rotations preserve distances and areas.

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Derivation
The columns of a transformation matrix are the images of the standard basis vectors. Rotating (1, 0) by θ gives (cos θ, sin θ) (column 1). Rotating (0, 1) by θ gives (-sin θ, cos θ) (column 2; either via a separate unit-circle argument or by applying $R_{90°}$ to column 1). Reading off the columns gives the matrix.
Example
$R_{30°}$ applied to (-4, -4√3): new x = (√3/2)(-4) + (-1/2)(-4√3) = -2√3 + 2√3 = 0; new y = (1/2)(-4) + (√3/2)(-4√3) = -2 - 6 = -8. Result: (0, -8). Sanity check: original point has magnitude √(16+48) = 8 at angle 240°, after rotation it's at 270°, magnitude unchanged → (0, -8). ✓
Picture
Unit circle with point P at angle θ, point Q at angle θ + π/2; the columns of the matrix are exactly P and Q.

Identity Matrix (2x2)

WS7 Rotation Matrix
$$ I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} $$

In words. The 2x2 identity matrix has ones on the main diagonal and zeros elsewhere; multiplying any vector by I leaves it unchanged.

Domain / range. Applies to any 2D vector (or 2x2 matrix). I is the zero-rotation case: $R_0$ = I.

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Derivation
By construction, I has the property that I·v = v for every column vector v. It serves as the rotation matrix for θ = 0: $R_0$ has cos(0) = 1 on the diagonal and -sin(0) = sin(0) = 0 off-diagonal.
Example
I · (x, y) = (x, y). Geometrically, rotating a point by 0° leaves it where it is.
Picture
Coordinate plane with point P; an arrow labeled 'I' from P back to P showing zero displacement.

Composition of Rotations

WS7 / WS8
$$ R_\alpha \cdot R_\beta = R_{\alpha + \beta} $$

In words. Rotating by β and then by α gives a single rotation by the sum α + β. This is the parent identity from which all sum-of-angles formulas descend.

Domain / range. α, β are any real numbers (in radians).

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Derivation
Geometrically obvious: applying two rotations in sequence about the same center adds the angles. The proof that this works for the matrix form is the matrix multiplication R_α · R_β, which produces a matrix whose entries match $R_{α+β}$ — this matrix multiplication IS the proof of the sum-of-angles identities.
Example
$R_{30°}$ · $R_{60°}$ should equal $R_{90°}$. Check: $R_{30°}$ · $R_{60°}$ has top-left entry cos(30°)cos(60°) - sin(30°)sin(60°) = (√3/2)(1/2) - (1/2)(√3/2) = 0 = cos(90°). ✓
Picture
Point on unit circle rotated by β to a new position, then rotated by α; the net angular displacement is α + β.

Pythagorean Identity (Primary)

WS8 Sum & Difference Identities
$$ \sin^2\theta + \cos^2\theta = 1 $$

In words. The sum of the squares of sine and cosine of any angle equals one.

Domain / range. θ is any real number.

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Derivation
On the unit circle, the point at angle θ has coordinates (cos θ, sin θ). The Pythagorean theorem applied to the right triangle formed by this point, the origin, and the foot of perpendicular gives x² + y² = r² = 1, i.e., cos²θ + sin²θ = 1.
Example
If sin θ = 1/3, then cos²θ = 1 - 1/9 = 8/9, so cos θ = ±(2√2/3). (Sign determined by quadrant of θ.)
Picture
Unit circle with point P at angle θ; right triangle dropped from P to x-axis with legs cos θ and sin θ and hypotenuse 1.

Pythagorean Identity (tan/sec)

WS8 Sum & Difference Identities
$$ \tan^2\theta + 1 = \sec^2\theta $$

In words. Tangent squared plus one equals secant squared.

Domain / range. Undefined where cos θ = 0, i.e., θ = π/2 + πk.

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Derivation
Start with the primary Pythagorean identity sin² + cos² = 1. Divide every term by cos²θ: sin²/cos² + 1 = 1/cos². Identify sin/cos = tan and 1/cos = sec to get tan² + 1 = sec².
Example
If tan θ = 3, then sec²θ = 9 + 1 = 10, so sec θ = ±√10. (Sign from quadrant.) Therefore cos θ = ±1/√10 = ±√10/10.
Picture
Same unit-circle right triangle; the secant arm extends from the origin along the x-axis until it meets the tangent line drawn at the point P.

Pythagorean Identity (cot/csc)

WS8 Sum & Difference Identities
$$ 1 + \cot^2\theta = \csc^2\theta $$

In words. One plus cotangent squared equals cosecant squared.

Domain / range. Undefined where sin θ = 0, i.e., θ = πk.

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Derivation
Start with sin² + cos² = 1. Divide every term by sin²θ: 1 + cos²/sin² = 1/sin². Identify cos/sin = cot and 1/sin = csc to get 1 + cot² = csc².
Example
If cot θ = √3, then csc²θ = 1 + 3 = 4, so csc θ = ±2. Therefore sin θ = ±1/2. (Sign from quadrant.)
Picture
Same unit-circle right triangle; cosecant is the reciprocal of sine — the length from origin to the y-axis along the tangent line.

Sum Identity (Cosine)

WS8 Sum & Difference Identities
$$ \cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta $$

In words. Cosine of the sum of two angles equals the product of cosines minus the product of sines.

Domain / range. α, β are any real numbers.

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Derivation
Take the point B = (cos β, sin β) on the unit circle at angle β. Apply the rotation matrix R_α: R_α · B gives the point A = (cos(α+β), sin(α+β)) at angle α + β. The top-row entry of the matrix product is cos α · cos β + (-sin α) · sin β = cos α cos β - sin α sin β, which therefore equals cos(α + β).
Example
cos(75°) = cos(45° + 30°) = cos 45° cos 30° - sin 45° sin 30° = (√2/2)(√3/2) - (√2/2)(1/2) = (√6 - √2)/4 ≈ 0.2588.
Picture
Unit circle with point B at angle β and point A at angle α + β; rotation matrix R_α takes B to A.

Sum Identity (Sine)

WS8 Sum & Difference Identities
$$ \sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta $$

In words. Sine of the sum of two angles equals sine-times-cosine plus cosine-times-sine (mixed cross-products, added).

Domain / range. α, β are any real numbers.

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Derivation
Same construction as the cosine sum: R_α · (cos β, sin β) = (cos(α+β), sin(α+β)). The bottom-row entry of the matrix product is sin α · cos β + cos α · sin β, which equals sin(α + β).
Example
sin(75°) = sin(45° + 30°) = sin 45° cos 30° + cos 45° sin 30° = (√2/2)(√3/2) + (√2/2)(1/2) = (√6 + √2)/4 ≈ 0.9659.
Picture
Same as cosine sum; reading off the bottom-row entry of R_α R_β.

Difference Identity (Cosine)

WS8 Sum & Difference Identities
$$ \cos(\alpha - \beta) = \cos\alpha\cos\beta + \sin\alpha\sin\beta $$

In words. Cosine of the difference of two angles equals the product of cosines plus the product of sines (note the sign flip from the sum formula).

Domain / range. α, β are any real numbers.

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Derivation
Replace β with -β in the cosine sum identity. Cosine is an even function (cos(-β) = cos β) and sine is odd (sin(-β) = -sin β). So cos(α - β) = cos(α + (-β)) = cos α cos β - sin α (-sin β) = cos α cos β + sin α sin β.
Example
cos(15°) = cos(45° - 30°) = cos 45° cos 30° + sin 45° sin 30° = (√2/2)(√3/2) + (√2/2)(1/2) = (√6 + √2)/4 ≈ 0.9659.
Picture
Sum identity with β replaced by its reflection across the x-axis; cosine is preserved under reflection, sine flips.

Difference Identity (Sine)

WS8 Sum & Difference Identities
$$ \sin(\alpha - \beta) = \sin\alpha\cos\beta - \cos\alpha\sin\beta $$

In words. Sine of the difference of two angles equals sine-times-cosine minus cosine-times-sine.

Domain / range. α, β are any real numbers.

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Derivation
Replace β with -β in the sine sum identity, using cos(-β) = cos β and sin(-β) = -sin β: sin(α - β) = sin(α + (-β)) = sin α cos β + cos α (-sin β) = sin α cos β - cos α sin β.
Example
sin(15°) = sin(45° - 30°) = sin 45° cos 30° - cos 45° sin 30° = (√2/2)(√3/2) - (√2/2)(1/2) = (√6 - √2)/4 ≈ 0.2588.
Picture
Sum identity with β replaced by -β; the second term flips sign.

Even/Odd (Negative-Angle) Identities

WS8 (and Trig Fundamentals)
$$ \sin(-\theta) = -\sin\theta, \quad \cos(-\theta) = \cos\theta, \quad \tan(-\theta) = -\tan\theta $$

In words. Sine is odd (sign flips under negation of input). Cosine is even (preserved under negation). Tangent is odd (a ratio of an odd over an even is odd).

Domain / range. θ is any real number for sine and cosine; tangent excludes π/2 + πk.

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Derivation
On the unit circle, replacing θ with -θ reflects the terminal-side point across the x-axis. Under this reflection, the x-coordinate (cosine) is preserved and the y-coordinate (sine) flips sign. Tangent = sin/cos flips sign because the numerator changes sign while the denominator doesn't.
Example
sin(-π/3) = -sin(π/3) = -√3/2. cos(-π/3) = cos(π/3) = 1/2. tan(-π/3) = -tan(π/3) = -√3.
Picture
Unit circle with point P at angle θ and point P' at angle -θ; P' is P reflected across the x-axis.

Co-function Identities

WS8 (Problems 1, 2)
$$ \sin\left(\tfrac{\pi}{2} - \alpha\right) = \cos\alpha, \quad \cos\left(\tfrac{\pi}{2} - \alpha\right) = \sin\alpha, \quad \tan\left(\tfrac{\pi}{2} - \alpha\right) = \cot\alpha $$

In words. Sine and cosine swap when the input is replaced by its complement (π/2 minus the angle). Similarly for tan/cot and sec/csc — each function equals its co-function applied to the complementary angle.

Domain / range. α is any real number (for sine/cosine); tan/cot version excludes the singularities of the relevant function.

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Derivation
Apply the sum/difference identities with the first angle = π/2: sin(π/2 - α) = sin(π/2)cos α - cos(π/2)sin α = (1)cos α - (0)sin α = cos α. Same for the cosine version, swapping the structure.
Example
sin(60°) = sin(90° - 30°) = cos(30°) = √3/2. ✓ The name 'cosine' literally means 'sine of complementary angle.'
Picture
Right triangle with two acute angles α and π/2 - α; the sine of one equals the cosine of the other (opposite/hypotenuse vs. adjacent/hypotenuse — these are the same side from different perspectives).

Reduction Formulas (π and 3π/2 Shifts)

WS8 (Problems 3, 4, 5)
$$ \cos(\pi - \alpha) = -\cos\alpha, \quad \sin(\pi + \alpha) = -\sin\alpha, \quad \cos(x + 3\pi/2) = \sin x $$

In words. Shifting the input by π or 3π/2 produces predictable sign flips or function swaps. These are special cases of the sum-of-angles identities.

Domain / range. α, x are any real numbers.

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Derivation
Apply the sum identity with the second angle equal to π, 3π/2, or similar. For example: cos(π - α) = cos π cos α + sin π sin α = (-1)cos α + (0)sin α = -cos α. For cos(x + 3π/2) = cos x cos(3π/2) - sin x sin(3π/2) = cos x · 0 - sin x · (-1) = sin x.
Example
cos(π - π/6) = cos(5π/6) = -cos(π/6) = -√3/2. sin(π + π/4) = sin(5π/4) = -sin(π/4) = -√2/2. cos(π/4 + 3π/2) = cos(7π/4) = sin(π/4) = √2/2.
Picture
Unit circle showing points at angles α, π-α (Q2 reflection), π+α (Q3 reflection), and 3π/2 + α (Q4 rotation); sign flips track quadrant.

Sum Identity (Tangent)

WS9 A Sum of Tangents
$$ \tan(\alpha + \beta) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha \tan\beta} $$

In words. Tangent of the sum of two angles equals the sum of tangents over one minus the product of tangents.

Domain / range. Undefined when tan α tan β = 1 (i.e., when α + β = π/2 + πk).

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Derivation
Start with tan(α + β) = sin(α + β) / cos(α + β). Substitute the sum identities for numerator and denominator. Divide every term in the resulting fraction by cos α cos β to convert sines to tangents. Result: (tan α + tan β) / (1 - tan α tan β).
Example
tan(105°) = tan(60° + 45°) = (√3 + 1)/(1 - √3 · 1) = (√3 + 1)/(1 - √3). Rationalize by multiplying by (1 + √3)/(1 + √3): = (√3+1)(1+√3)/(1-3) = (4 + 2√3)/(-2) = -2 - √3 ≈ -3.73.
Picture
Two lines crossing through the origin at angles α and β; the angle between them is α - β, and the tangent of that angle reduces to (m₁ - m₂)/(1 + m₁m₂).

Difference Identity (Tangent)

WS9 A Sum of Tangents
$$ \tan(\alpha - \beta) = \frac{\tan\alpha - \tan\beta}{1 + \tan\alpha \tan\beta} $$

In words. Tangent of the difference of two angles equals the difference of tangents over one plus the product of tangents.

Domain / range. Undefined when tan α tan β = -1 (i.e., when α - β = π/2 + πk).

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Derivation
Replace β with -β in the tangent sum identity, using tan(-β) = -tan β: tan(α - β) = tan(α + (-β)) = (tan α + tan(-β))/(1 - tan α tan(-β)) = (tan α - tan β)/(1 + tan α tan β).
Example
tan(15°) = tan(45° - 30°) = (1 - √3/3)/(1 + √3/3) = (3 - √3)/(3 + √3). Rationalize: = (3-√3)²/(9 - 3) = (12 - 6√3)/6 = 2 - √3 ≈ 0.27.
Picture
Same as the sum, but with the second angle reflected.

Slope-Inclination Relation

WS9 A Sum of Tangents
$$ m = \tan\alpha, \quad \alpha \in [0, \pi) $$

In words. The slope of a line equals the tangent of its angle of inclination (the angle measured counter-clockwise from the positive x-axis to the line, restricted to [0°, 180°)).

Domain / range. α ∈ [0, π) by convention (a line and its reversal have the same inclination). For vertical lines, α = π/2 and the slope is undefined.

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Derivation
Slope is rise over run between two points on the line. If the line makes angle α with the x-axis, then rise/run for any segment along the line equals the tangent of α (opposite/adjacent of the slope-triangle formed by horizontal and vertical components).
Example
Line y = 3x + 4: slope m = 3, so tan α = 3, α = arctan(3) ≈ 71.57°. Line y = -2x - 6: slope -2, so α = 180° - arctan(2) ≈ 116.57° (must be in Q2 since the inclination is in [0°, 180°)).
Picture
Coordinate axes with a line at angle α above the horizontal; right triangle with horizontal leg 'run' and vertical leg 'rise' showing tan α = rise/run = slope.

Angle Between Two Lines

WS9 A Sum of Tangents
$$ \tan\theta = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right| $$

In words. The tangent of the acute angle between two lines with slopes m₁ and m₂ equals the absolute value of their slope difference divided by one plus their slope product.

Domain / range. Defined whenever 1 + m₁m₂ ≠ 0 (i.e., when the lines are NOT perpendicular). If 1 + m₁m₂ = 0, then m₁m₂ = -1 and the lines are perpendicular (angle = 90°).

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Derivation
The angle between two lines is the difference of their angles of inclination, α₁ - α₂. Apply the tangent-difference identity with tan α₁ = m₁ and tan α₂ = m₂: tan(α₁ - α₂) = (m₁ - m₂)/(1 + m₁m₂). Take absolute value to get the acute angle.
Example
Lines y = 3x + 4 and y = -x - 2: tan θ = |(3 - (-1))/(1 + (3)(-1))| = |4/(-2)| = 2. So θ = arctan(2) ≈ 63.43°.
Picture
Two intersecting lines; the four angles at the intersection are α, 180°-α, α, 180°-α — pick the acute one.

Perpendicular Slopes (Special Case)

WS9 A Sum of Tangents
$$ m_1 \cdot m_2 = -1 \iff \text{lines are perpendicular} $$

In words. Two non-vertical lines are perpendicular if and only if the product of their slopes is -1. This is the denominator-zero case of the angle-between-lines formula.

Domain / range. Both lines must be non-vertical (otherwise slopes don't exist; perpendicularity is handled by inspection — horizontal and vertical lines are perpendicular).

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Derivation
If the lines are perpendicular, the angle between them is 90°, so tan(angle) is undefined — meaning the formula's denominator 1 + m₁m₂ must equal zero. Solve: m₁m₂ = -1.
Example
Lines y = 2x + 3 and y = -x/2 + 5: m₁ · m₂ = 2 · (-1/2) = -1. They're perpendicular. ✓
Picture
Two lines crossing at right angles; one steep upward (slope 2), the other shallow downward (slope -1/2).

Double-Angle Identity (Sine)

WS10 Double Trouble
$$ \sin(2\alpha) = 2\sin\alpha\cos\alpha $$

In words. Sine of double an angle equals twice the product of sine and cosine of that angle.

Domain / range. α is any real number.

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Derivation
Set β = α in the sine sum identity sin(α + β) = sin α cos β + cos α sin β: sin(2α) = sin α cos α + cos α sin α = 2 sin α cos α.
Example
Given sin α = 1/3 with α ∈ (0, π/2): cos α = 2√2/3 (positive, Q1). sin(2α) = 2(1/3)(2√2/3) = 4√2/9.
Picture
Unit circle with points at angles α and 2α; the rectangle-area interpretation of 2 sin α cos α matches the sine of the doubled angle.

Double-Angle Identity (Cosine — Three Forms)

WS10 Double Trouble
$$ \cos(2\alpha) = \cos^2\alpha - \sin^2\alpha = 2\cos^2\alpha - 1 = 1 - 2\sin^2\alpha $$

In words. Cosine of double an angle has three equivalent forms: cosine squared minus sine squared, or twice cosine squared minus one, or one minus twice sine squared.

Domain / range. α is any real number.

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Derivation
Set β = α in the cosine sum identity: cos(2α) = cos α cos α - sin α sin α = cos²α - sin²α. Then use the Pythagorean identity sin² + cos² = 1: replacing sin²α with 1 - cos²α gives 2cos²α - 1; replacing cos²α with 1 - sin²α gives 1 - 2sin²α. Choose whichever form has the unknown that you already have a value for.
Example
Find cos(120°) using cos²(60°) - sin²(60°) = (1/2)² - (√3/2)² = 1/4 - 3/4 = -1/2. ✓
Picture
Three equivalent boxed formulas with arrows showing the Pythagorean swap that connects them.

Double-Angle Identity (Tangent)

WS10 Double Trouble
$$ \tan(2\alpha) = \frac{2\tan\alpha}{1 - \tan^2\alpha} $$

In words. Tangent of double an angle equals twice the tangent over one minus tangent squared.

Domain / range. Undefined when tan²α = 1 (i.e., α = π/4 + π/2 · k).

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Derivation
Set β = α in the tangent sum identity: tan(2α) = (tan α + tan α)/(1 - tan α · tan α) = 2 tan α / (1 - tan²α).
Example
If tan α = 1/2: tan(2α) = 2(1/2)/(1 - 1/4) = 1/(3/4) = 4/3.
Picture
Same line-rotation interpretation: doubling the inclination angle of a line.

Half-Angle Formula (Cosine, Power-Reduced)

WS10 / WS11 (Problem 6)
$$ \cos^2\alpha = \frac{1 + \cos(2\alpha)}{2} $$

In words. Cosine squared of an angle equals one-half plus half the cosine of double the angle. Used to reduce powers of cosine to first-order expressions in cosine of a doubled angle.

Domain / range. α is any real number.

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Derivation
Start with cos(2α) = 2cos²α - 1. Solve for cos²α: 2cos²α = 1 + cos(2α), so cos²α = (1 + cos 2α)/2. This is the 'power-reduction' rearrangement of the double-angle formula.
Example
cos²(15°) = (1 + cos(30°))/2 = (1 + √3/2)/2 = (2 + √3)/4. So cos(15°) = √((2+√3)/4) = √(2+√3)/2.
Picture
Cosine-squared graph and (1 + cos 2α)/2 graph overlaid — same curve, different formula.

Half-Angle Formula (Sine, Power-Reduced)

WS10 / WS11 (Problem 7)
$$ \sin^2\alpha = \frac{1 - \cos(2\alpha)}{2} $$

In words. Sine squared of an angle equals one-half minus half the cosine of double the angle.

Domain / range. α is any real number.

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Derivation
Start with cos(2α) = 1 - 2sin²α. Solve for sin²α: 2sin²α = 1 - cos(2α), so sin²α = (1 - cos 2α)/2.
Example
sin²(15°) = (1 - cos(30°))/2 = (1 - √3/2)/2 = (2 - √3)/4. So sin(15°) = √(2-√3)/2.
Picture
Sine-squared graph and (1 - cos 2α)/2 graph overlaid — same curve.

Half-Angle Formula (Tangent Squared)

WS11 (Problem 10)
$$ \tan^2\alpha = \frac{1 - \cos(2\alpha)}{1 + \cos(2\alpha)} $$

In words. Tangent squared of an angle equals one minus cosine of the doubled angle, divided by one plus cosine of the doubled angle.

Domain / range. Undefined when 1 + cos(2α) = 0, i.e., when 2α = π + 2πk (α = π/2 + πk — where tangent itself is undefined).

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Derivation
Divide the half-angle formula for sin² by the half-angle formula for cos²: tan²α = sin²α/cos²α = [(1 - cos 2α)/2] / [(1 + cos 2α)/2] = (1 - cos 2α)/(1 + cos 2α). Both halves of the fraction cancel.
Example
tan²(15°) = (1 - cos 30°)/(1 + cos 30°) = (1 - √3/2)/(1 + √3/2) = (2 - √3)/(2 + √3). Rationalize: = (2-√3)²/(4-3) = 7 - 4√3 ≈ 0.072. So tan(15°) ≈ 0.27 ≈ 2 - √3. ✓
Picture
Ratio of the sine-squared and cosine-squared half-angle formulas.

Triple-Angle Identity (Sine)

WS11 (Problem 7)
$$ \sin(3x) = 3\sin x - 4\sin^3 x $$

In words. Sine of triple an angle equals three times sine minus four times sine cubed.

Domain / range. x is any real number.

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Derivation
Write sin(3x) = sin(2x + x) and apply the sine sum identity: sin 2x cos x + cos 2x sin x. Substitute the double-angle formulas sin(2x) = 2 sin x cos x and cos(2x) = 1 - 2sin²x: (2 sin x cos x)(cos x) + (1 - 2sin²x)(sin x) = 2 sin x cos²x + sin x - 2sin³x. Use Pythagorean (cos²x = 1 - sin²x): 2 sin x (1 - sin²x) + sin x - 2sin³x = 2sin x - 2sin³x + sin x - 2sin³x = 3 sin x - 4 sin³x.
Example
sin(3 · 30°) = sin(90°) = 1. Check: 3 sin(30°) - 4 sin³(30°) = 3(1/2) - 4(1/8) = 3/2 - 1/2 = 1. ✓
Picture
Stacking three identities: sum identity → double-angle substitution → Pythagorean substitution. Three identities collapse into one formula.

Sine of Double-Angle (Tangent Form)

WS11 (Problem 9)
$$ \sin(2\theta) = \frac{2\tan\theta}{1 + \tan^2\theta} $$

In words. Sine of double an angle can be written as twice the tangent divided by one plus tangent squared.

Domain / range. Undefined where tan θ is undefined (θ = π/2 + πk). Useful when only tangent of the angle is known.

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Derivation
Start with sin(2θ) = 2 sin θ cos θ. Divide numerator and denominator by cos²θ: 2 sin θ cos θ / 1 = (2 sin θ / cos θ) · cos²θ = 2 tan θ · cos²θ. Use cos²θ = 1/(1+tan²θ) (from Pythagorean identity 1 + tan² = sec² = 1/cos²): sin(2θ) = 2 tan θ / (1 + tan²θ).
Example
If tan θ = 1, then sin(2θ) = 2(1)/(1+1) = 1. ✓ (since 2θ = π/2 and sin(π/2) = 1.)
Picture
Same as sin(2θ) = 2 sin θ cos θ, but expressed solely in terms of tan θ; common in Weierstrass substitution.

tan + cot = 2csc(2θ) Identity

WS11 (Problem 8)
$$ \tan\theta + \cot\theta = 2\csc(2\theta) $$

In words. The sum of tangent and cotangent of an angle equals twice the cosecant of the doubled angle.

Domain / range. θ ≠ πk/2 (excludes points where sin or cos is zero).

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Derivation
Convert to sin/cos: tan θ + cot θ = sin θ/cos θ + cos θ/sin θ. Common denominator: (sin²θ + cos²θ)/(sin θ cos θ) = 1/(sin θ cos θ) by Pythagorean. Multiply numerator and denominator by 2: 2/(2 sin θ cos θ) = 2/sin(2θ) = 2 csc(2θ).
Example
θ = π/4: tan(π/4) + cot(π/4) = 1 + 1 = 2. Check: 2 csc(π/2) = 2(1) = 2. ✓
Picture
Identity-proof chain showing the steps from tan + cot through Pythagorean and double-angle.

Periodicity Template (Sine)

WS12 Trig Equations
$$ \sin\theta = c \iff \theta = \arcsin c + 2\pi k \;\text{or}\; \theta = \pi - \arcsin c + 2\pi k $$

In words. All solutions of sin θ = c come in two families: the principal value plus full revolutions, and the supplement of the principal value plus full revolutions.

Domain / range. c ∈ [-1, 1]; k ∈ ℤ. If |c| > 1, no solutions.

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Derivation
Sine has period 2π and takes the same value at θ and π - θ (supplementary angles have equal sine). The arcsin function returns one of these in [-π/2, π/2]; the other family is its supplement π - arcsin(c). Adding 2πk to each captures the full periodicity.
Example
sin θ = 1/2: arcsin(1/2) = π/6. Solutions: θ = π/6 + 2πk OR θ = π - π/6 + 2πk = 5π/6 + 2πk.
Picture
Horizontal line y = c crossing the sine graph; the intersections cluster in pairs (one increasing, one decreasing) repeating every 2π.

Periodicity Template (Cosine)

WS12 Trig Equations
$$ \cos\theta = c \iff \theta = \pm\arccos c + 2\pi k $$

In words. All solutions of cos θ = c are the principal value (±) plus full revolutions.

Domain / range. c ∈ [-1, 1]; k ∈ ℤ. If |c| > 1, no solutions.

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Derivation
Cosine has period 2π and is an even function: cos(-θ) = cos θ. So if arccos(c) = θ₀ is one solution, then -θ₀ is another. Adding 2πk to each captures full periodicity.
Example
cos θ = 1/2: arccos(1/2) = π/3. Solutions: θ = ±π/3 + 2πk. For 4cos²x = 1, cos x = ±1/2, giving four families: ±π/3 + 2πk and ±2π/3 + 2πk.
Picture
Horizontal line y = c crossing cosine graph; intersections come in pairs symmetric about the y-axis, repeating every 2π.

Periodicity Template (Tangent)

WS12 Trig Equations
$$ \tan\theta = c \iff \theta = \arctan c + \pi k $$

In words. All solutions of tan θ = c are the principal value plus half-revolutions (period of tangent is π, not 2π).

Domain / range. c is any real number; k ∈ ℤ.

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Derivation
Tangent has period π (since tan(θ + π) = sin(θ+π)/cos(θ+π) = (-sin θ)/(-cos θ) = tan θ). Arctan returns one solution in (-π/2, π/2); adding πk captures the rest.
Example
tan θ = √3: arctan(√3) = π/3. Solutions: θ = π/3 + πk. For tan²x = 3, tan x = ±√3, giving two families: ±π/3 + πk.
Picture
Horizontal line y = c crossing the tangent graph; intersections occur once per period of length π.

sin α = sin β Two-Case Identity

WS12 (Problem 17)
$$ \sin\alpha = \sin\beta \iff \alpha = \beta + 2\pi k \;\text{or}\; \alpha = \pi - \beta + 2\pi k $$

In words. Two sines are equal exactly when the angles differ by a full revolution or are supplementary (mod 2π).

Domain / range. α, β are any real numbers; k ∈ ℤ.

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Derivation
From the unit circle: two points have the same y-coordinate (sine) iff they're at the same angle (mod 2π) or at supplementary angles (mirror images across the y-axis). This is the same logic as the periodicity template for sin θ = c, applied with the constant c replaced by sin β.
Example
sin(3.4x) = sin x: either 3.4x = x + 2πk → 2.4x = 2πk → x = 5πk/6; or 3.4x = π - x + 2πk → 4.4x = π + 2πk → x = π/4.4 + 2πk/4.4 ≈ 0.71 + 1.43k.
Picture
Two unit-circle points with equal y-values; either same angle or mirror-images across the y-axis.

Harmonic Combination (A cos + B sin → R sin)

WS12 (Problem 16)
$$ A\cos x + B\sin x = R\sin(x + \phi), \text{ where } R = \sqrt{A^2 + B^2}, \; \tan\phi = A/B $$

In words. Any linear combination of sine and cosine of the same argument can be rewritten as a single sinusoid with amplitude √(A² + B²) and phase shift φ.

Domain / range. x is any real number. A and B are any real numbers (not both zero).

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Derivation
Expand R sin(x + φ) using the sum identity: R(sin x cos φ + cos x sin φ) = R cos φ · sin x + R sin φ · cos x. Match coefficients to B sin x + A cos x: R cos φ = B and R sin φ = A. Then R² = A² + B² (Pythagorean) and tan φ = (R sin φ)/(R cos φ) = A/B.
Example
3 cos x - sin x: A = 3, B = -1. R = √10. tan φ = 3/(-1) = -3, φ = arctan(-3). Solving 3 cos x = 2 + sin x means √10 sin(x + φ) = 2, so x + φ = arcsin(2/√10) and x = arcsin(2/√10) - φ. Sweep over (-2π, 2π) for all solutions.
Picture
Two sinusoids added pointwise (one cosine, one sine) producing a single sinusoid with different amplitude and shifted peak.

Radian-Degree Conversion

Trig Fundamentals / Throughout
$$ \text{rad} = \text{deg} \cdot \frac{\pi}{180°}, \quad \text{deg} = \text{rad} \cdot \frac{180°}{\pi} $$

In words. To convert degrees to radians, multiply by π/180. To convert radians to degrees, multiply by 180/π.

Domain / range. Any real angle measure. Mental anchors: 180° = π rad, 90° = π/2, 60° = π/3, 45° = π/4, 30° = π/6.

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Derivation
By definition, the arc length of a full circle of radius 1 is 2π (radians) which corresponds to one full revolution of 360°. So 360° = 2π rad, hence 180° = π rad. All other conversions are linear scaling.
Example
45° = 45 · π/180 = π/4 rad. 5π/12 rad = (5π/12)(180/π) = 75°. 1 rad ≈ 57.30°.
Picture
Full circle labeled with both '360°' and '2π rad'; quadrant markers at 90°/π/2, 180°/π, 270°/3π/2.

Degrees-Minutes-Seconds (DMS) Notation

WS4 (notation only)
$$ x° \, y' \, z'' = x + \frac{y}{60} + \frac{z}{3600} \text{ decimal degrees} $$

In words. An angle in degrees, minutes, seconds is converted to decimal degrees by adding the degrees, the minutes divided by 60, and the seconds divided by 3600.

Domain / range. x integer (degrees), 0 ≤ y < 60 (minutes), 0 ≤ z < 60 (seconds). 1° = 60' = 3600''.

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Derivation
Minutes and seconds follow a base-60 (sexagesimal) system: 1° is divided into 60 minutes, each minute into 60 seconds. So a measurement like 42°20'45'' is 42 + 20/60 + 45/3600 ≈ 42.346°.
Example
78°30' = 78 + 30/60 = 78.5°. 83°15' = 83.25°. Convert 42.346° back: 42° + (0.346 · 60)' = 42°20.76' = 42°20' + (0.76 · 60)'' = 42°20'45.6''.
Picture
Protractor scale showing degrees with finer subdivisions in minutes and seconds.
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Vectors

66 equations

Vector component form (2D)

WS1
$$ \vec{v} = \langle a, b\rangle = a\mathbf{i} + b\mathbf{j} = \begin{bmatrix} a \\ b \end{bmatrix} $$

In words. A 2D vector v has two equivalent notations: angle-bracket components, basis-vector form using i and j, or a column matrix. The first component is the x-component (how far east); the second is the y-component (how far north).

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Derivation
Once a vector's tail is anchored at the origin, its tip's (x, y) coordinates ARE its components. The basis vectors i = ⟨1, 0⟩ and j = ⟨0, 1⟩ point east and north with unit length, so any vector a*i + b*j is just 'go a east, then b north' — the same instruction as ⟨a, b⟩.
Example
If v points from the origin to (3, 4), then v = ⟨3, 4⟩ = 3i + 4j. Going 3 units east and 4 units north lands you at the tip.
Picture
Arrow from origin to point (a, b). The right-triangle legs are a (horizontal) and b (vertical); the hypotenuse is the vector itself.

Magnitude of a 2D vector

WS1
$$ |\vec{v}| = \sqrt{a^2 + b^2} $$

In words. The magnitude (length) of vector ⟨a, b⟩ equals the square root of the sum of squared components. It is the Pythagorean distance from the tail to the tip.

Domain / range. Always non-negative; equals 0 iff v is the zero vector.

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Derivation
The components a and b are the legs of a right triangle whose hypotenuse is the vector. By the Pythagorean theorem, a² + b² = |v|². Take the positive square root.
Example
For v = ⟨3, 4⟩: |v| = sqrt(9 + 16) = sqrt(25) = 5 (the classic 3-4-5 Pythagorean triple).
Picture
Right triangle with horizontal leg a, vertical leg b, hypotenuse |v|. Pythagoras applied directly.

Angle from positive x-axis (standard math angle)

WS1
$$ \theta = \arctan\!\left(\dfrac{b}{a}\right)\quad\text{(with quadrant adjustment)} $$

In words. For vector ⟨a, b⟩, the angle theta measured counterclockwise from the positive x-axis has tangent equal to b/a — BUT you must sketch first and add 180° if the vector lies in Q2 or Q3, because arctan only outputs angles in (−90°, 90°).

Domain / range. Defined for nonzero vectors; result in [0°, 360°) or (−180°, 180°] depending on convention.

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Derivation
Cosine and sine of theta give the horizontal and vertical legs of a unit vector in that direction; the ratio sin/cos = tan equals b/a. The calculator's arctan returns only the principal value, so a sketch tells you which quadrant the vector actually occupies, and you correct by +180° if needed.
Example
For t = ⟨−1, 3⟩ (Q2): arctan(3/−1) = arctan(−3) ≈ −71.57°. Vector is in Q2, so add 180°: theta ≈ 108.43°.
Picture
Counterclockwise arc from +x-axis to the vector. Always sketch the quadrant first before trusting the calculator.

Polar to rectangular (2D)

WS1
$$ \vec{v} = \langle r\cos\theta,\, r\sin\theta\rangle $$

In words. A vector with magnitude r at angle theta (measured counterclockwise from the positive x-axis) has rectangular components ⟨r·cos(theta), r·sin(theta)⟩. Magnitude times direction.

Domain / range. r >= 0; theta any real angle (typically [0°, 360°) or [0, 2π)).

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Derivation
The unit vector at angle theta is ⟨cos(theta), sin(theta)⟩ — a point on the unit circle. Scaling this by r stretches it to length r without changing its direction. So v = r·⟨cos(theta), sin(theta)⟩.
Example
For r = 10, theta = 45°: v = ⟨10·cos 45°, 10·sin 45°⟩ = ⟨5√2, 5√2⟩ ≈ ⟨7.07, 7.07⟩.
Picture
Point on a circle of radius r at angle theta from +x. The (x, y) coordinates of that point are the components.

Universal bearing formula (clockwise from North)

WS1
$$ \vec{v} = \langle r\sin B,\, r\cos B\rangle $$

In words. For a vector with magnitude r and bearing B (measured clockwise from due North), the x-component (east) is r·sin(B) and the y-component (north) is r·cos(B). Sine and cosine swap roles relative to the standard math convention.

Domain / range. r >= 0; B any real angle, typically [0°, 360°).

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Derivation
Bearing B = 0° means due North (+y); B = 90° means due East (+x). The bearing B and the math angle theta from +x satisfy theta = 90° − B. Using cos(90° − B) = sin(B) and sin(90° − B) = cos(B): x = r·cos(theta) = r·sin(B), y = r·sin(theta) = r·cos(B).
Example
For r = 10 knots, bearing 45° (NE): v = ⟨10·sin 45°, 10·cos 45°⟩ = ⟨5√2, 5√2⟩ ≈ ⟨7.07, 7.07⟩. Equal east and north components, as NE demands.
Picture
Compass rose with N at top. Bearing measured clockwise. Vector points r units in that direction.

Compass direction N α° E

WS1
$$ \vec{v} = \langle +r\sin\alpha,\, +r\cos\alpha\rangle $$

In words. A vector of magnitude r at compass direction 'N alpha degrees E' means: start pointing North, rotate alpha degrees toward East. The x-component is +r·sin(alpha); the y-component is +r·cos(alpha). Both positive because the vector lies in the NE quadrant (Q1).

Domain / range. r >= 0; alpha in [0°, 90°].

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Derivation
Alpha is measured from the y-axis (north), so swap the trig functions relative to standard. Since rotation is toward east (+x), both components are positive.
Example
v = (55 km/h, N15°E): x = 55·sin 15° ≈ 14.24, y = 55·cos 15° ≈ 53.13. Mostly north with a small lean east.
Picture
Compass rose. Arrow leans alpha degrees from due North toward East. Tip lands in Q1.

Compass direction N α° W

WS1
$$ \vec{v} = \langle -r\sin\alpha,\, +r\cos\alpha\rangle $$

In words. A vector of magnitude r at compass direction 'N alpha degrees W' means: start pointing North, rotate alpha degrees toward West. The x-component is −r·sin(alpha) (negative because west); the y-component is +r·cos(alpha) (positive because north).

Domain / range. r >= 0; alpha in [0°, 90°].

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Derivation
Alpha is measured from north toward west. Same magnitudes as NαE, but x flips sign because we're going west, not east.
Example
v = (25 km/h, N30°W): x = −25·sin 30° = −12.5, y = 25·cos 30° = 25√3/2 ≈ 21.65.
Picture
Compass rose. Arrow leans alpha degrees from due North toward West. Tip lands in Q2.

Compass direction S α° E

WS1
$$ \vec{v} = \langle +r\sin\alpha,\, -r\cos\alpha\rangle $$

In words. A vector of magnitude r at compass direction 'S alpha degrees E' means: start pointing South, rotate alpha degrees toward East. The x-component is +r·sin(alpha) (positive — east); the y-component is −r·cos(alpha) (negative — south).

Domain / range. r >= 0; alpha in [0°, 90°].

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Derivation
Alpha measured from south toward east. The 'south' reference makes y negative; the 'east' rotation makes x positive.
Example
v = (70 m/s, S40°E): x = 70·sin 40° ≈ 45.00, y = −70·cos 40° ≈ −53.62.
Picture
Compass rose. Arrow leans alpha degrees from due South toward East. Tip lands in Q4.

Compass direction S α° W

WS1
$$ \vec{v} = \langle -r\sin\alpha,\, -r\cos\alpha\rangle $$

In words. A vector of magnitude r at compass direction 'S alpha degrees W' means: start pointing South, rotate alpha degrees toward West. Both components are negative — the vector lies in Q3.

Domain / range. r >= 0; alpha in [0°, 90°].

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Derivation
Same logic as the other compass directions, but both reference directions (south and west) give negative components.
Example
v = (80 m/s, SW) is the same as S45°W: x = −80·sin 45° = −40√2 ≈ −56.57, y = −80·cos 45° = −40√2 ≈ −56.57.
Picture
Compass rose. Arrow leans alpha degrees from due South toward West. Tip lands in Q3.

Bearing ↔ math angle conversion

WS1
$$ \theta = 90° - B \quad\Leftrightarrow\quad B = 90° - \theta $$

In words. The math angle theta (counterclockwise from +x) and the bearing B (clockwise from +y) for the same vector are complementary in a flipped sense: theta = 90° − B.

Domain / range. Any angle pair.

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Derivation
Both measure direction, but from different reference axes and with opposite rotational senses. Starting at +x and rotating CCW by 90° lands on +y; equivalently, starting at +y and rotating CW by 90° lands on +x. So the two systems differ by a 90° shift and a sign flip on rotation.
Example
A vector at math angle 60° (60° CCW from east) has bearing B = 90° − 60° = 30° (30° CW from north — pointing northeast).
Picture
Two overlapping protractors: one measures CCW from east (math); the other measures CW from north (bearing). They sum to 90° for the same vector.

Vector addition (component-wise / tip-to-tail)

WS2
$$ \vec{v_1} + \vec{v_2} = \langle a_1 + a_2,\, b_1 + b_2\rangle $$

In words. To add two vectors, add their components separately. Geometrically: place v2's tail at v1's tip; the sum is the arrow from v1's tail to v2's tip (tip-to-tail). Equivalent: build a parallelogram with v1 and v2 as adjacent sides; the diagonal from the shared corner is the sum.

Domain / range. Always defined; commutative and associative.

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Derivation
A vector is a displacement. Walking v1 then v2 is the same total trip as the straight line from start to end. Coordinates are bookkeeping for displacement, so the components add. Parallelogram method is just doing it two ways: v1+v2 vs v2+v1.
Example
⟨−5, 1⟩ + ⟨2, 6⟩ = ⟨−3, 7⟩. Add x-components (−5+2=−3) and y-components (1+6=7) separately.
Picture
Two arrows placed tip-to-tail forming an open chain. The resultant is the straight arrow closing the chain.

Vector subtraction

WS2
$$ \vec{v} - \vec{w} = \vec{v} + (-\vec{w}) = \langle a_v - a_w,\, b_v - b_w\rangle $$

In words. To subtract vectors, subtract their components. Equivalent: add v to the opposite of w. Geometrically: v − w points from the tip of w to the tip of v.

Domain / range. Always defined.

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Derivation
Subtraction is addition of the opposite. The opposite of w is −w = ⟨−$a_w$, −$b_w$⟩ — same magnitude, reversed direction. Then use component-wise addition.
Example
If v = ⟨1, 2⟩ and w = ⟨−3, 4⟩, then v − w = ⟨1−(−3), 2−4⟩ = ⟨4, −2⟩, and w − v = −(v − w) = ⟨−4, 2⟩.
Picture
Place v and w with tails together. The vector from w's tip to v's tip is v − w.

Opposite (negation) of a vector

WS2
$$ -\vec{v} = \langle -a, -b\rangle $$

In words. The opposite of v has the same magnitude as v but points in the reverse direction. Negate every component.

Domain / range. Always defined.

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Derivation
Scalar multiplication by −1 reverses direction while preserving magnitude: |−v| = |−1|·|v| = |v|, and the direction is flipped (rotated 180°).
Example
If v = ⟨3, −2⟩, then −v = ⟨−3, 2⟩.
Picture
An arrow and its 180°-rotated twin, head-to-head.

Magnitude of a vector difference

WS2
$$ |\vec{v} - \vec{w}| = \sqrt{(a_v - a_w)^2 + (b_v - b_w)^2} $$

In words. The magnitude of v − w is the distance between the tips of v and w (when both share a tail). You must subtract the vectors FIRST, then take the magnitude; |v − w| is generally NOT equal to |v| − |w|.

Domain / range. Always non-negative.

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Derivation
v − w is itself a vector; its magnitude is computed by Pythagoras on its components. The triangle inequality guarantees |v − w| ≠ ||v| − |w|| unless v and w are parallel.
Example
$v_2$ = ⟨4, 1⟩, $v_4$ = ⟨−4, −2⟩. $v_2$ − $v_4$ = ⟨8, 3⟩, so |$v_2$ − $v_4$| = sqrt(64 + 9) = sqrt(73) ≈ 8.544. (Compare: |$v_2$| − |$v_4$| = sqrt(17) − sqrt(20) ≈ −0.35 — totally different!)
Picture
Triangle with sides v, w, and v − w. The third side closes the triangle from the tip of w to the tip of v.

Ground velocity equation (wind/current)

WS2
$$ \vec{v}_{\text{ground}} = \vec{v}_{\text{heading}} + \vec{v}_{\text{wind/current}} $$

In words. The actual velocity of a vehicle relative to the ground equals its velocity relative to the air/water (heading) PLUS the air/water's velocity relative to the ground (wind or current). Solve for $v_h$eading when you know the desired ground velocity: $v_h$eading = $v_g$round − $v_w$ind.

Domain / range. All velocity vectors in the same frame.

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Derivation
Relative velocities add. The vehicle moves at $v_h$eading through the medium; the medium moves at $v_w$ind through the ground; combine to get ground motion. This is vector addition of two displacements that happen simultaneously.
Example
Plane heads South at 400 mph; wind blows West at 20 mph. $v_g$round = ⟨0, −400⟩ + ⟨−20, 0⟩ = ⟨−20, −400⟩. Speed ≈ 400.5 mph at S2.86°W.
Picture
Two arrows tip-to-tail: the heading vector and the wind vector. The resultant is the actual ground path.

Scalar multiplication

WS3
$$ c\vec{v} = \langle ca,\, cb\rangle $$

In words. Multiplying a vector by a scalar c scales every component by c. The magnitude scales by |c|; if c > 0 the direction is preserved; if c < 0 the direction is reversed; if c = 0 you get the zero vector.

Domain / range. c any real scalar; v any vector.

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Derivation
Component-wise multiplication preserves the direction (when c > 0) because each component scales by the same factor. The magnitude scales because |cv| = sqrt((ca)² + (cb)²) = |c|·sqrt(a² + b²) = |c|·|v|.
Example
If v = ⟨3, −2⟩, then 3v = ⟨9, −6⟩, (1/2)v = ⟨3/2, −1⟩, −v = ⟨−3, 2⟩.
Picture
An arrow stretched (|c| > 1), shrunk (0 < |c| < 1), or flipped (c < 0). Always along the same line through the origin.

Linear combination

WS3
$$ \vec{v} = a\vec{x} + b\vec{y} $$

In words. A linear combination of vectors x and y with scalar coefficients a and b. Any 2D vector v can be written uniquely as a linear combination of two non-parallel vectors x and y — they form a basis for the plane. Solve by equating components to get a system of two equations in two unknowns.

Domain / range. Requires x and y to be linearly independent (not parallel) for a unique solution.

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Derivation
Set ax + by = v and equate component-by-component: a·$x_1$ + b·$y_1$ = $v_1$ and a·$x_2$ + b·$y_2$ = $v_2$. A 2x2 linear system. Unique solution iff the determinant $x_1$·$y_2$ − $x_2$·$y_1$ is nonzero (vectors not parallel).
Example
Express v = ⟨5, −3⟩ as a·x + b·y with x = ⟨−4, 1⟩, y = ⟨3, −1⟩. System: −4a + 3b = 5, a − b = −3. Solve: a = 4, b = 7. Check: 4⟨−4,1⟩ + 7⟨3,−1⟩ = ⟨−16+21, 4−7⟩ = ⟨5, −3⟩. ✓
Picture
A grid of parallelograms with sides x and y covering the plane; v lands at the (a, b) cell of that grid.

Midpoint Connector Theorem (via vectors)

WS3
$$ M,\,N \text{ midpoints of } AB,\,AC \;\Longrightarrow\; \vec{MN} = \tfrac{1}{2}\vec{BC} $$

In words. In a triangle, the segment joining the midpoints of two sides is parallel to the third side and exactly half its length. Vector proof reduces to a one-line calculation: MN = (1/2)·BC.

Domain / range. Any triangle.

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Derivation
Let u = AQ and v = QB so that AB = 2u (since A is the midpoint) and QR = 2v (since R is the midpoint). Then AB = u + v and PR = AB + BR... in the standard setup: MN = MA + AC + CN... pure vector arithmetic shows MN = (1/2)(BC), equivalently the third side scaled by 1/2.
Example
Triangle PQR. A = midpoint PQ, B = midpoint QR. Let u = AQ, v = QB. Then AB = u + v, and PR = 2u + 2v = 2(u+v) = 2·AB. So PR is parallel to AB and twice as long — equivalently AB is parallel to and half of PR.
Picture
Triangle with three midpoints connected to form a smaller similar triangle (1:2 scale). The midpoint segment is parallel to the opposite side.

Parallelogram diagonals bisect each other

WS3
$$ E = \text{midpoint of } \overline{AC} = \text{midpoint of } \overline{BD} $$

In words. In parallelogram ABCD, the diagonals AC and BD cross at a point E that is the midpoint of BOTH diagonals. Equivalently: AE = (1/2)·AC and BE = (1/2)·BD.

Domain / range. Any parallelogram.

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Derivation
In parallelogram ABCD, AC = AB + BC = AB + AD (opposite sides equal as vectors). Setting AE = (1/2)·AC and verifying BE = (1/2)·BD yields the same point.
Example
Parallelogram with AB = 3v and AD = 2u. Then AC = 2u + 3v and BD = 2u − 3v. Midpoint of AC from A: AE = u + (3/2)v. Midpoint of BD from B: BE = u − (3/2)v. Both reach the same point E.
Picture
A parallelogram with both diagonals drawn, meeting at a center point E that bisects each.

Unit vector (definition)

WS4
$$ \hat{v} = \dfrac{\vec{v}}{|\vec{v}|} $$

In words. The unit vector v-hat in the direction of v is v divided by its own magnitude. The result has length 1 and points the same way as v. The 'pure direction' of v with magnitude stripped out.

Domain / range. Defined for any nonzero vector v.

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Derivation
Scalar multiplication by 1/|v| scales the magnitude by 1/|v|·|v| = 1, leaving direction unchanged.
Example
If v = ⟨3, 4⟩ with |v| = 5, then v-hat = (1/5)⟨3, 4⟩ = ⟨3/5, 4/5⟩. Check magnitude: sqrt(9/25 + 16/25) = sqrt(1) = 1. ✓
Picture
Original arrow shrunk down to length 1 along the same line.

Magnitude-direction decomposition

WS4
$$ \vec{v} = |\vec{v}|\,\hat{v} $$

In words. Every vector splits as the product of its magnitude (how long) and its unit vector (which way). This is THE foundational identity for the entire unit — every other formula sits on top of it.

Domain / range. Any nonzero vector.

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Derivation
Direct from the definition v-hat = v/|v|: multiply both sides by |v|.
Example
For v = ⟨3, 4⟩: |v| = 5 and v-hat = ⟨3/5, 4/5⟩. Reconstruct: 5·⟨3/5, 4/5⟩ = ⟨3, 4⟩ = v. ✓
Picture
An arrow labeled |v| above the line and v-hat for the unit-length copy of the arrow.

Unit vector at angle θ from +x axis

WS4
$$ \hat{u}(\theta) = \langle\cos\theta,\, \sin\theta\rangle $$

In words. The unit vector at angle theta (measured CCW from the positive x-axis) has components ⟨cos(theta), sin(theta)⟩. This is the bridge between trigonometry (angles) and vectors (direction): every unit vector in 2D is a point on the unit circle.

Domain / range. theta any real angle.

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Derivation
By definition, cos(theta) and sin(theta) are the x and y coordinates of the unit-circle point at angle theta. That point, viewed as a position vector from the origin, IS the unit vector at angle theta. Magnitude check: sqrt(cos² + sin²) = sqrt(1) = 1. ✓
Example
At theta = 45°: u-hat = ⟨cos 45°, sin 45°⟩ = ⟨√2/2, √2/2⟩ ≈ ⟨0.707, 0.707⟩.
Picture
Unit circle with a radius drawn to the point (cos(theta), sin(theta)) at angle theta from the positive x-axis.

Standard basis vectors (2D)

WS4
$$ \mathbf{i} = \langle 1, 0\rangle,\;\; \mathbf{j} = \langle 0, 1\rangle $$

In words. The two standard unit vectors. i points east (along +x), j points north (along +y). Every 2D vector ⟨a, b⟩ = a·i + b·j.

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Derivation
These are the unit vectors at angles 0° and 90°. By definition i = ⟨cos 0°, sin 0°⟩ = ⟨1, 0⟩ and j = ⟨cos 90°, sin 90°⟩ = ⟨0, 1⟩.
Example
v = ⟨3, 4⟩ = 3i + 4j. The component 3 is 'how many i' and 4 is 'how many j' you stack to build v.
Picture
Two perpendicular unit-length arrows along the +x and +y axes.

Dot product (algebraic / component form)

WS5
$$ \vec{v}\cdot\vec{w} = v_1 w_1 + v_2 w_2 \;\;(+\; v_3 w_3\text{ in 3D}) $$

In words. The dot product of two vectors is the sum of products of corresponding components. It returns a scalar (single number), not a vector. Extends to 3D by adding the third component product.

Domain / range. Defined for two vectors of the same dimension; output is a real scalar.

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Derivation
Defined component-wise. Justified by the equivalent geometric form |v|·|w|·cos(theta), which emerges from the Law of Cosines applied to the triangle with sides v, w, v−w. The cross terms in the expansion of |v−w|² are exactly the dot product.
Example
For v = ⟨−2, 6⟩ and w = ⟨3, 4⟩: v·w = (−2)(3) + (6)(4) = −6 + 24 = 18.
Picture
Two vectors with the formula written below: multiply matching components, sum the products. The result is a single number.

Dot product (geometric form)

WS5
$$ \vec{v}\cdot\vec{w} = |\vec{v}|\,|\vec{w}|\,\cos\theta $$

In words. The dot product also equals the product of the magnitudes times the cosine of the angle between the vectors. Combined with the algebraic form, this lets you recover the angle from purely algebraic data (and vice versa).

Domain / range. theta in [0°, 180°]: the geometric angle between v and w when placed tail-to-tail.

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Derivation
From the Law of Cosines applied to the triangle with sides v, w, v−w: |v−w|² = |v|² + |w|² − 2·|v|·|w|·cos(theta). Expanding |v−w|² in components: ($v_1$−$w_1$)² + ($v_2$−$w_2$)² = $v_1$² + $w_1$² + $v_2$² + $w_2$² − 2($v_1$·$w_1$ + $v_2$·$w_2$). The squared-magnitude terms cancel, leaving $v_1$·$w_1$ + $v_2$·$w_2$ = |v|·|w|·cos(theta).
Example
v = ⟨3, 0⟩, w = ⟨0, 4⟩: v·w = 0 (perpendicular). Algebraic check: 0 = 3·4·cos 90° = 0. ✓ The dot product being zero IS the perpendicularity.
Picture
Two vectors with the angle theta between them; the dot product encodes how much they 'agree' in direction.

Angle between two vectors

WS5
$$ \cos\theta = \dfrac{\vec{v}\cdot\vec{w}}{|\vec{v}|\,|\vec{w}|} $$

In words. The cosine of the angle between two vectors equals their dot product divided by the product of their magnitudes. Take arccos to recover theta. Works in any dimension. Range of theta is [0°, 180°].

Domain / range. Defined for nonzero vectors; theta in [0°, 180°].

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Derivation
Solve the geometric dot product formula v·w = |v|·|w|·cos(theta) for cos(theta). Arccos handles both acute and obtuse angles correctly (unlike arcsin from the cross product, which can't distinguish).
Example
v = ⟨3, 2⟩, w = ⟨4, 1⟩. v·w = 14. |v| = √13, |w| = √17. cos(theta) = 14/√221 ≈ 0.9417. theta = arccos(0.9417) ≈ 19.65°.
Picture
Two vectors meeting at a point with arc showing theta; arccos lookup table or unit circle for evaluating.

Perpendicularity test (dot product)

WS5
$$ \vec{v}\perp\vec{w} \;\Longleftrightarrow\; \vec{v}\cdot\vec{w} = 0 $$

In words. Two nonzero vectors are perpendicular if and only if their dot product is zero. This is the single most useful application of the dot product: a one-line algebraic test for orthogonality.

Domain / range. Both vectors nonzero.

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Derivation
From v·w = |v|·|w|·cos(theta): for nonzero vectors, the dot product is zero iff cos(theta) = 0 iff theta = 90°.
Example
Find a so ⟨2, 6⟩ ⊥ ⟨−2, a⟩: solve (2)(−2) + (6)(a) = 0 → −4 + 6a = 0 → a = 2/3.
Picture
Two arrows meeting at a right angle (small square symbol). The dot product is the algebraic detector for that 90° corner.

Sign of dot product → angle classification

WS5
$$ \text{sign}(\vec{v}\cdot\vec{w}) \;\leftrightarrow\; \begin{cases} + & \text{acute }(\theta<90°)\\ 0 & \text{right }(\theta=90°)\\ - & \text{obtuse }(\theta>90°)\end{cases} $$

In words. You can classify the angle between two vectors without computing it: positive dot product means acute, zero means perpendicular, negative means obtuse. (Caveat: positive can also mean parallel/same direction, theta = 0°; negative can mean anti-parallel, theta = 180°. Always check parallelism before declaring 'acute' or 'obtuse.')

Domain / range. Both vectors nonzero.

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Derivation
Since v·w = |v|·|w|·cos(theta) and magnitudes are positive, the sign of the dot product matches the sign of cos(theta). Cosine is positive in (0°, 90°), zero at 90°, negative in (90°, 180°).
Example
$v_1$ = ⟨2, 5⟩, $v_2$ = ⟨−3, 1⟩: dot product = −6 + 5 = −1 < 0 → obtuse. $v_1$ = −11i − 3j, $v_2$ = −2i − j: dot product = 22 + 3 = 25 > 0 → acute. ⟨4, 9⟩ and ⟨−18, 8⟩: dot product = −72 + 72 = 0 → right angle.
Picture
Three vector pairs with arcs labeled <90°, =90°, >90° and corresponding dot-product signs.

Parallel test for vectors

WS5
$$ \vec{v}\parallel\vec{w} \;\Longleftrightarrow\; \vec{w} = k\vec{v}\text{ for some scalar }k\neq 0 $$

In words. Two nonzero vectors are parallel if and only if one is a scalar multiple of the other. If k > 0, they point the same direction (theta = 0°); if k < 0, anti-parallel (theta = 180°). Equivalent component-wise test in 2D: $v_1$/$w_1$ = $v_2$/$w_2$ (proportional components).

Domain / range. Both vectors nonzero.

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Derivation
Scalar multiplication preserves the line of a vector. Conversely, if two vectors lie on the same line through the origin, one is a stretched/flipped version of the other. In 3D, the cleanest test is v × w = 0.
Example
⟨−4, 6⟩ and ⟨8, −12⟩: ⟨8, −12⟩ = −2·⟨−4, 6⟩, so parallel (anti-parallel, k = −2, theta = 180°). To make ⟨3, 5⟩ parallel to ⟨a+3, 20⟩: need 20/5 = 4 = (a+3)/3, so a = 9.
Picture
Two arrows on the same line through the origin, possibly of different lengths or pointing opposite ways.

Line equation as dot product (normal-vector form)

WS5
$$ \vec{a}\cdot\vec{v} = c \;\;\Leftrightarrow\;\; a_1 x + a_2 y = c $$

In words. The equation a·v = c (where v = ⟨x, y⟩ is variable and a = ⟨$a_1$, $a_2$⟩ is fixed) defines a LINE. The vector a is the NORMAL VECTOR — perpendicular to the line. Changing c shifts the line parallel to itself without changing its normal direction. Every linear equation ax + by = c hides this dot product structure: the coefficients form the normal vector.

Domain / range. a nonzero; c any real.

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Derivation
Pick any two points $P_1$, $P_2$ on the line: a·$P_1$ = c and a·$P_2$ = c. Subtract: a·($P_2$ − $P_1$) = 0. Since $P_2$ − $P_1$ is a vector along the line, the equation says a ⊥ (any vector on the line). So a is the line's normal vector.
Example
For ⟨2, 1⟩·⟨x, y⟩ = 4: this is 2x + y = 4, slope −2, y-intercept 4. The vector ⟨2, 1⟩ is perpendicular to the line. In 3D this generalizes: ax + by + cz = d is a PLANE with normal ⟨a, b, c⟩.
Picture
A family of parallel lines with the same normal vector a; varying c shifts which line you have.

Dot product properties

WS5
$$ \vec{v}\cdot\vec{w} = \vec{w}\cdot\vec{v},\;\; \vec{u}\cdot(\vec{v}+\vec{w}) = \vec{u}\cdot\vec{v} + \vec{u}\cdot\vec{w},\;\; (c\vec{v})\cdot\vec{w} = c(\vec{v}\cdot\vec{w}) $$

In words. The dot product is commutative (order doesn't matter), distributive over vector addition, and compatible with scalar multiplication (the scalar can move outside).

Domain / range. Any vectors and scalars.

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Derivation
All three follow directly from the component-wise definition and the corresponding properties of real-number addition and multiplication.
Example
(2v)·w = 2(v·w): if v = ⟨1, 2⟩, w = ⟨3, 4⟩, then (2v)·w = ⟨2, 4⟩·⟨3, 4⟩ = 22 = 2·(1·3 + 2·4) = 2·11 = 22. ✓
Picture
Algebraic shorthand on a whiteboard, no figure needed.

Self dot product = magnitude squared

WS5
$$ \vec{v}\cdot\vec{v} = |\vec{v}|^2 $$

In words. The dot product of a vector with itself equals the square of its magnitude. Convenient identity used inside the projection formula and elsewhere.

Domain / range. Any vector v.

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Derivation
Algebraic: v·v = $v_1$² + $v_2$² (+ $v_3$²) = |v|². Geometric: v·v = |v|·|v|·cos(0°) = |v|².
Example
v = ⟨3, 4⟩: v·v = 9 + 16 = 25 = 5² = |v|². So |v| = sqrt(v·v) = 5.
Picture
Same vector dotted with itself — angle 0°, cosine 1, product is the square of length.

Projection (length × direction form)

WS6
$$ \text{proj}_w v = (|\vec{v}|\cos\theta)\,\hat{w} $$

In words. The projection of v onto w equals the scalar (|v|·cos(theta)) — which is the signed length of the shadow — multiplied by the unit vector w-hat — which gives it direction. 'Length-times-direction' decomposition of the projection.

Domain / range. w nonzero; theta = angle between v and w.

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Derivation
Drop a perpendicular from the tip of v to the line of w. In the resulting right triangle, the adjacent side (along w) has length |v|·cos(theta). The direction along w is given by w-hat. Multiply them.
Example
If v has magnitude 5 and makes a 60° angle with w, then |pro$j_w$ v| = 5·cos 60° = 2.5, pointing along w-hat.
Picture
Right triangle with v as hypotenuse, perpendicular dropped from v's tip to the line of w. The 'shadow' lies on w's line.

Projection formula (algebraic / dot-product form)

WS6
$$ \text{proj}_w v = \left(\dfrac{\vec{v}\cdot\vec{w}}{\vec{w}\cdot\vec{w}}\right)\vec{w} $$

In words. The projection of v onto w equals a scalar coefficient (the ratio of two dot products) times the vector w itself. NO cosine, NO arccos, NO explicit magnitudes — just dot products. This is the workhorse formula. Same form in 2D and 3D.

Domain / range. w nonzero.

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Derivation
Start from pro$j_w$ v = (|v|·cos(theta))·w-hat. Substitute cos(theta) = (v·w)/(|v|·|w|) and w-hat = w/|w|. The |v| factors cancel and one |w| combines with the other |w| to form |w|² = w·w in the denominator. Result: ((v·w)/(w·w))·w.
Example
For v = ⟨3, 1⟩, w = ⟨2, 6⟩: v·w = 12, w·w = 40, so pro$j_w$ v = (12/40)·⟨2, 6⟩ = (3/10)·⟨2, 6⟩ = ⟨3/5, 9/5⟩.
Picture
v projected onto the line through w; the foot of the perpendicular from v's tip lies at the tip of the projection vector.

Projection works for all angles (sign of dot product)

WS6
$$ \text{sign}(\text{coefficient of }\vec{w}) = \text{sign}(\vec{v}\cdot\vec{w}) $$

In words. The projection formula handles acute, right, and obtuse angles uniformly. The sign of v·w determines whether pro$j_w$ v points along w (positive, acute angle), is zero (perpendicular), or points opposite to w (negative, obtuse angle). The formula has no triangle in it — purely algebraic, valid in all three cases.

Domain / range. w nonzero; theta in [0°, 180°].

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Derivation
The denominator w·w = |w|² is always positive, so the sign of the coefficient (v·w)/(w·w) matches sign(v·w). For acute theta, v·w > 0 → projection along w. For theta = 90°, v·w = 0 → zero projection. For obtuse theta, v·w < 0 → projection opposite to w (the 'shadow falls on the negative side of the line').
Example
v = 3i + 4j, w = −2i − j (angle ≈ 153° obtuse). v·w = −10 < 0. pro$j_w$ v = (−10/5)·⟨−2, −1⟩ = −2·⟨−2, −1⟩ = ⟨4, 2⟩ — points along +x and +y, OPPOSITE to w = ⟨−2, −1⟩, as expected for the obtuse case.
Picture
Three side-by-side cases: acute (projection along w), perpendicular (zero projection), obtuse (projection points opposite w). All from the same formula.

When projection is zero

WS6
$$ \text{proj}_w v = \vec{0} \;\Longleftrightarrow\; \vec{v}\perp\vec{w} $$

In words. The projection of v onto w is the zero vector if and only if v and w are perpendicular. The shadow of v has zero length because v is hitting w's line edge-on.

Domain / range. w nonzero.

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Derivation
The projection coefficient (v·w)/(w·w) is zero iff v·w = 0, which is the perpendicularity condition.
Example
v = ⟨1, 0⟩, w = ⟨0, 1⟩ (perpendicular). v·w = 0 → pro$j_w$ v = 0·⟨0, 1⟩ = ⟨0, 0⟩. ✓
Picture
v sticking out perpendicular to w; no shadow falls on w's line.

When projection equals the original vector

WS6
$$ \text{proj}_w v = \vec{v} \;\Longleftrightarrow\; \vec{v}\parallel\vec{w} $$

In words. The projection of v onto w equals v itself if and only if v is parallel to w. The shadow IS the vector — because v already lies along w's line.

Domain / range. w nonzero.

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Derivation
If v = k·w, then pro$j_w$ v = ((k·w)·w/(w·w))·w = (k·(w·w)/(w·w))·w = k·w = v.
Example
v = ⟨6, 3⟩, w = ⟨2, 1⟩ (v = 3w). pro$j_w$ v = ((6·2+3·1)/(4+1))·⟨2, 1⟩ = (15/5)·⟨2, 1⟩ = 3·⟨2, 1⟩ = ⟨6, 3⟩ = v. ✓
Picture
v drawn along the same line as w; the projection is just v.

2D rotation matrix

WS7
$$ R(\theta) = \begin{bmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta\end{bmatrix} $$

In words. The 2x2 matrix that rotates any 2D vector counterclockwise by theta. Apply via matrix-vector multiplication: R(theta)·v gives v rotated by theta. The columns of R are the rotated basis vectors: first column = where i lands, second column = where j lands.

Domain / range. theta any real angle.

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Derivation
Any 2D linear transformation is determined by where it sends i and j. Rotation sends i = ⟨1, 0⟩ to ⟨cos(theta), sin(theta)⟩ (the unit vector at angle theta — from WS4) and sends j = ⟨0, 1⟩ to ⟨−sin(theta), cos(theta)⟩ (unit vector at angle theta+90°). Place these images as columns of the matrix.
Example
R(90°) = [[0, −1], [1, 0]]. Apply to ⟨1, 0⟩: [[0,−1],[1,0]]·⟨1,0⟩ = ⟨0, 1⟩ — i has rotated to j, exactly 90° CCW.
Picture
Original vector and its rotated image, with a curved arrow showing the angle theta of rotation.

Rotation matrix properties

WS7
$$ \det R(\theta)=1,\;\;|R(\theta)\vec{v}|=|\vec{v}|,\;\;R(\theta)^{-1}=R(-\theta)=R(\theta)^T $$

In words. Rotation matrices preserve magnitudes (isometries), preserve angles, have determinant 1 (area-preserving and orientation-preserving), and their inverse equals their transpose (which is the same as rotation by negative theta, i.e., clockwise rotation by the same angle).

Domain / range. Any theta.

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Derivation
det R = cos²(theta) + sin²(theta) = 1 by the Pythagorean identity. Magnitude preservation: expand |R·v|² in components and show it collapses to $v_1$² + $v_2$² = |v|². The transpose has cos's on the diagonal and the sines with flipped signs — exactly R(−theta).
Example
R(45°)·R(−45°) = R(0°) = I (identity). The transpose of R(45°) has top-right sin(45°) instead of −sin(45°), giving R(−45°).
Picture
A square rotated and unchanged in size; arrow lengths preserved; orientations preserved.

Rotation composition (matrix product = angle sum)

WS7
$$ R(\alpha)\,R(\beta) = R(\alpha+\beta) $$

In words. Composing two rotations is equivalent to a single rotation by the sum of the angles. The matrix product encodes the trig angle-sum identities: when you multiply out R(alpha)·R(beta), the resulting entries are cos(alpha+beta), −sin(alpha+beta), sin(alpha+beta), cos(alpha+beta).

Domain / range. Any angles alpha, beta. 2D rotations commute; 3D rotations in general do not.

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Derivation
Direct matrix multiplication. The (1,1) entry of R(alpha)·R(beta) is cos(alpha)·cos(beta) − sin(alpha)·sin(beta) = cos(alpha+beta) by the cosine sum identity. Similarly for the other entries.
Example
R(30°)·R(60°) = R(90°). The 90° rotation matrix [[0, −1], [1, 0]] should equal the product of the 30° and 60° matrices — verifies the angle sum identity cos(90°) = cos(30°)cos(60°) − sin(30°)sin(60°) = (√3/2)(1/2) − (1/2)(√3/2) = 0.
Picture
Two curved arrows added head-to-tail, total rotation equals the sum.

Inclined-plane tangential component (sliding force)

WS8
$$ |\vec{w}_T| = |\vec{w}|\sin\theta $$

In words. On a ramp inclined at angle theta above horizontal, the component of weight parallel to the ramp surface (the part that tries to slide the object down) equals weight times sin(theta).

Domain / range. theta in [0°, 90°].

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Derivation
Decompose weight vector into components along and perpendicular to the ramp. By alternate-interior-angles, the ramp angle theta also appears in the force triangle between w (hypotenuse) and $w_N$ (adjacent). Opposite-to-theta is $w_T$, hence |$w_T$| = |w|·sin(theta).
Example
433 N object on 17° ramp: |$w_T$| = 433·sin 17° ≈ 126.59 N. Sanity check at theta = 0° (flat): $w_T$ = 0, no sliding force. ✓ At theta = 90° (vertical wall): $w_T$ = w, all weight tries to slide. ✓
Picture
Block on a ramp with the weight arrow decomposed into 'along ramp' ($w_T$, sliding) and 'into ramp' ($w_N$, normal) components.

Inclined-plane normal component (pressing force)

WS8
$$ |\vec{w}_N| = |\vec{w}|\cos\theta $$

In words. On a ramp at angle theta, the component of weight perpendicular to the ramp (pressing the object into the surface) equals weight times cos(theta).

Domain / range. theta in [0°, 90°].

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Derivation
In the force triangle, $w_N$ is adjacent to theta with hypotenuse |w|, so |$w_N$| = |w|·cos(theta). Pythagorean check: $w_T$² + $w_N$² = |w|²·(sin²+cos²) = |w|². ✓
Example
415 N block on 23° ramp: |$w_N$| = 415·cos 23° ≈ 382 N (force into the plank).
Picture
Same ramp diagram as tangential; the normal component points into the ramp surface (perpendicular to it).

Inclined-plane ratio (find angle from forces)

WS8
$$ \tan\theta = \dfrac{|\vec{w}_T|}{|\vec{w}_N|} $$

In words. The ratio of tangential to normal force component equals the tangent of the ramp angle. Lets you find theta without knowing the actual weight — the weight cancels.

Domain / range. theta in [0°, 90°); $w_N$ > 0.

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Derivation
Divide |$w_T$| = |w|·sin(theta) by |$w_N$| = |w|·cos(theta). The |w| cancels and sin/cos = tan.
Example
Tangential 117.6 N, normal 49 N: tan(theta) = 117.6/49 = 2.4. theta = arctan(2.4) ≈ 67.38°.
Picture
Right triangle with legs labeled |$w_T$| (opposite) and |$w_N$| (adjacent); theta at the corner; tan = opp/adj.

Equilibrium of forces

WS8
$$ \sum_i \vec{F}_i = \vec{0} $$

In words. An object is in equilibrium (not accelerating) when the vector sum of all forces on it equals the zero vector. In 2D this single vector equation becomes TWO scalar equations: sum of x-components = 0, sum of y-components = 0. With two unknowns, the system is solvable.

Domain / range. Any system of force vectors.

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Derivation
Newton's first law: an object at rest stays at rest if net force is zero. The vector equation breaks into one scalar equation per dimension.
Example
20 kg mass hanging from cords at 30° and 45° above horizontal. Weight = 196 N. x-equation: |$T_1$|·cos 150° + |$T_2$|·cos 45° = 0. y-equation: |$T_1$|·sin 150° + |$T_2$|·sin 45° = 196. Solve: |$T_2$| ≈ 175.73 N, |$T_1$| ≈ 143.5 N.
Picture
An object with multiple force arrows (tensions, weight) that form a closed polygon when laid tip-to-tail.

Weight vector

WS8
$$ \vec{w} = -mg\,\mathbf{j},\quad |\vec{w}| = mg $$

In words. An object of mass m has weight vector pointing straight down (negative y) with magnitude m·g, where g ≈ 9.8 m/s² is gravitational acceleration. Units: kilograms times m/s² = Newtons.

Domain / range. m > 0; g ≈ 9.8 m/s² near Earth's surface.

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Derivation
Force = mass × acceleration. Gravity gives every object near Earth's surface a downward acceleration of g, so the force is m·g pointing down.
Example
20 kg mass: |w| = 20·9.8 = 196 N; weight vector = ⟨0, −196⟩. 10 kg mass: |w| = 98 N; weight vector = ⟨0, −98⟩.
Picture
Arrow pointing straight down from an object's center, labeled mg in Newtons.

Cord tension components

WS8
$$ \vec{T} = \langle |\vec{T}|\cos\alpha,\, |\vec{T}|\sin\alpha\rangle $$

In words. A cord tension with magnitude |T|, oriented at angle alpha measured CCW from the positive x-axis, has components |T|·cos(alpha) and |T|·sin(alpha). For a cord pulling up-left at angle beta above horizontal, alpha = 180° − beta. For up-right at angle beta, alpha = beta.

Domain / range. |T| >= 0; alpha any angle (typically chosen so the cord pulls 'up' away from the object).

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Derivation
A cord pulls along its length. The pulling vector has magnitude |T| and direction given by the unit vector ⟨cos(alpha), sin(alpha)⟩ at angle alpha from +x. Multiply magnitude by direction.
Example
Cord pulling up-right at 45° above horizontal with tension 100 N: T = ⟨100·cos 45°, 100·sin 45°⟩ = ⟨50√2, 50√2⟩ ≈ ⟨70.7, 70.7⟩.
Picture
Cord drawn at an angle; the tension vector along the cord has horizontal and vertical components.

3D vector component form

WS9
$$ \vec{v} = \langle v_1, v_2, v_3\rangle = v_1\mathbf{i} + v_2\mathbf{j} + v_3\mathbf{k} $$

In words. A 3D vector has three components corresponding to the three coordinate axes. The new basis vector k = ⟨0, 0, 1⟩ points in the +z direction (up). Same notation conventions as 2D, just one more component.

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Derivation
Natural extension: add a third axis perpendicular to x and y. k is the third unit basis vector.
Example
v = ⟨5, 3, 2⟩ = 5i + 3j + 2k. Tip at point (5, 3, 2). Draw as a 3D arrow with a dashed-line box showing the (5, 3, 2) corner.
Picture
Three perpendicular axes labeled x, y, z. A dashed rectangular box from origin to the point ($v_1$, $v_2$, $v_3$); the vector is the diagonal.

Standard basis vectors (3D)

WS9
$$ \mathbf{i}=\langle 1,0,0\rangle,\;\mathbf{j}=\langle 0,1,0\rangle,\;\mathbf{k}=\langle 0,0,1\rangle $$

In words. The three standard 3D unit vectors: i along +x (east), j along +y (north), k along +z (up). Mutually perpendicular, each of length 1.

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Derivation
Definitions. Mutually perpendicular: i·j = j·k = i·k = 0. Each unit length.
Example
Cross-product identities: i × j = k, j × k = i, k × i = j (cyclic). Useful for sanity-checking cross products.
Picture
Three short arrows from origin, one along each positive axis, forming a right-handed coordinate system.

Magnitude of a 3D vector

WS9
$$ |\vec{v}| = \sqrt{v_1^2 + v_2^2 + v_3^2} $$

In words. The magnitude of a 3D vector is the square root of the sum of the three squared components. Pythagoras applied twice (stacked): once for the base diagonal in the xy-plane, again for the full 3D diagonal.

Domain / range. Always non-negative; equals 0 only for the zero vector.

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Derivation
The 2D projection of v in the xy-plane has length sqrt($v_1$² + $v_2$²). The full vector is the hypotenuse of a right triangle with legs (that 2D length) and $v_3$. Pythagoras again: |v| = sqrt(($v_1$² + $v_2$²) + $v_3$²) = sqrt($v_1$² + $v_2$² + $v_3$²). 'Pythagoras stacks.'
Example
v = ⟨2, 2, 1⟩: |v| = sqrt(4 + 4 + 1) = sqrt(9) = 3. (A '2-2-1' Pythagorean-style triple in 3D.) Or v = ⟨3, 4, 12⟩: |v| = sqrt(9 + 16 + 144) = sqrt(169) = 13.
Picture
A 3D box with the diagonal as the vector; first form the floor diagonal (Pythagoras in xy), then stand up to the top corner (Pythagoras with z).

Dot product (3D)

WS9
$$ \vec{v}\cdot\vec{w} = v_1 w_1 + v_2 w_2 + v_3 w_3 $$

In words. The 3D dot product is the sum of the three products of corresponding components. The geometric identity v·w = |v|·|w|·cos(theta) holds unchanged in 3D — same angle interpretation as 2D.

Domain / range. Both vectors 3D.

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Derivation
Direct extension of the 2D formula. The Law-of-Cosines derivation goes through identically in 3D (just one more squared term in |v−w|² that cancels).
Example
v = ⟨3, 2, 4⟩, w = ⟨−1, 3, −2⟩: v·w = −3 + 6 − 8 = −5 (obtuse angle).
Picture
Three pairs of matching components multiplied and summed — same picture as 2D, just with three terms instead of two.

3D linear combination (change of basis)

WS9
$$ \vec{v} = a\vec{x} + b\vec{y} + c\vec{z} $$

In words. Any 3D vector can be written uniquely as a linear combination of three linearly independent basis vectors x, y, z. Component-wise equality produces a 3x3 linear system in unknowns a, b, c.

Domain / range. Requires x, y, z linearly independent — equivalent to det[x | y | z] ≠ 0.

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Derivation
Equate component-wise: three scalar equations in three unknowns. Solve by substitution or matrix methods. Unique solution exists iff the determinant of the basis matrix is nonzero.
Example
v = ⟨3, −1, 4⟩, x = ⟨1, 2, 1⟩, y = ⟨0, 4, −1⟩, z = ⟨−3, −2, 3⟩. System: a − 3c = 3; 2a + 4b − 2c = −1; a − b + 3c = 4. Solve: a = 75/28, b = −23/14, c = −3/28.
Picture
Three non-coplanar vectors forming an oblique basis; v expressed as a stretch-and-combine of these three.

Direction cosines

WS9
$$ \cos\alpha = \dfrac{v_1}{|\vec{v}|},\;\cos\beta = \dfrac{v_2}{|\vec{v}|},\;\cos\gamma = \dfrac{v_3}{|\vec{v}|};\;\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1 $$

In words. The angles a 3D vector makes with each coordinate axis have cosines equal to the corresponding components of the unit vector. The squared direction cosines sum to 1 — this is just the magnitude-of-unit-vector identity in disguise.

Domain / range. v nonzero; alpha, beta, gamma in [0°, 180°].

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Derivation
cos(alpha) is the cosine of the angle between v and the x-axis. From the angle formula: cos(alpha) = (v·i)/(|v|·|i|) = $v_1$/|v|. The sum of squared direction cosines is ($v_1$² + $v_2$² + $v_3$²)/|v|² = |v|²/|v|² = 1.
Example
v = ⟨2, 2, 1⟩, |v| = 3. Direction cosines: 2/3, 2/3, 1/3. Check: 4/9 + 4/9 + 1/9 = 9/9 = 1. ✓
Picture
A 3D vector with three angles drawn to each coordinate axis; their cosines are the unit-vector components.

Cross product (determinant form)

WS10
$$ \vec{v}\times\vec{w} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \\ v_1 & v_2 & v_3 \\ w_1 & w_2 & w_3\end{vmatrix} $$

In words. The cross product of two 3D vectors is computed as the symbolic determinant of a 3x3 matrix with the basis vectors i, j, k in the first row and the components of v and w in the second and third rows. Expand along the first row with sign pattern +, −, +.

Domain / range. Defined only in 3D. Output is a 3D vector.

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Derivation
Set up two perpendicularity conditions: v·(v × w) = 0 and w·(v × w) = 0. Eliminate one component of the output and the result has the structure of the 2x2 sub-determinants. The determinant mnemonic packages this cleanly.
Example
v × w = ($v_2$·$w_3$ − $v_3$·$w_2$)·i − ($v_1$·$w_3$ − $v_3$·$w_1$)·j + ($v_1$·$w_2$ − $v_2$·$w_1$)·k. For v = ⟨5, 3, 1⟩, w = ⟨2, 5, 6⟩: i: (3·6 − 1·5) = 13; j: −(5·6 − 1·2) = −28; k: (5·5 − 3·2) = 19. Result: ⟨13, −28, 19⟩.
Picture
A 3x3 matrix with the top row being symbols i, j, k and the bottom two rows being the vector components. Expand with alternating signs.

Cross product (component form)

WS10
$$ \vec{v}\times\vec{w} = \langle v_2 w_3 - v_3 w_2,\, v_3 w_1 - v_1 w_3,\, v_1 w_2 - v_2 w_1\rangle $$

In words. Direct component formula for the cross product. Each output component is a 2x2 determinant of the 'other two columns' from the input vectors. Equivalent to the determinant form, just written out.

Domain / range. v, w in 3D.

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Derivation
Expand the determinant formula along the first row. The i-component comes from the minor of column 1 (uses y- and z-columns). The j-component is the negated minor of column 2 (uses x- and z-columns). The k-component is the minor of column 3 (uses x- and y-columns).
Example
⟨1, 2, 1⟩ × ⟨−2, −1, 3⟩: i: 2·3 − 1·(−1) = 7; j: 1·(−2) − 1·3 = −5; k: 1·(−1) − 2·(−2) = 3. Result: ⟨7, −5, 3⟩.
Picture
Three boxed terms, each a 2x2 sub-determinant of the input matrix, with the middle term negated.

Cross product magnitude = parallelogram area

WS10
$$ |\vec{v}\times\vec{w}| = |\vec{v}|\,|\vec{w}|\sin\theta = \text{area of the parallelogram spanned by }\vec{v},\vec{w} $$

In words. The magnitude of the cross product equals the product of the input magnitudes times the sine of the angle between them — which is exactly the area of the parallelogram with sides v and w. This is the geometric companion to the dot product (which extracts cosine).

Domain / range. theta in [0°, 180°]; the magnitude is always non-negative.

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Derivation
For vectors in the xy-plane, the cross product is ($v_1$·$w_2$ − $v_2$·$w_1$)·k. The bounding-rectangle-minus-triangles calculation in WS10 page 10.4 verifies that |$v_1$·$w_2$ − $v_2$·$w_1$| equals the parallelogram's area. The general 3D case reduces to this by viewing the spanned plane.
Example
v = ⟨3, 0, 0⟩ and w = ⟨0, 4, 0⟩ (perpendicular). v × w = ⟨0, 0, 12⟩. |v × w| = 12, which equals the rectangle area 3 × 4 = 12. ✓
Picture
A parallelogram with sides v and w (in 3D); the area equals |v × w|, and the cross product vector sticks straight out of the parallelogram's plane.

Right-hand rule for cross product direction

WS10
$$ \widehat{\vec{v}\times\vec{w}} = \text{thumb when right-hand fingers curl from }\vec{v}\text{ to }\vec{w} $$

In words. Point your right hand's fingers in the direction of v. Curl them toward w (through the shorter angle). Your thumb points in the direction of v × w. The result is always perpendicular to both v and w; the right-hand rule disambiguates which of the two perpendicular directions you get.

Domain / range. Defined whenever v × w is nonzero.

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Derivation
Convention. Equivalent statements: (v, w, v×w) forms a right-handed coordinate system; det[v | w | v×w] > 0 when nonzero; matches the standard i × j = k orientation.
Example
i × j = k. Point right hand along +x (i), curl toward +y (j); thumb points along +z (k). ✓. Reverse: j × i = −k.
Picture
A right hand with fingers curling from v to w and thumb sticking up, labeled as the direction of v × w.

Cross product anti-commutativity

WS10
$$ \vec{w}\times\vec{v} = -(\vec{v}\times\vec{w}) $$

In words. Swapping the order of the cross product flips the result. The right-hand rule's thumb points the opposite way when you curl from w to v instead of v to w. This is the defining feature that distinguishes cross product from addition, multiplication, and dot product (all of which are commutative).

Domain / range. Any 3D v, w.

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Derivation
Swapping the second and third rows of the determinant changes its sign (basic determinant property). The component formula confirms: v × w has $v_2$·$w_3$ − $v_3$·$w_2$, while w × v has $w_2$·$v_3$ − $w_3$·$v_2$ = −($v_2$·$w_3$ − $v_3$·$w_2$).
Example
If v × w = ⟨13, −28, 19⟩, then w × v = ⟨−13, 28, −19⟩.
Picture
Two cross-product vectors pointing in exactly opposite directions, like a thumb flipped.

Cross product perpendicular to inputs

WS10
$$ (\vec{v}\times\vec{w})\cdot\vec{v} = 0 = (\vec{v}\times\vec{w})\cdot\vec{w} $$

In words. By construction the cross product v × w is perpendicular to BOTH v and w. Dotting the cross product with either input always gives zero. Useful as a self-check after computing a cross product.

Domain / range. Any 3D v, w.

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Derivation
Algebraic verification: (v×w)·v = ($v_2$·$w_3$ − $v_3$·$w_2$)·$v_1$ + ($v_3$·$w_1$ − $v_1$·$w_3$)·$v_2$ + ($v_1$·$w_2$ − $v_2$·$w_1$)·$v_3$. Expand and watch every term cancel in pairs. Same for w.
Example
For p × q = ⟨1, 1, 1⟩ from p = ⟨−1, 0, 1⟩, q = ⟨0, −1, 1⟩: (p×q)·p = −1 + 0 + 1 = 0 ✓ and (p×q)·q = 0 − 1 + 1 = 0 ✓.
Picture
Cross-product arrow sticking straight up from the plane of v and w, forming right angles with both.

Cross product = 0 iff parallel

WS10
$$ \vec{v}\times\vec{w} = \vec{0} \;\Longleftrightarrow\; \vec{v}\parallel\vec{w}\text{ (or either vector is zero)} $$

In words. Two nonzero 3D vectors are parallel if and only if their cross product is the zero vector. The 'parallel detector' in 3D — analog of the dot product's perpendicularity detector. Also: v × v = 0 always.

Domain / range. Both vectors 3D.

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Derivation
|v × w| = |v|·|w|·sin(theta). For nonzero vectors, sin(theta) = 0 iff theta = 0° or 180° iff parallel (or anti-parallel). Geometrically: the parallelogram spanned by parallel vectors has zero area.
Example
v × v = 0 for any v (the parallelogram degenerates). v = ⟨1, 2, 3⟩, w = ⟨2, 4, 6⟩ = 2v: cross product is ⟨2·6 − 3·4, 3·2 − 1·6, 1·4 − 2·2⟩ = ⟨0, 0, 0⟩. ✓
Picture
Two arrows on the same line; the 'parallelogram' is degenerate (just a line segment with zero area).

Triangle area (from two side vectors)

WS10
$$ \text{Area}_{\triangle} = \tfrac{1}{2}|\vec{v}\times\vec{w}| $$

In words. The area of the triangle formed by two vectors v and w (originating at a common vertex) is half the magnitude of their cross product. A triangle is exactly half of the parallelogram spanned by its two sides.

Domain / range. v, w nonzero and not parallel.

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Derivation
The parallelogram with sides v and w has area |v × w|. Cut it along its diagonal — get two congruent triangles. Each has area (1/2)·|v × w|.
Example
Triangle with sides ⟨3, 0, 0⟩ and ⟨5, 4, 0⟩: cross product = ⟨0, 0, 12⟩, magnitude 12, triangle area = 6. Sanity via 2D shoelace on vertices (0,0), (3,0), (5,4): (1/2)·|3·4 − 0·5| = 6. ✓
Picture
A parallelogram split by a diagonal into two congruent triangles; the triangle is half of the parallelogram.

Always use dot product (not cross) for angles

WS10
$$ \theta = \arccos\!\left(\dfrac{\vec{v}\cdot\vec{w}}{|\vec{v}|\,|\vec{w}|}\right)\;\;\text{(handles all }[0°,180°]) $$

In words. To find the angle between two vectors, ALWAYS use the dot product formula with arccos. The cross-product formula sin(theta) = |v×w|/(|v|·|w|) seems analogous, but arcsin only outputs angles in [−90°, 90°] and cannot distinguish acute from obtuse — you can get an answer 180° off the truth.

Domain / range. theta in [0°, 180°] for the angle between vectors.

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Derivation
arccos has range [0°, 180°], which exactly matches the natural range of angles between two vectors — so arccos is one-to-one and gives the correct angle. arcsin has range [−90°, 90°]; if the true angle is obtuse (90°−180°), arcsin returns its supplementary acute angle instead. Workaround: use cross-product sine for the reference angle, then check sign of dot product to determine acute vs. obtuse.
Example
m = ⟨1,1,1⟩, w = ⟨−2,3,−6⟩. Dot: m·w = −5; |m| = √3, |w| = 7. cos(theta) = −5/(7√3) ≈ −0.412; theta ≈ 114.4° (obtuse, correct). Cross: |m × w| = √122 ≈ 11.045; sin(theta) ≈ 0.911; arcsin gives 65.6° (WRONG — that's the supplementary acute angle). 114.4° + 65.6° = 180° confirms they're supplementary. Always trust the dot product.
Picture
Two side-by-side angles summing to 180° — one obtuse (correct), one acute (the wrong arcsin answer). The dot product picks the right one.

Scalar triple product

WS11
$$ \vec{u}\cdot(\vec{v}\times\vec{w}) = \det\begin{bmatrix}u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \\ w_1 & w_2 & w_3\end{bmatrix} $$

In words. The scalar triple product takes three 3D vectors and produces a scalar equal to the determinant of the 3x3 matrix with the vectors as rows. Its absolute value equals the volume of the parallelepiped spanned by the three vectors.

Domain / range. u, v, w in 3D.

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Derivation
Start with |v × w| = area of base parallelogram. The triple product u·(v×w) = |u|·|v×w|·cos(theta), where theta is the angle between u and the cross product. |u|·cos(theta) is the height of the parallelepiped above the base. Area times height = volume.
Example
u = ⟨1, 0, 0⟩, v = ⟨0, 1, 0⟩, w = ⟨0, 0, 1⟩ (standard basis): triple product = 1 (unit cube volume). For arbitrary vectors, expand the 3x3 determinant.
Picture
A parallelepiped (slanted box) with edges u, v, w; the triple product's absolute value is its volume.

Parallelepiped volume

WS11
$$ V_{\text{parallelepiped}} = |\vec{u}\cdot(\vec{v}\times\vec{w})| $$

In words. The volume of the parallelepiped with edges u, v, w is the absolute value of the scalar triple product.

Domain / range. u, v, w in 3D, linearly independent (else volume = 0).

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Derivation
Base area = |v × w|. Height = component of u perpendicular to base = |u|·|cos(angle between u and v×w)|. Volume = base·height = |v × w|·|u|·|cos| = |u·(v × w)|.
Example
Standard unit cube: u, v, w = i, j, k. Triple product = 1. Volume = 1. For u = ⟨2, 0, 0⟩, v = ⟨0, 3, 0⟩, w = ⟨0, 0, 4⟩: triple product = 24. Volume = 24.
Picture
A 3D slanted box (parallelepiped); the triple product's absolute value is the box's volume.

Tetrahedron volume (1/6 triple product)

WS11
$$ V_{\text{tetra}} = \tfrac{1}{6}|\vec{u}\cdot(\vec{v}\times\vec{w})| $$

In words. The volume of a tetrahedron with three edge vectors u, v, w meeting at one vertex equals one-sixth of the absolute value of the scalar triple product. A tetrahedron is 1/6 of the corresponding parallelepiped.

Domain / range. u, v, w in 3D, linearly independent.

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Derivation
The parallelepiped spanned by u, v, w can be sliced into 6 congruent tetrahedra (each sharing the common vertex). Triangular prism = 1/2 of parallelepiped; pyramid (tetrahedron) = 1/3 of that prism. Combine: (1/2)·(1/3) = 1/6.
Example
Tetrahedron with vertices $P_1$=(1,4,−2), $P_2$=(2,0,−1), $P_3$=(3,5,3), $P_4$=(4,−2,2). Edge vectors from $P_1$: ⟨1,−4,1⟩, ⟨2,1,5⟩, ⟨3,−6,4⟩. Triple product = −9. Volume = |−9|/6 = 3/2 cubic units.
Picture
A pyramid (tetrahedron) inside the corresponding parallelepiped; the tetrahedron occupies 1/6 of the box.

Fourth vertex of a parallelogram

WS11
$$ D = A + C - B \quad\text{(if AC is the diagonal)} $$

In words. Given three vertices A, B, C of a parallelogram, the fourth vertex D is found using the fact that the diagonals bisect each other. If A and C are opposite (diagonal AC), then midpoint of AC equals midpoint of BD, so D = A + C − B. Three different pairings of 'diagonal' give three possible fourth vertices.

Domain / range. Any three points in 2D or 3D not all collinear.

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Derivation
Diagonals of a parallelogram bisect each other (vector property). If A and C are opposite vertices, midpoint of AC = (A+C)/2 = midpoint of BD = (B+D)/2, so D = A + C − B.
Example
A = (1, −2, 5), B = (2, 3, 6), C = (4, 1, −2). Three possible fourth vertices: A+C−B = (3, −4, −3); A+B−C = (−1, 0, 13); B+C−A = (5, 6, −1).
Picture
Three given points; the missing fourth vertex completes the parallelogram. Three different choices depending on which pair is treated as the diagonal.

Dot vs. cross duality (geometric companion)

WS10
$$ \vec{v}\cdot\vec{w} = |\vec{v}|\,|\vec{w}|\cos\theta;\quad |\vec{v}\times\vec{w}| = |\vec{v}|\,|\vec{w}|\sin\theta $$

In words. Dot and cross product are the geometric companions of cosine and sine. Dot extracts the cosine (and is scalar, defined in all dimensions, commutative, zero iff perpendicular). Cross extracts the sine and a perpendicular direction (and is a 3D-only vector, anti-commutative, zero iff parallel). Together they characterize the entire geometric relationship between two vectors.

Domain / range. theta in [0°, 180°].

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Derivation
Both formulas come from the same geometric construction (parallelogram-with-angle-theta). Cosine is the 'shadow length' coefficient; sine is the 'perpendicular height' coefficient.
Example
v = ⟨3, 0, 0⟩, w = ⟨0, 4, 0⟩ (perpendicular, theta = 90°). Dot = 0 (cos 90° = 0). Cross = ⟨0, 0, 12⟩, magnitude 12 (sin 90° = 1, |v|·|w| = 12).
Picture
A side-by-side table: dot extracts cosine (scalar), cross extracts sine (vector); zero iff perpendicular (dot) vs parallel (cross).
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Polar Coordinates

66 equations

Foundation (W1)

Plotting Rule for $(r, \theta)$

WS1
$$ \text{Spin to angle } \theta, \text{ then walk } |r| \text{ units: forward if } r > 0, \text{ backward if } r < 0. $$
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Derivation
The two-step recipe for plotting a single polar point. Negative $r$ is the moment students first lose Cartesian intuition: $r=-3$ at angle $\pi/2$ is 3 units BELOW the origin, not above.
Example
$(2, \pi/6)$: rotate 30° CCW, walk forward 2. $(-2, \pi/6)$: rotate 30° CCW, walk backward 2 (ends up in Q3 at angle $7\pi/6$).

Negative-$r$ Equivalence

WS1
$$ (-r,\, \theta) \equiv (r,\, \theta + \pi) $$
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Derivation
The single sentence that converts negative-$r$ confusion into a mechanical rule: walking backward at angle $\theta$ is identical to walking forward at angle $\theta + \pi$. Lets you always work with positive $r$ if you prefer.
Example
$(-3, \pi/4) \equiv (3, 5\pi/4)$ — same Cartesian point $(-\tfrac{3\sqrt{2}}{2}, -\tfrac{3\sqrt{2}}{2})$.

The Two Families of Polar Representations

WS1
$$ \text{Same point as }(r,\theta):\quad (r,\, \theta + 2\pi k) \quad \text{or} \quad (-r,\, \theta + \pi + 2\pi k), \quad k \in \mathbb{Z} $$
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Derivation
Conceptual Peak #1. A single Cartesian point has INFINITE polar representations. Cartesian is unique; polar is not. The two families together generate every alias of a given point. This is the trade for circular elegance.
Example
The point $(\sqrt{3}, 1)$ in Cartesian is $(2, \pi/6)$, $(2, 13\pi/6)$, $(-2, 7\pi/6)$, $(2, -11\pi/6)$, ... — all the same point.

The Pole's Special Representation

WS1
$$ (0,\, \theta) = \text{pole, for any } \theta $$
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Derivation
When $r=0$ the angle is undefined — every $\theta$ works. This is why $\tan\theta = y/x$ is undefined at the origin (both numerator and denominator are zero). The pole has uncountably many representations even in $\theta \in [0, 2\pi)$.
Example
$(0, 0), (0, \pi/4), (0, 7\pi/3)$ all represent the same point — the origin.

Foundation (W1, W4)

The 4-Flavor Polar Representation

WS1, WS4
$$ \text{For each point: } (r^+, \theta^+),\;(r^+, \theta^-),\;(r^-, \theta^+),\;(r^-, \theta^-) $$
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Derivation
Every Cartesian point has 4 canonical polar representations within standard ranges: positive $r$ with positive or negative $\theta$, and negative $r$ with positive or negative $\theta$. Drilled in W4 Problem 5. The 'four-flavor' label is unique to this worksheet.
Example
$(6, -6) \to r = 6\sqrt{2}$, Q4. Four flavors: $(6\sqrt{2}, 7\pi/4), (6\sqrt{2}, -\pi/4), (-6\sqrt{2}, 3\pi/4), (-6\sqrt{2}, -5\pi/4)$.

Conversion (W1)

Polar → Rectangular: $x$

WS1
$$ x = r\cos\theta $$
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Derivation
Comes from the right triangle: $\cos\theta = x/r$. Direct substitution — the easy direction. Works in all four quadrants because $\cos\theta$ automatically carries the right sign per quadrant.
Example
$(4, \pi/3) \to x = 4\cos(\pi/3) = 4 \cdot \tfrac{1}{2} = 2$.

Polar → Rectangular: $y$

WS1
$$ y = r\sin\theta $$
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Derivation
Comes from the right triangle: $\sin\theta = y/r$. Direct substitution. Like the $x$ formula, works in all quadrants via the signs of $\sin\theta$.
Example
$(4, \pi/3) \to y = 4\sin(\pi/3) = 4 \cdot \tfrac{\sqrt{3}}{2} = 2\sqrt{3}$.

Pythagoras in Polar: $r^2$

WS1
$$ r^2 = x^2 + y^2 $$
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Derivation
Polar's Pythagorean identity. The squared form avoids the sign choice on $r$. Most polar-to-rectangular conversions go through $r^2$ rather than $r$ directly because squaring kills the sign ambiguity.
Example
Polar circle $r = 5$ → $r^2 = 25$ → $x^2 + y^2 = 25$. One step.

Slope from Origin: $\tan\theta$

WS1
$$ \tan\theta = \frac{y}{x} $$
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Derivation
Recovers the angle from the slope of the radial line, but doesn't pin down which quadrant — both $(x,y)$ and $(-x,-y)$ give the same ratio. You must check signs of $x$ and $y$ to disambiguate.
Example
$(-2, 2) \to \tan\theta = -1$. Calculator gives $-\pi/4$ (Q4), but the point is in Q2, so correct $\theta = 3\pi/4$.

Quadrant Disambiguation Cheatsheet

WS1
$$ \theta = \begin{cases} \alpha & \text{Q1 } (+,+) \\ \pi - \alpha & \text{Q2 } (-,+) \\ \pi + \alpha & \text{Q3 } (-,-) \\ 2\pi - \alpha \text{ or } -\alpha & \text{Q4 } (+,-) \end{cases} $$
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Derivation
Calculator $\arctan$ only returns values in $(-\pi/2, \pi/2)$ — Q1 and Q4 only. For Q2 and Q3, you must hand-adjust. This is the highest-yield Cartesian-to-polar trap, tested every time.
Example
$(-2, 2)$: reference angle $\alpha = \arctan|2/(-2)| = \arctan(1) = \pi/4$. Q2 → $\theta = \pi - \pi/4 = 3\pi/4$. $r = 2\sqrt{2}$.

Rectangular → Polar: $r$

WS1
$$ r = \sqrt{x^2 + y^2}\quad(\text{positive root by convention}) $$
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Derivation
Take the positive square root to get the canonical $r$. The negative root gives an equivalent representation $(-r, \theta + \pi)$, which is sometimes preferred but standard practice is $r > 0$.
Example
$(-3, -3\sqrt{3}) \to r = \sqrt{9 + 27} = \sqrt{36} = 6$.

Conversion (W4)

Cardioid → Rectangular Form

WS4
$$ r = a + a\cos\theta \;\Longleftrightarrow\; x^2 + y^2 - 2a\sqrt{x^2+y^2} = -2ax\;(\text{after manipulation}) $$
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Derivation
Cardioids have messy rectangular forms because the $\sqrt{x^2+y^2}$ term appears (from isolating $r$). The W4 Problem 4b example walks through this in reverse: given $x^2 + y^2 - 2\sqrt{x^2+y^2} = 2x$, isolate $r$ and divide to get $r = 2 + 2\cos\theta$.
Example
$x^2 + y^2 - 2\sqrt{x^2+y^2} = 2x \to r^2 - 2r = 2r\cos\theta \to r - 2 = 2\cos\theta \to r = 2 + 2\cos\theta$.

Conversion Technique (W2-W4)

Polar → Rectangular Multiply-by-$r$ Trick

WS2-WS4
$$ \text{Multiply both sides by } r:\;\;r\cdot[\text{eqn}]\;\to\;\text{substitute } r^2 \to x^2+y^2,\; r\cos\theta \to x,\; r\sin\theta \to y $$
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Derivation
The universal polar-to-rectangular technique. When the polar equation has a leftover $r$ or trig term, multiplying both sides by $r$ creates an $r^2$ on one side and substitution-ready products $r\cos\theta, r\sin\theta$ on the other. Squaring resolves sign ambiguities.
Example
$r = -10\cos\theta$: multiply by $r$ → $r^2 = -10r\cos\theta$ → $x^2 + y^2 = -10x$ → $(x+5)^2 + y^2 = 25$.

Rect ↔ Polar (W4)

Rectangular-Polar Function Correspondence

WS4
$$ y = f(x) \;\leftrightarrow\; r = f(\theta)\;\text{(replace }y \to r, x \to \theta\text{)} $$
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Derivation
Conceptual Peak #4 setup. Every rectangular sinusoid has a polar twin. Wave features map directly to polar features: max $y$ ↔ max distance from pole, $y = 0$ crossing ↔ pole crossing, $y < 0$ region ↔ inner loop. Memorize the table.
Example
$y = 2 - 2\cos x$ (max 4 at $x = \pi$, min 0 at $x = 0$) ↔ $r = 2 - 2\cos\theta$ (cardioid bulging left to $(-4,0)$, cusp at pole opening right).

Inner Loop ↔ Wave Below Axis

WS4
$$ y = d + a\sin x \text{ dips below zero} \;\Longleftrightarrow\; r = d + a\sin\theta \text{ has inner loop} $$
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Derivation
The polar inner loop is created exactly by negative-$r$ values, which correspond to the rectangular wave dipping below the $x$-axis. Condition: $a > d$ (i.e., amplitude exceeds midline). Cardioid case ($a = d$): wave just touches zero (no inner loop, but pole crossing). Convex/dimpled ($d > a$): wave stays above zero.
Example
$y = 2 - 4\sin x$: max 6, min $-2$, dips below for $x \in (\pi/6, 5\pi/6)$. Polar $r = 2 - 4\sin\theta$: inner-loop limaçon, inner loop traced as $\theta$ sweeps $(\pi/6, 5\pi/6)$.

Symmetry (W3)

Cosine Form Symmetry Axis

WS3
$$ r = D + A\cos\theta\;\text{is symmetric about the }x\text{-axis (polar axis)} $$
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Derivation
Because $\cos(-\theta) = \cos\theta$, the curve is invariant under $\theta \to -\theta$. Geometrically: cosine-form limaçons bulge along the $x$-axis (right if $A > 0$, left if $A < 0$).
Example
$r = 5 + 3\cos\theta$: dimpled limaçon symmetric about $x$-axis. Max on $+x$-axis, min on $-x$-axis.

Sine Form Symmetry Axis

WS3
$$ r = D + A\sin\theta\;\text{is symmetric about the }y\text{-axis} $$
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Derivation
Because $\sin(\pi - \theta) = \sin\theta$, the curve is invariant under $\theta \to \pi - \theta$ (reflection across $y$-axis). Sine-form limaçons bulge along the $y$-axis (up if $A > 0$, down if $A < 0$).
Example
$r = 4 + 4\sin\theta$: cardioid pointing up to $(0, 8)$, cusp at pole opening down.

Edge Cases (W2)

Absolute Value Doubles the Circle

WS2
$$ r = |\sin\theta| \;\Longrightarrow\; \text{two tangent circles (radius 1, centers }(0, \pm 1)\text{)} $$
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Derivation
Taking absolute value forces $r \geq 0$ for all $\theta$, which kills the retracing collapse that turns $r = \sin\theta$ into a single circle. The lower-half-plane gets its own mirror-image circle. Illustrates how subtle equation changes produce dramatic graph changes.
Example
$r = \sin\theta$ is ONE circle (radius 1/2 centered at $(0, 1/2)$, traced over $[0, \pi]$). $r = 2|\sin\theta|$ is TWO circles tangent at origin.

Retracing Rule for Simple Circles

WS2
$$ r = A\sin\theta \text{ or } r = A\cos\theta\;\text{traces full circle over any }\theta\text{-interval of length }\pi $$
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Derivation
The full $[0, 2\pi]$ traces the circle TWICE because the negative-$r$ rows (in the second $\pi$) plot at the same Cartesian points as the positive-$r$ rows (in the first $\pi$). This is why $A\sin\theta$ and $A\cos\theta$ are circles and not weird two-loop figures.
Example
$r = 2\sin\theta$: full circle traced over $[0, \pi]$. The interval $[\pi, 2\pi]$ retraces — the 'Be Careful!!' warning in tables is about this.

Big-Picture (W2)

Polar Function vs. Cartesian Non-function

WS2
$$ r = f(\theta) \text{ single-valued in } \theta \;\text{ but the same curve in Cartesian may fail VLT.} $$
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Derivation
Conceptual Peak #2. A polar function $r = f(\theta)$ assigns exactly one $r$ to each $\theta$ — it IS a function. But the same curve drawn in Cartesian space can fail the vertical-line test (multiple $y$ for one $x$). This is the structural advantage of polar for closed/self-intersecting curves.
Example
$r = 1 + 2\cos\theta$ is a polar function. Same curve in rectangular: $(x^2+y^2-2x)^2 = x^2+y^2$. At $x=0$: $y^4 = y^2 \Rightarrow y = 0, +1, -1$. Three points on $x=0$ — fails VLT.

Algebraic Tool (W3)

Sine-Cosine Rotation Identity

WS3
$$ \sin\theta = \cos\left(\theta - \tfrac{\pi}{2}\right) \quad\Longleftrightarrow\quad \sin(\theta - \pi/2) = -\cos\theta $$
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Derivation
The algebraic root of 'sine = cosine rotated 90°' for polar curves. Lets you collapse a sine-form into a cosine-form with a phase shift, or convert a phase-shifted sine into a reflected cosine.
Example
$r = 4\sin(\theta - \pi/2) = -4\cos\theta$ — a sine-circle phase-shifted by $\pi/2$ is the cosine-circle reflected across the $y$-axis (center at $(-2, 0)$).

Derivation (W3)

Polar Circle Derivation ($r = A\cos\theta$)

WS3 (checklist item)
$$ r = A\cos\theta \overset{\times r}{\to} r^2 = Ar\cos\theta \to x^2+y^2 = Ax \to (x-\tfrac{A}{2})^2 + y^2 = (\tfrac{A}{2})^2 $$
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Derivation
The on-demand derivation that turns a one-line polar equation into a rectangular circle equation with explicit center and radius. Memorize this derivation pattern — it's the most-tested polar derivation on the final.
Example
$r = 4\cos\theta$: $(x-2)^2 + y^2 = 4$ — radius 2, center $(2,0)$.

Meta (W3 convention bridge)

The Unifying Classification Question

Unit study guide
$$ \text{Is the trig coefficient bigger, smaller, or equal to the constant?} $$
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Derivation
Bypasses the worksheet/video convention confusion ($A$/$D$ vs $a$/$b$). The classification rule is invariant under letter naming — it's only about the relative size of the trig coefficient vs. the constant offset. Memorize the question, not the letters.
Example
$r = 5 + 2\sin\theta$: trig coefficient (2) < constant (5), and $5 < 2 \cdot 2 = 4$? No, $5 > 4$. So convex limaçon (since $D \geq 2A$).

Canonical Shapes (W2)

Circle Centered at Pole

WS2
$$ r = a \quad\Longleftrightarrow\quad x^2 + y^2 = a^2 $$
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Derivation
The cleanest example of polar's elegance. 'Every point at distance $|a|$ from the pole' is one symbol in polar; the Cartesian form needs two squared terms. $r = a$ and $r = -a$ trace the same circle.
Example
$r = 3$ is the circle of radius 3 at the origin; $r = -3$ is the same circle.

Line Through the Pole

WS2
$$ \theta = \alpha \quad\Longleftrightarrow\quad y = \tan(\alpha)\cdot x $$
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Derivation
All radii in a single direction. Because $r$ ranges over both positive and negative reals, the polar 'ray' becomes a full line through the origin in both directions.
Example
$\theta = 5\pi/6$ is the line $y = -x/\sqrt{3}$ (extends into Q2 and Q4).

Archimedean Spiral

WS2
$$ r = \theta $$
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Derivation
The simplest spiral: radius equals angle. Each full revolution adds $2\pi$ to the radius, so successive 'turns' are uniformly spaced. For $\theta < 0$, the negative-$r$ rule produces a mirror-image spiral, not a clockwise one on the same side.
Example
On $[0, 2\pi]$: a CCW spiral expanding uniformly. On $[-2\pi, 2\pi]$: two interlocking spirals mirrored across the $y$-axis, meeting at the pole.

Canonical Shapes (W2-W3)

Cosine-Circle (off-axis)

WS2-WS3
$$ r = A\cos\theta \quad\Longleftrightarrow\quad \left(x - \tfrac{A}{2}\right)^2 + y^2 = \left(\tfrac{A}{2}\right)^2 $$
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Derivation
A circle of diameter $|A|$ sitting on the $x$-axis, tangent to the $y$-axis at the pole. Center $(A/2, 0)$, radius $|A|/2$. Derivation: multiply both sides by $r$, substitute, complete the square. Full circle is traced for $\theta \in [0, \pi]$ — the second $\pi$ retraces.
Example
$r = 4\cos\theta$: circle of radius 2 centered at $(2,0)$. $r = -10\cos\theta$: circle of radius 5 centered at $(-5, 0)$.

Sine-Circle (off-axis)

WS2-WS3
$$ r = A\sin\theta \quad\Longleftrightarrow\quad x^2 + \left(y - \tfrac{A}{2}\right)^2 = \left(\tfrac{A}{2}\right)^2 $$
Show more
Derivation
A circle of diameter $|A|$ sitting on the $y$-axis, tangent to the $x$-axis at the pole. Center $(0, A/2)$, radius $|A|/2$. Sine = cosine rotated 90° CCW. Sign of $A$ flips above/below the $x$-axis.
Example
$r = 2\sin\theta$: circle radius 1 centered at $(0, 1)$. $r = -4\sin\theta$: circle radius 2 centered at $(0, -2)$.

Canonical Shapes (W3)

Rotated Cosine-Circle (phase shift)

WS3
$$ r = A\cos(\theta - C) $$
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Derivation
Replacing $\theta$ with $\theta - C$ rotates the entire circle by $C$ radians CCW about the pole (since $\theta$ is an angle, not a coordinate). The center moves to polar $(A/2, C)$. Same shape, twisted frame.
Example
$r = 6\cos(\theta - 5\pi/6)$: circle of diameter 6 with center at polar $(3, 5\pi/6)$ (upper-left). Equivalent to $r = 6\sin(\theta - \pi/3)$ via $\sin\theta = \cos(\theta - \pi/2)$.

Rotated Sine-Circle: Center Location

WS3
$$ r = A\sin(\theta - C)\;\text{has center at polar}\;\left(\tfrac{A}{2},\, C + \tfrac{\pi}{2}\right) $$
Show more
Derivation
Base sine-circle has center straight up (at angle $\pi/2$). Phase shift $C$ rotates that center by $C$ more. Useful for reading rotated-circle equations off the graph.
Example
$r = 4\sin(\theta - \pi/4)$: center at polar $(2, 3\pi/4)$ = Cartesian $(-\sqrt{2}, \sqrt{2})$.

Canonical Shapes (W4)

General Circle in Polar (through pole)

WS4
$$ r = a\cos\theta + b\sin\theta \quad\Longleftrightarrow\quad \left(x - \tfrac{a}{2}\right)^2 + \left(y - \tfrac{b}{2}\right)^2 = \tfrac{a^2 + b^2}{4} $$
Show more
Derivation
Any circle passing through the origin has center at $(a/2, b/2)$ and radius $\tfrac{1}{2}\sqrt{a^2 + b^2}$. Combines the cosine-circle and sine-circle into one form. Multiply both sides by $r$ to derive.
Example
$r = 2\cos\theta - 4\sin\theta$: center $(1, -2)$, radius $\sqrt{5}$. Polar to rectangular: $(x-1)^2 + (y+2)^2 = 5$.

Parabola in Polar (special form)

WS4
$$ r = \tan\theta\sec\theta \quad\Longleftrightarrow\quad y = x^2 $$
Show more
Derivation
Specific worksheet example: $r = \sin\theta/\cos^2\theta$ becomes $r\cos^2\theta = \sin\theta$, then multiply by $r$ once more to get $(r\cos\theta)^2 = r\sin\theta$, i.e., $x^2 = y$. Useful for spotting unusual polar-to-rectangular transforms.
Example
$r = \tan\theta\sec\theta$ traces the standard upward parabola $y = x^2$.

Lines (W4)

Vertical Line in Polar

WS4
$$ r = a\sec\theta \quad\Longleftrightarrow\quad x = a $$
Show more
Derivation
Derive: $r = a/\cos\theta \Rightarrow r\cos\theta = a \Rightarrow x = a$. Any vertical line written cleanly in polar uses secant. Sign of $a$ tells you left or right of $y$-axis.
Example
$r = 2\sec\theta$ is the line $x = 2$. $r = -2\sec\theta$ is the line $x = -2$.

Horizontal Line in Polar

WS4
$$ r = a\csc\theta \quad\Longleftrightarrow\quad y = a $$
Show more
Derivation
Derive: $r = a/\sin\theta \Rightarrow r\sin\theta = a \Rightarrow y = a$. The cosecant version of the secant-line formula. Together they give vertical and horizontal lines in compact polar form.
Example
$r = 3\csc\theta$ is the line $y = 3$.

General Line in Polar

WS4
$$ r = \frac{C}{A\sin\theta + B\cos\theta} \quad\Longleftrightarrow\quad Ay + Bx = C $$
Show more
Derivation
Every non-pole-passing straight line has this form. Derive: multiply both sides by the denominator, then substitute $r\sin\theta = y$ and $r\cos\theta = x$. Highest-yield 'line in polar' pattern.
Example
$r = 12/(4\sin\theta - 7\cos\theta) \to 4y - 7x = 12$, slope $7/4$, $y$-intercept $3$.

Limaçons (W3)

Limaçon General Form

WS3
$$ r = D + A\cos\theta \quad\text{or}\quad r = D + A\sin\theta $$
Show more
Derivation
The umbrella family. Constant $D$ adds 'baseline radius'; the trig term $A\,\text{trig}(\theta)$ modulates that baseline. The shape (cardioid, dimpled, convex, or inner-loop) depends entirely on the relationship between $|A|$ and $|D|$.
Example
$r = 4 + 3\cos\theta$ has $D=4, A=3$, ratio $D/A = 4/3 \approx 1.33$ → dimpled limaçon.

Cardioid Condition

WS3
$$ |A| = |D| \;\Longrightarrow\; \text{cardioid (heart with cusp at pole)} $$
Show more
Derivation
When $|A| = |D|$, the minimum value of $r$ is exactly zero — the curve touches the pole at exactly one $\theta$. Mnemonic: 'meant for each other → forms a heart.' Max radius is $2|A|$.
Example
$r = 2 + 2\cos\theta$ ($A=D=2$): cardioid bulging right to $(4,0)$, cusp at pole opening left. Max radius $4 = 2|A|$.

Cardioid Standard Form

WS3
$$ r = A(1 \pm \cos\theta) \quad\text{or}\quad r = A(1 \pm \sin\theta) $$
Show more
Derivation
Cardioid written with the cardioid condition built in. The $\pm$ inside the trig flips the bulge direction (cosine: right vs left; sine: up vs down). Factored form makes it obvious that $r=0$ when the trig term equals $\mp 1$.
Example
$r = 5(1 + \sin\theta)$: cardioid pointing up, max at $(0, 10)$, cusp at pole.

Cardioid Max Radius

WS3
$$ r_{\max} = 2|A| $$
Show more
Derivation
Maximum extent of the cardioid: hit when the trig term equals $+1$ (same sign as $A$). The bulge tip is at distance $2|A|$ from the pole, on the axis of symmetry.
Example
$r = 3 + 3\cos\theta$: max at $\theta=0$, $r = 6$. Bulge tip at $(6, 0)$.

Cardioid Pole-Crossing Angle

WS3-WS4
$$ r = 0 \;\text{when}\; \cos\theta = -1 \text{ (or } \sin\theta = -1\text{)} $$
Show more
Derivation
Cardioid touches the pole exactly once. For $r = A(1 + \cos\theta)$: $r=0$ at $\theta = \pi$. For $r = A(1 + \sin\theta)$: $r=0$ at $\theta = 3\pi/2$. This is the cusp — a single point, not a transit.
Example
$r = 9 + 9\cos\theta$: cusp at $(0, \pi)$ in polar coordinates (the pole itself, reached at $\theta = \pi$).

Dimpled Limaçon Condition

WS3
$$ |A| < |D| < 2|A| \;\Longleftrightarrow\; 1 < \tfrac{|D|}{|A|} < 2 $$
Show more
Derivation
Radius stays strictly positive (no pole crossing) but the modulation is strong enough to produce a visible dent on the side opposite the bulge. Mnemonic: 'the tip moved on, just a dimple left.' No inner loop.
Example
$r = 3 + 2\cos\theta$: $D/A = 1.5$ — dimpled. Max $(5, 0)$, min $(-1, 0)$, dimple on the left.

Convex Limaçon Condition

WS3-WS4
$$ |D| \geq 2|A| \;\Longleftrightarrow\; \tfrac{|D|}{|A|} \geq 2 $$
Show more
Derivation
When $D$ dominates by at least a factor of 2, the modulation is too small to dent the shape. The curve is a smooth blob — looks almost like an off-center oval, no dimple visible. Mnemonic: 'fully moved on, no trace.'
Example
$r = 12 + 6\cos\theta$: $D/A = 2$, exactly at boundary → convex. Max $(18, 0)$, min $(-6, 0)$.

Dimpled/Convex Max and Min Radius

WS3
$$ r_{\max} = |D| + |A|, \quad r_{\min} = |D| - |A| > 0 $$
Show more
Derivation
For limaçons with $|D| > |A|$, $r$ is always positive: max at the bulge ($D+A$), min at the dent ($D-A$). Both are distances from the pole on the axis of symmetry.
Example
$r = 5 + 3\cos\theta$: max $8$ (right), min $2$ (left). Curve extends from $(2, 0)$ inward... wait, from $(-2, 0)$ to $(8, 0)$.

Inner-Loop Limaçon Condition

WS3
$$ |A| > |D| \;\Longrightarrow\; \text{inner-loop limaçon (loop on the bulge side)} $$
Show more
Derivation
When the trig coefficient exceeds the constant, $r$ goes negative for some $\theta$. Those negative-$r$ values flip the points through the pole, drawing a small inner loop on the same side as the bulge. Mnemonic: 'hole in the heart.'
Example
$r = 1 + 2\cos\theta$ ($A=2 > D=1$): outer curve from $(-1,0)$ to $(3,0)$, inner loop with tip at $(1, 0)$ on the positive-$x$ side.

Inner-Loop Outer Tip Distance

WS3-WS4
$$ \text{Outer tip distance} = |A| + |D| $$
Show more
Derivation
Max value of $r = D + A\cos\theta$ (cosine form) occurs at $\theta = 0$ where $\cos\theta = 1$, giving $r = D + A$. The outer tip sits on the bulge axis at this distance from the pole.
Example
$r = 6 + 12\cos\theta$: outer tip at $(18, 0)$, distance $|A|+|D| = 18$.

Inner-Loop Length

WS3
$$ \text{Inner loop length} = |A| - |D| $$
Show more
Derivation
Min value of $r = D + A\cos\theta$ is $D - A < 0$ at $\theta = \pi$. The inner loop tip is at distance $|D - A| = A - D$ from the pole (since $A > D$), plotted on the opposite side via the negative-$r$ rule — which puts it on the SAME side as the bulge.
Example
$r = 1 + 2\cos\theta$: inner loop length $= 2 - 1 = 1$. Tip at $(1, 0)$ (positive side, same as bulge).

Inner-Loop Pole-Crossing Angles

WS3
$$ \cos\theta = -\tfrac{D}{A} \;\text{(for cosine form)} $$
Show more
Derivation
The two angles where the inner loop enters and exits the pole. Solve $D + A\cos\theta = 0$ for $\theta$. Yields two solutions in $[0, 2\pi)$, symmetric about $\theta = \pi$. The negative-$r$ portion of the curve (between these angles) traces the inner loop.
Example
$r = 1 + 2\cos\theta$: $\cos\theta = -1/2 \Rightarrow \theta = 2\pi/3, 4\pi/3$. Inner loop is drawn as $\theta$ sweeps through $(2\pi/3, 4\pi/3)$.

Perpendicular-Axis Intercepts (Inner-Loop)

WS3
$$ (0, \pm D)\;\text{for cosine form}\;(\text{or}\;(\pm D, 0)\;\text{for sine form}) $$
Show more
Derivation
On the axis perpendicular to the bulge axis, the trig term is zero, so only the constant $D$ contributes. For cosine-form limaçons, these are the $y$-intercepts. Direct read-off from $D$, no algebra.
Example
$r = 1 + 2\cos\theta$: $y$-intercepts at $(0, 1)$ and $(0, -1)$ — exactly $\pm D = \pm 1$.

Sign Flip = 180° Rotation

WS3-WS4
$$ r = D - A\cos\theta \;\equiv\; r = D + A\cos(\theta + \pi) $$
Show more
Derivation
Negating the sign of $A$ (or $D$) rotates the curve by $\pi$ about the pole. For cosine-form limaçons, this is a reflection across the $y$-axis (bulge swaps sides). For sine-form, reflection across $x$-axis. Same shape, opposite orientation.
Example
$r = 4 + 3\cos\theta$ bulges right; $r = 4 - 3\cos\theta$ bulges left to $(-7, 0)$ with the same shape.

Roses (W3)

Rose Curve General Form

WS3
$$ r = A\cos(B\theta) \quad\text{or}\quad r = A\sin(B\theta) $$
Show more
Derivation
Flower curves. $B$ controls petal count; $|A|$ controls petal length. No constant offset, so the curve always passes through the pole multiple times. Sine vs. cosine determines orientation.
Example
$r = 3\cos(2\theta)$: 4-petal rose along axes, each petal length 3.

Rose Petal Count: $B$ Odd

WS3
$$ B \text{ odd integer} \;\Longrightarrow\; B \text{ petals} $$
Show more
Derivation
Conceptual Peak #3. The most-tested polar fact on the final. Why: for odd $B$, the negative-$r$ petals overlap perfectly with the positive-$r$ petals, so each petal gets traced twice and only $B$ distinct petals appear.
Example
$r = 3\cos(3\theta)$: 3 petals (one along $+x$-axis, others at $120°, 240°$).

Rose Petal Count: $B$ Even

WS3
$$ B \text{ even integer} \;\Longrightarrow\; 2B \text{ petals} $$
Show more
Derivation
Conceptual Peak #3 continued. For even $B$, the negative-$r$ petals fall in DIFFERENT angular positions from the positive-$r$ petals, so all $2B$ potential petals appear distinctly.
Example
$r = 3\cos(2\theta)$: 4 petals. $r = 2\sin(4\theta)$: 8 petals.

Rose Petal Length

WS3
$$ \text{Petal length} = |A| $$
Show more
Derivation
Each petal extends from the pole out to distance $|A|$ at the petal tip (where the trig term hits $\pm 1$). All petals have identical length — that's what makes them a rose.
Example
$r = 4\sin(3\theta)$: 3 petals, each length 4.

Rose Petal Spacing

WS3 / yt-digest
$$ \Delta\theta = \frac{2\pi}{\text{petal count}} = \frac{360°}{\text{petal count}} $$
Show more
Derivation
Petals are equally spaced around the pole. Once you find the anchor petal, just step around by this angle to find all the others. Reviewer's emphasized formula.
Example
5-petal rose: $360°/5 = 72°$ between adjacent petals. 4-petal cosine rose: $90°$ apart (on the axes).

Anchor-Petal Workflow

WS3 / yt-digest
$$ B\theta = \tfrac{\pi}{2} \;\Longrightarrow\; \theta_{\text{anchor}} = \tfrac{\pi}{2B}\;\text{(if leading sign positive)} $$
Show more
Derivation
Find ONE petal by making the trig argument equal $\pi/2$ (so trig = +1, $r = +A$). If the leading sign of the equation is negative, anchor sits at $\theta_{\text{anchor}} + \pi$ (flipped through pole). Then space the rest evenly. Reviewer's method.
Example
$r = -4\sin(5\theta)$: $5\theta = \pi/2 \Rightarrow \theta = \pi/10$, but leading sign is negative, so anchor at $\pi/10 + \pi = 11\pi/10$. Other 4 petals at $72°$ steps.

Cosine Rose Axis Orientation

WS3
$$ r = A\cos(B\theta): \text{ petal along positive } x\text{-axis at } \theta = 0 $$
Show more
Derivation
Cosine peaks at $\theta = 0$ (where $\cos(0) = 1$), so a petal automatically sits on the positive $x$-axis. Other petals spaced from there. For even $B$: petals on both axes. For odd $B$: one petal on $+x$-axis, others at multiples of $2\pi/B$.
Example
$r = 3\cos(2\theta)$: petals at $0°, 90°, 180°, 270°$ — all four axes.

Sine Rose Rotation Offset

WS3
$$ r = A\sin(B\theta) = A\cos\!\left(B\left(\theta - \tfrac{\pi}{2B}\right)\right) $$
Show more
Derivation
Sine rose is the cosine rose rotated by $\pi/(2B)$ radians. First petal of a sine rose is at $\theta = \pi/(2B)$, not at $\theta = 0$.
Example
$r = A\sin(3\theta)$: first petal at $\theta = \pi/6 = 30°$, others at $150°$ and $270°$ (one points straight down).

Roses (W3) — TRAP

Even-$B$ Sine Rose: No Axis Petals

WS3 / yt-digest
$$ r = A\sin(B\theta), B\text{ even}:\; \text{NO petals on } x\text{- or } y\text{-axes} $$
Show more
Derivation
All $2B$ petals fall on diagonals between the axes — petal tips at $\theta = \pi/(2B), 3\pi/(2B), 5\pi/(2B), \ldots$. None lands on a coordinate axis. Likely test trap, flagged by reviewer.
Example
$r = 3\sin(4\theta)$: 8 petals at $\theta = \pi/8, 3\pi/8, 5\pi/8, \ldots$ — every $\pi/4$, but offset from axes by $\pi/8$.

Negative-Sign Rose Reflection

WS3 / yt-digest
$$ r = -A\sin(B\theta)\;\text{reflects every petal through the pole} $$
Show more
Derivation
A leading negative sign flips every petal to the opposite side of the pole. Crucial when finding the anchor petal — forgetting the negative puts all petals in wrong positions. Reviewer emphasized this is a high-yield trap.
Example
$r = -4\sin(5\theta)$: anchor petal computed at $\theta = \pi/10$ but actually lands at $\pi/10 + \pi = 11\pi/10$.

Roses (W4)

Non-Integer-$B$ Domain Rule

WS4
$$ r = f\!\left(\tfrac{p}{q}\theta\right),\;\gcd(p,q)=1\;\Longrightarrow\; \theta \in [0,\, 2\pi q] $$
Show more
Derivation
For rational non-integer $B = p/q$ in lowest terms, the curve doesn't close at $\theta = 2\pi$. You must extend the domain to $2\pi q$ (or sometimes $\pi q$ if the trig function's own symmetry suffices) for a complete graph. Calculator-discipline rule.
Example
$r = \sin(2\theta/3)$: $q=3$, full domain $[0, 6\pi]$, but $\sin$'s $2\pi$ symmetry lets $[0, 3\pi]$ work. $r = \cos(4\theta/5)$: $q=5$, full domain $[0, 10\pi]$.

Composite Curves (W4)

Composite Curve: Rose + Baseline

WS4
$$ r = 1 + \sin(n\theta):\; n\text{ connected lobes, each reaching } r=2 $$
Show more
Derivation
Adding constant 1 to a rose equation eliminates negative $r$ (since the rose has min $-1$, the offset puts min at $0$). The result: instead of $n$ or $2n$ separate petals, you get $n$ connected lobes joined at the pole. 'Limaçonification' of a rose.
Example
$r = 1 + \sin(2\theta)$: 2 connected lobes (figure-eight-like). $r = 1 + \sin(3\theta)$: 3 lobes meeting at pole. $r = 1 + \sin\theta$: cardioid (the $n=1$ case).

Intersections (W4)

Polar Intersection (Algebraic Step)

WS4
$$ r_1(\theta) = r_2(\theta) \;\Longrightarrow\; \text{solve for } \theta $$
Show more
Derivation
Step 1 of polar intersection. Set the two radial functions equal and solve. Each solution $\theta_0$ gives an intersection point at polar $(r_1(\theta_0), \theta_0)$.
Example
$r_1 = 3 - 2\cos\theta$ and $r_2 = 2$: $3 - 2\cos\theta = 2 \Rightarrow \cos\theta = 1/2 \Rightarrow \theta = \pi/3, 5\pi/3$. Two intersections at $(2, \pi/3)$ and $(2, 5\pi/3)$.

Full Polar Intersection Procedure

WS4
$$ \text{1. Set } r_1 = r_2,\; \text{solve}\\\\ \text{2. Check: do BOTH curves hit pole (any }\theta\text{ with }r=0\text{)?}\\\\ \text{3. If yes: pole is an additional intersection.} $$
Show more
Derivation
The two-step procedure that beats the pole trap. Step 1 is the algebraic intersection. Step 2 is the pole check. Skip step 2 and you systematically lose a point on every pole-trap problem on the final.
Example
For two cardioids that both have cusps at the pole, the pole is always an intersection — regardless of whether algebraic intersection finds it.

Intersections (W4) — TRAP

The Pole Trap (Intersection)

WS4
$$ \text{If both curves have any } \theta \text{ with } r=0,\;\text{the pole is an additional intersection.} $$
Show more
Derivation
Conceptual Peak #4. Setting $r_1(\theta) = r_2(\theta)$ requires the SAME $\theta$ at both curves. But polar curves typically hit the pole at DIFFERENT $\theta$ values — so algebra misses the pole intersection. Always check the pole separately as Step 2.
Example
$r_1 = 2 - 2\sin\theta$ (cardioid, hits pole at $\theta = \pi/2$) and $r_2 = 2\sin\theta$ (circle, hits pole at $\theta = 0$). Algebra gives 2 intersections at $(1, \pi/6)$ and $(1, 5\pi/6)$. The pole is the missed 3rd.

Calculus Preview (W2)

Polar Area Formula (BC Calc preview)

WS2 (forward link)
$$ A = \frac{1}{2}\int_{\alpha}^{\beta} r^2\, d\theta $$
Show more
Derivation
Mentioned as a forward-link to BC Calc in W2 Problem 6(f). For the annulus in that problem, integrating $\tfrac{1}{2}(r_{\text{outer}}^2 - r_{\text{inner}}^2)$ over $[0, 2\pi]$ matches the elementary area computation. This is the polar analogue of $\int y\,dx$.
Example
Annulus between $r=3$ and $r=4$: $A = \pi(16) - \pi(9) = 7\pi$. Via integral: $\tfrac{1}{2}\int_0^{2\pi}(16-9)\,d\theta = \tfrac{7}{2}(2\pi) = 7\pi$. ✓

Reference (yt-digest)

Lemniscate (figure-8)

yt-digest (reference)
$$ r^2 = a^2\cos(2\theta)\;(\text{horizontal})\quad\text{or}\quad r^2 = a^2\sin(2\theta)\;(\text{diagonal}) $$
Show more
Derivation
Mentioned in the YouTube review video as a 'for reference' shape, not deep-tested in the worksheets but worth knowing. Figure-eight / infinity symbol. The squared $r$ means $r$ can be positive or negative (two branches per loop). Horizontal axis for cosine, diagonal for sine.
Example
$r^2 = 4\cos(2\theta)$: horizontal lemniscate with lobes along the $x$-axis, each reaching $r = 2$ at $\theta = 0, \pi$.
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Complex Numbers

40 equations

Warm-up

Quadratic formula (warm-up recall)

W1 warm-up
$$ x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$

Domain / range. Roots of a$x^2$ + bx + c = 0 with a, b, c real (or complex) and a $\neq$ 0.

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Derivation
Completing the square on a$x^2$ + bx + c = 0. Standard Algebra 2 derivation; assumed background.
Example
For $x^2$ + 1 = 0: a=1, b=0, c=1 gives x = $\pm$ $\sqrt{-4}$/2 = $\pm$ i. Roots are non-real because discriminant $b^2$ - 4ac = -4 < 0.
Picture
Parabola y = a$x^2$ + bx + c crossing (real roots) vs. floating off the x-axis (complex roots).
Category
warm-up / prerequisite

Discriminant condition for complex roots

W1 warm-up
$$ b^2 - 4ac < 0 \iff ax^2 + bx + c = 0 \text{ has two non-real (complex conjugate) roots.} $$

Domain / range. Real coefficients a, b, c with a $\neq$ 0.

Show more
Derivation
From the quadratic formula: the $\sqrt{b^2 - 4ac}$ term is imaginary precisely when $b^2$ - 4ac is negative, producing a conjugate pair x = -b/(2a) $\pm$ i$\sqrt{4ac - b^2}$/(2a).
Example
$x^2$ - 2x + 5 = 0: $b^2$ - 4ac = 4 - 20 = -16 < 0, so roots are 1 $\pm$ 2i (a conjugate pair).
Picture
Parabola whose vertex sits above (or below, if a<0) the x-axis with no real intersection.
Category
criterion

Condition for $z^n$ = 1 on the unit circle

W2 P5
$$ \text{If } z = \mathrm{cis}(\theta) \text{ (unit modulus), then } z^n = 1 \iff n\theta \equiv 0 \pmod{360°}. $$

Domain / range. z on the unit circle; n a positive integer.

Show more
Derivation
By DeMoivre, $z^n$ = cis(n$\theta$). cis equals 1 = cis(0°) precisely when its angle is a multiple of 360°. So we need n$\theta$ = 360°k for some integer k.
Example
z = cis(15°): need 15n ≡ 0 (mod 360), i.e., 15n = 360k. Smallest positive n is 24.
Picture
Angular wrap: the cumulative angle n$\theta$ has to land back on the positive real axis (an integer multiple of a full turn).
Category
criterion

Foundations

Fundamental Theorem of Algebra

W1 warm-up
$$ \text{Every degree-}n \text{ polynomial with complex coefficients has exactly } n \text{ roots in } \mathbb{C}, \text{ counted with multiplicity.} $$

Domain / range. Polynomials of positive degree n over $\mathbb{C}$.

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Derivation
Stated without proof at this level (proofs use complex analysis / topology). Used in the warm-up to justify that $x^n$ = 1 has exactly n complex solutions.
Example
$x^4$ - 1 = 0 factors as (x-1)(x+1)(x-i)(x+i) — exactly 4 roots in $\mathbb{C}$: \{1, -1, i, -i\}.
Picture
n roots arrange on the complex plane; for $x^n$ = 1 they sit on a regular n-gon inscribed in the unit circle.
Category
theorem

Solutions of $x^n$ = 1 (warm-up case studies)

W1 warm-up
$$ n=2: x \in \{1, -1\}. \quad n=4: x \in \{1, i, -1, -i\}. \quad n=8: x = \mathrm{cis}(45° \cdot k), \; k=0,1,\dots,7. $$

Domain / range. Positive integer n.

Show more
Derivation
Factor $x^n$ - 1 or apply FTA + symmetry: the n solutions sit evenly on the unit circle at angles 360°k/n.
Example
n=8: \{1, $\mathrm{cis}$(45°), i, $\mathrm{cis}$(135°), -1, $\mathrm{cis}$(225°), -i, $\mathrm{cis}$(315°)\}. The four diagonal points equal $\pm$ $\tfrac${$\sqrt{2}$}{2} $\pm$ $\tfrac${$\sqrt{2}$}{2} i.
Picture
Regular n-gon inscribed in the unit circle with one vertex anchored at z=1.
Category
worked-case

Defining identity for i

W1 (foundational, used throughout)
$$ i^2 = -1 \quad (\text{equivalently } i = \sqrt{-1}\text{ as a defining choice}). $$

Domain / range. Foundational identity of $\mathbb{C}$.

Show more
Derivation
Postulated: introduce a symbol i with $i^2$ = -1, extend the reals to $\mathbb{C}$ = \{a + bi : a, b $\in$ $\mathbb{R}$\} with the usual ring operations.
Example
$i^3$ = $i^2$ $\cdot$ i = -i. $\quad$ $i^4$ = ($i^2$)^2 = 1. Powers of i cycle with period 4: 1, i, -1, -i.
Picture
On the Argand plane, i is the point (0,1) — the unit step in the imaginary direction. Geometrically: a 90° CCW rotation operator.
Category
definition

Rectangular (Cartesian) form of a complex number

W1 P1 (setup)
$$ z = a + bi, \quad a, b \in \mathbb{R}. $$

Domain / range. Every z $\in$ $\mathbb{C}$.

Show more
Derivation
Direct: a is the real part, b is the imaginary part. The pair (a, b) places z as a point in the Argand plane.
Example
z = -3 + 3$\sqrt{3}$$\,$i sits at the point (-3, 3$\sqrt{3}$) in the plane (quadrant II).
Picture
Argand plane: horizontal axis = real, vertical axis = imaginary. Each complex number is a 2D point / vector from the origin.
Category
definition
Related

Definition of cis(θ)

W1 P1 / Study guide §1.1
$$ \mathrm{cis}(\theta) \;:=\; \cos\theta + i \sin\theta. $$

Domain / range. All real $\theta$ (angles in degrees or radians).

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Derivation
A piece of notation introduced for compactness. Later you can connect it to Euler's formula $e^{i\theta}$ = $\cos$$\theta$ + i$\sin$$\theta$, of which cis is just shorthand.
Example
$\mathrm{cis}$(60°) = $\cos$ 60° + i $\sin$ 60° = $\tfrac{1}{2}$ + $\tfrac${$\sqrt{3}$}{2} i. $\quad$ $\mathrm{cis}$(90°) = i.
Picture
cis($\theta$) is the unit-length point on the unit circle at angle $\theta$ from the positive real axis.
Category
definition
Related
polar-form unit-circle

Polar / cis form of a complex number

W1 P1 / Study guide §1.1
$$ z = r\,\mathrm{cis}(\theta) = r(\cos\theta + i \sin\theta), \quad r = |z| \ge 0, \;\; \theta = \arg(z). $$

Domain / range. Every z $\in$ $\mathbb{C}$; r $\in$ [0, $\infty$); $\theta$ $\in$ $\mathbb{R}$ (taken mod 360° or 2$\pi$).

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Derivation
Treat z = (a,b) as a 2D vector. r is its length, $\theta$ is the angle it makes with the positive real axis. Then a = r$\cos$$\theta$, b = r$\sin$$\theta$ gives z = r($\cos$$\theta$ + i$\sin$$\theta$) = r$\,$$\mathrm{cis}$($\theta$).
Example
z = -3 + 3$\sqrt{3}$$\,$i has r = 6, $\theta$ = 120°, so z = 6$\,$$\mathrm{cis}$(120°).
Picture
Same point on the Argand plane, described by length r from the origin and angle $\theta$ — the natural language for rotations.
Category
definition / representation

Modulus (magnitude) of a complex number

W1 P1, P2 (used throughout)
$$ |z| = r = \sqrt{a^2 + b^2} \quad \text{for } z = a + bi. $$

Domain / range. Every z $\in$ $\mathbb{C}$; |z| $\ge$ 0 with equality iff z = 0.

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Derivation
Pythagorean theorem applied to the right triangle with legs a and b in the Argand plane. Equivalently, |z|^2 = z $\overline{z}$.
Example
z = -4 - 4$\sqrt{3}$$\,$i: |z| = $\sqrt{16 + 48}$ = $\sqrt{64}$ = 8.
Picture
Distance from the origin to the point (a, b). The length of the position vector representing z.
Category
definition / formula

Argument of a complex number (with quadrant fix)

W1 P1, P2 / Study guide §1.1
$$ \theta = \arg(z) = \arctan\!\left(\dfrac{b}{a}\right) \;+\; \text{quadrant correction}. $$

Domain / range. z = a + bi $\neq$ 0. $\theta$ is defined modulo 360° (or 2$\pi$); principal value typically in [0°, 360°) or (-180°, 180°].

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Derivation
$\theta$ is the angle the position vector of z makes with the positive real axis. arctan only returns values in (-90°, 90°), so it sees the ratio b/a and can't distinguish quadrant II from IV (or quadrant III from I). Always plot the point first, then pick $\theta$ in the correct quadrant.
Example
z = -3 + 3$\sqrt{3}$$\,$i: quadrant II. Reference angle $\alpha$ = $\arctan$($\sqrt{3}$) = 60°. So $\theta$ = 180° - 60° = 120°.
Picture
Sweep CCW from the positive real axis to the position vector of z.
Category
definition / formula

Quadrant-fix table for the argument

W1 teaching notes / Study guide §1.1
$$ \text{Let } \alpha = \arctan(|b|/|a|) \in (0°, 90°). \;\; \mathrm{I}\!: \theta = \alpha. \;\; \mathrm{II}\!: \theta = 180° - \alpha. \;\; \mathrm{III}\!: \theta = 180° + \alpha. \;\; \mathrm{IV}\!: \theta = 360° - \alpha. $$

Domain / range. Nonzero z = a + bi with a, b $\in$ $\mathbb{R}$; $\alpha$ is the positive acute reference angle.

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Derivation
Each quadrant of the Argand plane corresponds to a different relationship between the reference angle and the actual argument. The signs of a and b determine which.
Example
z = -1 + $\sqrt{3}$$\,$i (quadrant II), reference angle $\alpha$ = $\arctan$($\sqrt{3}$) = 60° $\Rightarrow$ $\theta$ = 180° - 60° = 120°.
Picture
Argand plane split into four quadrants with reference triangles drawn in each.
Category
lookup / procedure
Related

Polar → rectangular conversion

W1 P1, P4 / Study guide §1.1
$$ a = r \cos\theta, \quad b = r \sin\theta \quad (\text{so } z = r\,\mathrm{cis}(\theta) = r\cos\theta + i \, r\sin\theta). $$

Domain / range. Every (r, $\theta$) with r $\ge$ 0; produces a unique z = a + bi.

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Derivation
Right-triangle trig: the leg adjacent to $\theta$ is r$\cos$$\theta$ (real part), the opposite leg is r$\sin$$\theta$ (imaginary part).
Example
15$\,$$\mathrm{cis}$(90°) = 15($\cos$ 90° + i $\sin$ 90°) = 15(0 + i) = 15i. $\quad$ 4$\,$$\mathrm{cis}$(30°) = 4($\tfrac{\sqrt 3}{2}$ + $\tfrac{1}{2}$ i) = 2$\sqrt$ 3 + 2i.
Picture
Drop a perpendicular from the tip of the position vector to the real axis; the foot and the height give the rectangular coordinates.
Category
formula

Complex conjugate

W1 P5 / W2 div-proof / Study guide §4.4
$$ \overline{z} = a - bi \quad \text{for } z = a + bi. $$

Domain / range. Every z $\in$ $\mathbb{C}$.

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Derivation
Reflect z across the real axis. Algebraically: flip the sign of the imaginary part. Polar form: $\overline${r$\,$$\mathrm{cis}$($\theta$)} = r$\,$$\mathrm{cis}$(-$\theta$).
Example
$\overline${1 + $\sqrt{3}$$\,$i} = 1 - $\sqrt{3}$$\,$i. $\quad$ $\overline${2$\,$$\mathrm{cis}$(60°)} = 2$\,$$\mathrm{cis}$(-60°) = 2$\,$$\mathrm{cis}$(300°).
Picture
Mirror reflection of z across the horizontal (real) axis.
Category
definition

Product of a complex number with its conjugate

W1 P5 (conjugate trick) / W2 div-proof
$$ z \, \overline{z} = (a + bi)(a - bi) = a^2 + b^2 = |z|^2. $$

Domain / range. Every z = a + bi $\in$ $\mathbb{C}$.

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Derivation
FOIL: $a^2$ - abi + abi - $b^2$ $i^2$ = $a^2$ - $b^2$(-1) = $a^2$ + $b^2$. Uses $i^2$ = -1 and difference-of-squares structure.
Example
(1 + $\sqrt{3}$$\,$i)(1 - $\sqrt{3}$$\,$i) = 1 - ($\sqrt{3}$$\,$i)^2 = 1 - 3$i^2$ = 1 + 3 = 4 = |1 + $\sqrt{3}$$\,$i|^2.
Picture
The conjugate trick: multiplying z by $\overline{z}$ collapses to a positive real number — the squared length. This is what rationalizes denominators in complex division.
Category
identity

Cycle period of powers of cis(θ)

W2 P5 / teaching notes / Study guide §4.9
$$ \text{Smallest positive } n \text{ with } \mathrm{cis}(\theta)^n = 1 \text{ is } n = \dfrac{360°}{\gcd(\theta, 360°)} \quad (\text{integer } \theta \text{ in degrees}). $$

Domain / range. $\theta$ a rational multiple of 360° (e.g., an integer number of degrees). For irrational $\theta$ no finite n exists.

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Derivation
From the congruence n$\theta$ ≡ 0 (mod 360), the smallest positive n is 360/gcd($\theta$, 360). Equivalently: the powers of cis($\theta$) trace out the cyclic subgroup of the unit circle generated by that rotation; its order is 360/gcd.
Example
$\theta$ = 15°: gcd(15, 360) = 15, so n = 24. Powers of cis(15°) cycle with period 24 — they form the 24th roots of unity. $\theta$ = 90°: gcd(90, 360) = 90, n = 4 (powers of i cycle with period 4). $\theta$ = 72°: gcd(72, 360) = 72, n = 5 (pentagonal roots of unity).
Picture
Powers trace a regular n-gon on the unit circle; n is exactly the cycle length.
Category
formula

Trig identities (load-bearing)

Pythagorean identity (used in division proof)

W2 division proof (denominator step)
$$ \sin^2\theta + \cos^2\theta = 1. $$

Domain / range. All real $\theta$.

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Derivation
Unit-circle definition: ($\cos$$\theta$, $\sin$$\theta$) lies on the unit circle $x^2$ + $y^2$ = 1.
Example
Used in division proof to collapse the denominator: $r_2$^2 $\cos$^2$\beta$ + $r_2$^2 $\sin$^2$\beta$ = $r_2$^2($\cos$^2$\beta$ + $\sin$^2$\beta$) = $r_2$^2.
Picture
Right triangle inscribed in the unit circle: legs cos$\theta$ and sin$\theta$, hypotenuse 1.
Category
trig identity (load-bearing)

Cosine angle-addition identity

W2 mult-proof step 5 / Study guide §1.3
$$ \cos(\alpha + \beta) = \cos\alpha \cos\beta - \sin\alpha \sin\beta. $$

Domain / range. All real $\alpha$, $\beta$.

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Derivation
Established in trig class (geometric or rotation-matrix derivation). In this unit it appears as the real part of (cos$\alpha$ + i$\sin$$\alpha$)(cos$\beta$ + i$\sin$$\beta$) — i.e., it IS the real-part output of complex multiplication.
Example
$\cos$(75°) = $\cos$(45° + 30°) = $\cos$ 45° $\cos$ 30° - $\sin$ 45° $\sin$ 30° = $\tfrac{\sqrt 2}{2}$ $\cdot$ $\tfrac{\sqrt 3}{2}$ - $\tfrac{\sqrt 2}{2}$ $\cdot$ $\tfrac{1}{2}$ = $\tfrac{\sqrt 6 - \sqrt 2}{4}$.
Picture
Side-by-side equality: the real part of cis($\alpha$) $\cdot$ cis($\beta$) literally writes out this identity.
Category
trig identity (load-bearing)

Sine angle-addition identity

W2 mult-proof step 5 / Study guide §1.3
$$ \sin(\alpha + \beta) = \sin\alpha \cos\beta + \cos\alpha \sin\beta. $$

Domain / range. All real $\alpha$, $\beta$.

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Derivation
Established in trig class. In this unit it appears as the imaginary part of (cos$\alpha$ + i$\sin$$\alpha$)(cos$\beta$ + i$\sin$$\beta$) after FOIL.
Example
$\sin$(75°) = $\sin$(45° + 30°) = $\tfrac{\sqrt 2}{2}$ $\cdot$ $\tfrac{\sqrt 3}{2}$ + $\tfrac{\sqrt 2}{2}$ $\cdot$ $\tfrac{1}{2}$ = $\tfrac{\sqrt 6 + \sqrt 2}{4}$.
Picture
Imaginary part of cis($\alpha$) $\cdot$ cis($\beta$) writes out this identity.
Category
trig identity (load-bearing)

Cosine angle-subtraction identity

W2 division proof
$$ \cos(\alpha - \beta) = \cos\alpha \cos\beta + \sin\alpha \sin\beta. $$

Domain / range. All real $\alpha$, $\beta$.

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Derivation
Replace $\beta$ with -$\beta$ in the cosine-addition identity, using cos(-$\beta$) = cos$\beta$ and sin(-$\beta$) = -sin$\beta$. Appears naturally in the division proof: it's the real part after FOIL of (cis$\alpha$)(cis(-$\beta$)).
Example
Real part of numerator in division proof: $r_1$ $r_2$($\cos$$\alpha$$\cos$$\beta$ + $\sin$$\alpha$$\sin$$\beta$) = $r_1$ $r_2$ $\cos$($\alpha$ - $\beta$).
Picture
Side-by-side with cosine-addition: same machinery, opposite sign on the sin$\cdot$ sin term.
Category
trig identity (load-bearing)

Sine angle-subtraction identity

W2 division proof
$$ \sin(\alpha - \beta) = \sin\alpha \cos\beta - \cos\alpha \sin\beta. $$

Domain / range. All real $\alpha$, $\beta$.

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Derivation
From sine-addition with $\beta$ $\to$ -$\beta$. Appears as the imaginary part after FOIL in the division proof.
Example
Imaginary part of numerator in division proof: $r_1$ $r_2$($\sin$$\alpha$$\cos$$\beta$ - $\cos$$\alpha$$\sin$$\beta$) = $r_1$ $r_2$ $\sin$($\alpha$ - $\beta$).
Picture
Side-by-side with sine-addition: same terms, opposite middle sign.
Category
trig identity (load-bearing)

Polar form & geometry

Trig identities ARE complex multiplication (the unification)

W2 mult-proof / Study guide §1.3
$$ (\cos\alpha + i\sin\alpha)(\cos\beta + i\sin\beta) = \cos(\alpha+\beta) + i\sin(\alpha+\beta). $$

Domain / range. All real $\alpha$, $\beta$.

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Derivation
FOIL the left side: $\cos$$\alpha$$\cos$$\beta$ + i$\cos$$\alpha$$\sin$$\beta$ + i$\sin$$\alpha$$\cos$$\beta$ + $i^2$ $\sin$$\alpha$$\sin$$\beta$. Use $i^2$ = -1 to convert the last term to a negative real. Group real and imaginary parts; recognize cos($\alpha$+$\beta$) and sin($\alpha$+$\beta$).
Example
Read the equation aloud: 'The real part is the cosine angle-addition identity; the imaginary part is the sine angle-addition identity. Trig identities you memorized in trig class were complex multiplication the whole time.'
Picture
Two equations stacked: (1) cos($\alpha$+$\beta$) = ... ; (2) cis($\alpha$) cis($\beta$) = cis($\alpha$+$\beta$). The right-hand equation contains the left-hand equation as its real part.
Category
key insight / equality of two languages

Complex multiplication rule (multiply radii, add angles)

W1 P1-P4 (discovery), W2 mult-proof (rigor)
$$ \bigl[r_1 \, \mathrm{cis}(\theta_1)\bigr] \cdot \bigl[r_2 \, \mathrm{cis}(\theta_2)\bigr] \;=\; r_1 r_2 \, \mathrm{cis}(\theta_1 + \theta_2). $$

Domain / range. Any $z_1$ = $r_1$$\,$cis$\,$$\theta$_1 and $z_2$ = $r_2$$\,$cis$\,$$\theta$_2 in $\mathbb{C}$; $r_1$, $r_2$ $\ge$ 0. Angles taken mod 360°.

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Derivation
Write each factor as $r_k$($\cos$$\theta$_k + i$\sin$$\theta$_k). FOIL. Use $i^2$ = -1. Group real and imaginary parts. Recognize cos($\theta$_1+$\theta$_2) and sin($\theta$_1+$\theta$_2). Result collapses to $r_1$ $r_2$ $\,$ cis($\theta$_1 + $\theta$_2). The trig angle-addition identities are exactly what makes this work.
Example
W1 P1: 6$\,$cis(120°) $\cdot$ 4$\,$cis(30°) = 24$\,$cis(150°). Verify via rectangular FOIL: (-3 + 3$\sqrt{3}$$\,$i)(2$\sqrt{3}$ + 2i) = -12$\sqrt{3}$ + 12i, which converts back to 24$\,$cis(150°). $\quad$ W1 P4: 3$\,$cis(20°) $\cdot$ 5$\,$cis(70°) = 15$\,$cis(90°) = 15i.
Picture
Two arrows from the origin; multiplying scales their lengths and adds their angles into a single new arrow.
Category
theorem (PEAK 1)

Multiplication & rotation

Complex division rule (divide radii, subtract angles)

W1 P5-P6 (discovery), W2 div-proof (rigor)
$$ \dfrac{r_1 \, \mathrm{cis}(\theta_1)}{r_2 \, \mathrm{cis}(\theta_2)} \;=\; \dfrac{r_1}{r_2} \, \mathrm{cis}(\theta_1 - \theta_2). $$

Domain / range. $z_2$ $\neq$ 0 (i.e., $r_2$ > 0). Otherwise any $z_1$, $z_2$ $\in$ $\mathbb{C}$; angles mod 360°.

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Derivation
Multiply numerator and denominator by the conjugate of the denominator, $r_2$$\cos$$\beta$ - i $r_2$$\sin$$\beta$. The denominator becomes $r_2$^2($\cos$^2$\beta$ + $\sin$^2$\beta$) = $r_2$^2 by the Pythagorean identity. The numerator FOILs to $r_1$ $r_2$ $\,$ cis($\alpha$ - $\beta$) using the cosine and sine SUBTRACTION identities. Divide.
Example
W1 P5: 12$\,$cis(90°) / 2$\,$cis(60°) = 6$\,$cis(30°), verified rectangularly as 12i / (1 + $\sqrt{3}$$\,$i) = 3$\sqrt{3}$ + 3i = 6$\,$cis(30°). $\quad$ W1 P6: 15$\,$cis(140°) / 5$\,$cis(35°) = 3$\,$cis(105°).
Picture
Inverse of multiplication: shrink length by the divisor's radius, rotate backwards by the divisor's angle.
Category
theorem (PEAK 2 — proof / technique)

Multiplication by i is a 90° rotation

W1 P7-P8
$$ i = 1\,\mathrm{cis}(90°), \quad \text{so } z \cdot i = r \, \mathrm{cis}(\theta + 90°) \;\; (\text{rotate } z \text{ by } 90° \text{ CCW about the origin}). $$

Domain / range. Every z $\in$ $\mathbb{C}$.

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Derivation
Write i in cis form: i = 0 + 1$\cdot$ i corresponds to (0, 1) on the unit circle, which is the point at angle 90°. Apply the multiplication rule with $r_2$ = 1, $\theta$_2 = 90°: the radius is unchanged, the angle increases by 90°.
Example
W1 P7: z = 4 + 4$\sqrt{3}$$\,$i = 8$\,$cis(60°). Then zi = -4$\sqrt{3}$ + 4i = 8$\,$cis(150°). Length unchanged (still 8), angle rotated +90°. $\quad$ W1 P8: For z = 6$\,$cis(40°), zi = 6$\,$cis(130°) — no algebra needed.
Picture
Plot z and zi: same distance from origin, separated by a 90° CCW arc.
Category
geometric corollary (PEAK 2 — concept)

Powers of i cycle with period 4

W1 P9 (implied) / teaching notes
$$ i^1 = i, \quad i^2 = -1, \quad i^3 = -i, \quad i^4 = 1, \quad i^{n+4} = i^n. $$

Domain / range. Integer exponents n (extends to negatives via $i^{-1}$ = -i).

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Derivation
Each multiplication by i = cis(90°) adds 90° to the angle on the unit circle. Four steps = 360°, returning to 1. This is the 'turn left, turn left, turn left, turn left = back where you started' picture.
Example
$i^{2024}$ = $i^{4 \cdot 506}$ = 1. $\quad$ $i^{2025}$ = i. $\quad$ $i^{-3}$ = $i^{-3+4}$ = i.
Picture
Four-point cycle on the unit circle: 1 → i → -1 → -i → 1. A finite group of rotations by 90°.
Category
consequence

The four special rotation multipliers

W1 P9 / Study guide §5
$$ \cdot i = \mathrm{cis}(90°)\;[+90°]; \quad \cdot(-1) = \mathrm{cis}(180°)\;[+180°]; \quad \cdot(-i) = \mathrm{cis}(270°)\;[+270°,\text{ or }-90°]; \quad \cdot 1 = \mathrm{cis}(0°)\;[\text{identity}]. $$

Domain / range. Multiplication applied to any z $\in$ $\mathbb{C}$.

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Derivation
Each multiplier has modulus 1 and a specific argument (0°, 90°, 180°, 270°). By the multiplication rule, multiplying by a unit complex number cis($\alpha$) is exactly a rotation by $\alpha$ with no scaling.
Example
W1 P9: 'What happens when z is multiplied by i, -1, or -i?' — 90°, 180°, 270° CCW rotations respectively, no change in length.
Picture
Four arrows around the unit circle at 0°/90°/180°/270°; multiplication by any one of them rotates z by that angle.
Category
geometric catalog

Multiplication by any unit complex number is pure rotation

W1 P9 insight box / Study guide §1.3
$$ \text{If } |w| = 1, \text{ i.e. } w = \mathrm{cis}(\alpha), \text{ then } z \mapsto z \cdot w \text{ is rotation by } \alpha \text{ about the origin.} $$

Domain / range. z $\in$ $\mathbb{C}$, w on the unit circle.

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Derivation
Apply the multiplication rule with $r_2$ = 1: $r_1$ $\cdot$ 1 = $r_1$ unchanged; angles add by $\alpha$. Length preserved, argument shifted by $\alpha$.
Example
Multiplying z = 5$\,$cis(40°) by cis(70°) gives 5$\,$cis(110°): same length, 70° more rotation.
Picture
z moves along a circular arc centered at the origin by exactly angle $\alpha$.
Category
geometric generalization

General multiplication = rotation + scaling

W1 P9 insight / Study guide §1.3
$$ z \mapsto z \cdot \bigl(r \, \mathrm{cis}(\alpha)\bigr) \;\; \text{is rotation by } \alpha \text{ about the origin } + \text{ scaling by factor } r. $$

Domain / range. z, w $\in$ $\mathbb{C}$ with w = r$\,$cis($\alpha$), r $\ge$ 0.

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Derivation
Multiplication rule decomposes a generic w into (scale by |w|) followed by (rotate by arg w). These two transformations commute and together describe the action of multiplication.
Example
z $\cdot$ (2$\,$cis(45°)) doubles z's distance from the origin and rotates it 45° CCW.
Picture
Two-step animation: first scale z by r along its existing direction, then sweep it through angle $\alpha$.
Category
geometric interpretation

Negative-radius cis is not standard form

W2 P1f / Study guide §4.8, anti-patterns
$$ -r \, \mathrm{cis}(\theta) = r \, \mathrm{cis}(\theta + 180°), \quad r \ge 0. $$

Domain / range. Any expression where a negative scalar appears in front of a cis term.

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Derivation
Multiplying by -1 = cis(180°) adds 180° to the angle (and the modulus must remain non-negative in standard form). So a 'negative radius' is rewritten as a positive radius with 180° added.
Example
W2 P1f: -4$\,$cis(50°) is not standard. Convert: 4$\,$cis(50° + 180°) = 4$\,$cis(230°). Now the multiplication rule applies cleanly.
Picture
Same point in the plane, but expressed via the position vector pointing in the opposite direction.
Category
normalization rule

Square of a complex number in cis form

W1 P10-P11 / W2 P2
$$ z^2 = r^2 \, \mathrm{cis}(2\theta), \quad \text{for } z = r\,\mathrm{cis}(\theta). $$

Domain / range. Every z $\in$ $\mathbb{C}$.

Show more
Derivation
Multiplication rule with $z_1$ = $z_2$ = z: radii multiply (r $\cdot$ r = $r^2$), angles add ($\theta$ + $\theta$ = 2$\theta$).
Example
W1 P10: z = 5 + 5$\sqrt{3}$$\,$i = 10$\,$cis(60°), so $z^2$ = 100$\,$cis(120°). Verify: $z^2$ = (5 + 5$\sqrt{3}$$\,$i)^2 = -50 + 50$\sqrt{3}$$\,$i = 100$\,$cis(120°). ✓
Picture
Squaring doubles the angle and squares the modulus.
Category
consequence of mult rule

Cube of a complex number in cis form

W1 P11 / W2 P2
$$ z^3 = r^3 \, \mathrm{cis}(3\theta). $$

Domain / range. Every z $\in$ $\mathbb{C}$.

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Derivation
$z^3$ = $z^2$ $\cdot$ z = $r^2$$\,$cis(2$\theta$) $\cdot$ r$\,$cis($\theta$) = $r^3$$\,$cis(3$\theta$) by the multiplication rule.
Example
If z = 2$\,$cis(30°), then $z^3$ = 8$\,$cis(90°) = 8i.
Picture
Triples the angle, cubes the modulus.
Category
consequence of mult rule

De Moivre & powers

DeMoivre's Theorem

W1 P11 (preview), W2 P2 + DeMoivre section + P3 (full)
$$ \bigl[r \, \mathrm{cis}(\theta)\bigr]^n \;=\; r^n \, \mathrm{cis}(n\theta), \quad n \in \mathbb{Z}_{\ge 0} \;\; (\text{extends to all integers}). $$

Domain / range. z = r$\,$cis($\theta$) $\in$ $\mathbb{C}$, r $\ge$ 0, n a non-negative integer. Extends to n < 0 via division ($z^{-1}$ = $r^{-1}$$\,$cis(-$\theta$)) and to n = 0 trivially ($z^0$ = 1).

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Derivation
Inductive proof. Base (n=1): trivially $z^1$ = $r^1$$\,$cis(1$\theta$). Step: assume $z^k$ = $r^k$$\,$cis(k$\theta$). Then $z^{k+1}$ = $z^k$ $\cdot$ z = $r^k$$\,$cis(k$\theta$) $\cdot$ r$\,$cis($\theta$) = $r^{k+1}$$\,$cis((k+1)$\theta$) by the multiplication rule. $\blacksquare$. Punchline: DeMoivre is not a new theorem — it is the multiplication rule applied n times.
Example
W2 P3a: (2$\,$cis(12°))^6 = $2^6$$\,$cis(72°) = 64$\,$cis(72°). $\quad$ W2 P3b: (4$\,$cis(23°))^3 = 64$\,$cis(69°). $\quad$ W2 P4: (2$\,$cis(45°))^5 = 32$\,$cis(225°).
Picture
A 'staircase' of powers: each step scales by r and rotates by $\theta$. Angles climb linearly with n, radii scale exponentially with n.
Category
named theorem (PEAK 3)

DeMoivre's theorem extended to negative exponents (and n=0)

Study guide §3.3 sanity check / W2 teaching notes
$$ z^{-n} = r^{-n} \, \mathrm{cis}(-n\theta) = \dfrac{1}{r^n} \, \mathrm{cis}(-n\theta), \quad z^0 = 1. $$

Domain / range. Nonzero z = r$\,$cis($\theta$), any integer n.

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Derivation
Use the division rule: $z^{-1}$ = (1$\,$cis(0°)) / (r$\,$cis($\theta$)) = (1/r)$\,$cis(-$\theta$). Then DeMoivre with -n gives the general negative-exponent form. For n=0, $z^0$ = 1 = $r^0$$\,$cis(0$\theta$) trivially.
Example
(2$\,$cis(30°))^{-3} = (1/8)$\,$cis(-90°) = -i/8.
Picture
Negative powers spiral the opposite way: angle rotates clockwise, modulus shrinks (for r > 1) or grows (for r < 1).
Category
extension of theorem

DeMoivre vs. binomial expansion (efficiency note)

W2 teaching notes / Study guide anti-patterns
$$ (r\cos\theta + i r \sin\theta)^n \;=\; r^n \, \mathrm{cis}(n\theta) \;=\; r^n \cos(n\theta) + i \, r^n \sin(n\theta). $$

Domain / range. Non-negative integer n; equates the binomial expansion with DeMoivre's compact answer.

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Derivation
Left side could be computed by the binomial theorem, producing the multiple-angle formulas as real and imaginary parts. Right side is the one-line DeMoivre output. Equating real and imaginary parts gives multiple-angle identities (e.g. cos 3$\theta$ = 4$\cos$^3$\theta$ - 3$\cos$$\theta$).
Example
(cos$\theta$ + i$\sin$$\theta$)^3 expanded gives $\cos$^3$\theta$ + 3i$\cos$^2$\theta$$\sin$$\theta$ - 3$\cos$$\theta$$\sin$^2$\theta$ - i$\sin$^3$\theta$. Compare to cis(3$\theta$) = cos(3$\theta$) + i$\sin$(3$\theta$). Real parts: $\cos$(3$\theta$) = $\cos$^3$\theta$ - 3$\cos$$\theta$$\sin$^2$\theta$ = 4$\cos$^3$\theta$ - 3$\cos$$\theta$.
Picture
Two routes to the same answer; cis route is one line, binomial route is half a page.
Category
computational identity

Nested-power exponent compounding (W2 P3e shortcut)

W2 P3e / Study guide §4.7
$$ \Bigl(\cdots\bigl((z_1^{a_1} z_2)^{a_2} z_3\bigr)^{a_3} \cdots z_k\Bigr)^{a_k} \Rightarrow \text{angle of } z_j \text{ contributes } \theta_j \cdot \prod_{i \ge j} a_i. $$

Domain / range. Nested products of cis-form expressions with integer exponents; particularly clean when all radii are 1.

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Derivation
DeMoivre commutes with products: each successive outer exponent multiplies every inner angle by that exponent. Track the cumulative product of outer exponents from each factor outward.
Example
W2 P3e: (((cis 5°)^2 cis 10°)^2 cis 20°)^2. Outer exponents from each factor outward: 5° appears inside three squarings → multiplier $2^3$ = 8; 10° inside two → multiplier 4; 20° inside one → multiplier 2. Total angle = 8·5° + 4·10° + 2·20° = 40° + 40° + 40° = 120°. Answer: cis(120°).
Picture
Tree diagram of nested parentheses; weight each leaf by the product of its ancestor exponents.
Category
calculation shortcut
Related

Powers trajectory: spiral or circle based on |z|

W2 P4 / Study guide §4.11
$$ |z| > 1 \Rightarrow \text{powers spiral outward (log spiral)}; \quad |z| = 1 \Rightarrow \text{powers march around the unit circle}; \quad |z| < 1 \Rightarrow \text{powers spiral inward to } 0. $$

Domain / range. Any z $\in$ $\mathbb{C}$, looking at the sequence $z^0$, $z^1$, $z^2$, $\dots$.

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Derivation
By DeMoivre, |$z^n$| = $r^n$. If r > 1, $r^n$ → $\infty$; if r = 1, $r^n$ = 1; if r < 1, $r^n$ → 0. Meanwhile arg($z^n$) = n$\theta$ cycles around the circle. Combining radial growth/decay with angular rotation traces a logarithmic spiral (or a circle when r = 1).
Example
W2 P4: z = 2$\,$cis(45°). Powers: 2, 4i, 8$\,$cis(135°), -16, 32$\,$cis(225°). Each step doubles length and rotates 45°. The points lie on a logarithmic spiral (same family as nautilus shells, hurricane bands, galaxy arms).
Picture
Logarithmic spiral on the Argand plane (r > 1 outward, r < 1 inward) or a circle (r = 1).
Category
qualitative behavior

Powers of cis(θ) reduce modulo the cycle period

W2 P5
$$ \text{If } \mathrm{cis}(\theta)^N = 1 \text{ has smallest positive solution } N, \text{ then } \mathrm{cis}(\theta)^k = \mathrm{cis}(\theta)^{k \bmod N}. $$

Domain / range. z = cis($\theta$) on the unit circle with rational angle (finite cycle); k any non-negative integer.

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Derivation
DeMoivre gives cis(k$\theta$). Reduce k$\theta$ mod 360°: write k = qN + r with 0 ≤ r < N, then k$\theta$ = qN$\theta$ + r$\theta$ and qN$\theta$ is a multiple of 360°.
Example
W2 P5: z = cis(15°), N = 24. $z^{30}$ = $z^{30 mod 24}$ = $z^6$ (since 30 - 24 = 6). $z^{100}$ = $z^{100 mod 24}$ = $z^4$ (since 100 = 4·24 + 4).
Picture
Wrapping around the n-gon: every exponent reduces modulo n to land on one of the n vertices.
Category
modular structure

Roots of unity

n-th roots of unity

W1 warm-up (cases), W2 P5 (general) / Study guide §4.10
$$ z^n = 1 \iff z \in \Bigl\{ \zeta_k = \mathrm{cis}\!\left(\dfrac{360° \cdot k}{n}\right) : k = 0, 1, 2, \ldots, n-1 \Bigr\}. $$

Domain / range. Positive integer n. Equation $z^n$ = 1 over $\mathbb{C}$ has exactly these n solutions.

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Derivation
Write z = r$\,$cis($\theta$). Then $z^n$ = $r^n$$\,$cis(n$\theta$) = 1 requires $r^n$ = 1 (so r = 1) and n$\theta$ ≡ 0 (mod 360°). The latter gives $\theta$ = 360°·k/n for n distinct k in 0..n-1. Equivalently, the n powers of $\zeta$_1 = cis(360°/n) are exactly the roots, and they form a cyclic group under multiplication.
Example
6th roots of unity: $\zeta$_k = cis(60° k), k = 0..5, i.e. \{1, cis 60°, cis 120°, -1, cis 240°, cis 300°\}. $\quad$ W2 P5 case: powers of cis(15°), n = 24, gives all 24 24th-roots of unity as a regular 24-gon.
Picture
Regular n-gon inscribed in the unit circle, one vertex at z = 1. Each successive vertex is a 360°/n CCW rotation of the previous.
Category
named family (PEAK 4)

n-th roots of an arbitrary complex number (preview / bridge)

W2 P5 closing / teaching notes preview / Study guide §4.10 generalization
$$ z^n = R \, \mathrm{cis}(\alpha) \;\Longrightarrow\; z = \sqrt[n]{R} \, \mathrm{cis}\!\left(\dfrac{\alpha + 360° \cdot k}{n}\right), \quad k = 0, 1, \ldots, n-1. $$

Domain / range. Any nonzero complex R$\,$cis($\alpha$); positive integer n. Produces n distinct n-th roots.

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Derivation
Set z = r$\,$cis($\theta$). Then $z^n$ = $r^n$$\,$cis(n$\theta$) must equal R$\,$cis($\alpha$). Match moduli: $r^n$ = R $\Rightarrow$ r = $R^{1/n}$. Match angles mod 360°: n$\theta$ = $\alpha$ + 360°k $\Rightarrow$ $\theta$ = ($\alpha$ + 360°k)/n for k = 0, 1, ..., n-1. The n choices of k give n distinct n-th roots spaced 360°/n apart around a circle of radius $R^{1/n}$.
Example
Cube roots of 8$\,$cis(60°): r = $8^{1/3}$ = 2, angles (60° + 360°k)/3 for k = 0, 1, 2 → 20°, 140°, 260°. Roots: 2$\,$cis(20°), 2$\,$cis(140°), 2$\,$cis(260°). They sit on the circle of radius 2, 120° apart. (This was foreshadowed in W2 teaching notes; full development is the next worksheet.)
Picture
Regular n-gon of radius $R^{1/n}$ centered at the origin, rotated so the first vertex sits at angle $\alpha$/n.
Category
generalization / preview
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Sequences & Series

71 equations

notation

Indexed sequence notation

WS2
$$ \{a_k\}_{k=1}^{n} = a_1, a_2, a_3, \ldots, a_n $$

In words. {$a_k$} from k=1 to n is the ordered list $a_1$, $a_2$, ..., $a_n$

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Derivation
A sequence is a function whose domain is a non-empty subset of the integers. Order matters — this is NOT set notation.
Key distinction
Curly-brace SET notation {2,5,7} discards order. Indexed-sequence notation {$a_k$}_{k=1}^{n} preserves order. Use indexed notation for sequences.
Examples
input: {$n^2$ + 2}_{n=0}^{3}  ·  output: 2, 3, 6, 11
input: {2n+1}_{n=3}^{8}  ·  output: 7, 9, 11, 13, 15, 17
input: {3k}_{k=1}^{4}  ·  output: 3, 6, 9, 12

Sequence vs. Series

WS2/WS6/WS10
$$ \text{Sequence: } \{a_n\}_{n=1}^{N} \quad\text{vs.}\quad \text{Series: } \sum_{n=1}^{N} a_n $$

In words. Sequence = ordered list. Series = sum of that list.

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Derivation
A sequence is a list; a series is a sum. They are NOT interchangeable. A sequence's terms can go to zero while the series still diverges (necessary but not sufficient).
Trap
Asking 'what is the limit of the sequence?' is different from 'what is the sum of the series?'

Explicit vs. recursive formulas

WS2
$$ \text{Explicit: } a_n = f(n) \quad\quad \text{Recursive: } a_n = g(a_{n-1}, \ldots) \text{ with starting value} $$

In words. Explicit: plug n directly. Recursive: define $a_n$ from previous term(s), need a starting value.

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Examples
input: Explicit: $a_n$ = 3 - 5n  ·  output: Gives -2, -7, -12, ...
input: Recursive: $a_0$ = 1, $a_n$ = n·$a_{n-1}$  ·  output: Gives 1, 1, 2, 6, 24, ... = n!

verification

Verification habit: plug n = 1

WS3/WS10
$$ \text{After deriving } S_n \text{ or } a_n: \text{ plug } n = 1, \text{ check it gives } a_1 $$

In words. Whenever you write a closed-form formula, plug n=1 first.

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Derivation
This single habit catches off-by-one errors (the most common slip in WS3) in 5 seconds.
Anti-pattern
Failing to check at n=1 leads to writing $r^n$ instead of r^(n-1), or $a_n$ = 4n + 3 instead of 4n - 1, etc. 7 such errors in the WS3 submission.
Examples
formula: n(n+1)/2 at n=1  ·  result: 1·2/2 = 1 ✓
formula: (n/2)(2a + (n-1)d) at n=1  ·  result: (1/2)(2a + 0) = a ✓
formula: a·r^(n-1) at n=1  ·  result: a·$r^0$ = a ✓

pattern

Quadratic-fit from second differences

WS1
$$ \text{If } \Delta^2 a_n = c \text{ (constant), leading coeff} = \dfrac{c}{2} $$

In words. Constant second difference c → quadratic with leading coefficient c/2

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Derivation
When first differences are linear and second differences are constant, the sequence is quadratic.
Generalization
p-th differences constant → polynomial of degree p with leading coefficient Δ^p / p!
Example
3, 8, 15, 24, 35, ... : first diffs 5, 7, 9, 11; second diff Δ² = 2; leading coeff = 1; $a_n$ = n² + 2n

Pattern recognition decision tree

WS1
$$ \text{Compute first differences and ratios:} $$

In words. Run differences AND ratios. Constant first diff → arithmetic. Constant ratio → geometric. Constant reciprocal-diff → harmonic. Constant p-th diff → polynomial degree p.

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Tests
test: First differences constant  ·  type: Linear/Arithmetic  ·  form: $a_n$ = $a_1$ + (n-1)d
test: Second differences constant  ·  type: Quadratic  ·  form: Leading coeff = Δ²/2
test: p-th differences constant  ·  type: Degree p polynomial  ·  form: Leading coeff = Δ^p/p!
test: Ratios constant  ·  type: Geometric  ·  form: $a_n$ = $a_1$ · r^(n-1)
test: Reciprocal differences constant  ·  type: Harmonic  ·  form: Flip → arithmetic → flip

arithmetic

Arithmetic sequence n-th term (explicit)

WS3
$$ a_n = a_1 + (n-1) \cdot d $$

In words. $a_n$ = $a_1$ + (n-1) · d

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Derivation
Each term differs from the previous by a constant called the common difference d. The graph of (n, $a_n$) is a straight line with slope d.
Trap
Off-by-one error: writing (n) or (n-2) in the coefficient of d instead of (n-1). The MOST common slip on this worksheet — appeared 7 times in WS3 submission.
Verification
Plug n=1: should give $a_1$. If $a_1$ + (n-1)d at n=1 gives $a_1$ + 0 = $a_1$, the formula is right.
Expansion
Equivalently $a_n$ = dn + ($a_1$ - d). The constant is $a_1$ - d, NOT $a_1$.
Variables
a_n: the n-th term
a_1: first term
d: common difference
n: term index (starting from 1)
Examples
input: $a_1$ = 3, d = 4  ·  output: $a_n$ = 3 + (n-1)(4) = 4n - 1
input: $a_1$ = 20, d = -7  ·  output: $a_n$ = 20 + (n-1)(-7) = -7n + 27

Arithmetic sequence (recursive)

WS3
$$ a_n = \begin{cases} a_1 & n = 1 \\ a_{n-1} + d & n > 1 \end{cases} $$

In words. $a_1$ given; $a_n$ = $a_{n-1}$ + d for n > 1

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Derivation
Recursive form: each term equals the previous plus the common difference d.
Equivalent form
$a_n$ - $a_{n-1}$ = d (constant first differences)

geometric

Geometric sequence n-th term (explicit)

WS3
$$ a_n = a_1 \cdot r^{n-1} $$

In words. $a_n$ = $a_1$ · r^(n-1)

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Derivation
Each term is a constant multiple of the previous, called the common ratio r. The graph is exponential.
Trap
Writing $r^n$ instead of r^(n-1) — at n=1 this gives $a_1$·r instead of $a_1$. Or writing r^(n-2). The exponent is ALWAYS (n-1).
Verification
Plug n=1: $a_1$ · $r^0$ = $a_1$. Always verify before moving on.
Variables
a_n: the n-th term
a_1: first term
r: common ratio
n: term index (starting from 1)
Examples
input: $a_1$ = 3, r = 5  ·  output: $a_n$ = 3 · 5^(n-1)
input: $a_1$ = 12, r = √2/2  ·  output: $a_n$ = 12 · (√2/2)^(n-1) = 12 · 2^(-(n-1)/2)

Geometric sequence (recursive)

WS3
$$ a_n = \begin{cases} a_1 & n = 1 \\ r \cdot a_{n-1} & n > 1 \end{cases} $$

In words. $a_1$ given; $a_n$ = r · $a_{n-1}$ for n > 1

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Derivation
Recursive form: each term equals the previous multiplied by the common ratio r.
Equivalent form
$a_n$ / $a_{n-1}$ = r (constant ratios)

harmonic

Harmonic sequence n-th term

WS4
$$ a_n = \dfrac{1}{a_1^{-1} + (n-1)d} $$

In words. Reciprocals of a harmonic sequence form an arithmetic sequence.

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Derivation
A sequence is harmonic iff its reciprocals form an arithmetic sequence.
The flip trick
Flip every term → work with arithmetic sequence → use $a_n$ = $a_1$ + (n-1)d → flip back. This is the ENTIRE harmonic technique.
Examples
input: 12, 6, 4, 3, ... (harmonic)  ·  output: Reciprocals 1/12, 1/6, 1/4, 1/3 are arithmetic with d = 1/12
input: 10th term of harmonic 1, 1/3, 1/5, 1/7, ...  ·  output: Reciprocals 1,3,5,7,... arithmetic with d=2, $a_1$0 = 19, so $h_1$0 = 1/19

means

Arithmetic Mean (two numbers)

WS4
$$ AM = \dfrac{a + b}{2} $$

In words. AM = (a + b) / 2

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Derivation
Default 'average.' The midpoint between two numbers.

Arithmetic Mean (n numbers)

WS4
$$ AM = \dfrac{a_1 + a_2 + \cdots + a_n}{n} $$

In words. AM = ($a_1$ + $a_2$ + ... + $a_n$) / n

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Derivation
Standard average of n numbers.

Geometric Mean (two numbers)

WS4
$$ GM = \sqrt{ab} $$

In words. GM = √(ab)

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Derivation
Multiplicative averaging. Shows up in compound interest, growth rates, altitude-to-hypotenuse problems.
For geometric sequence
For a geometric sequence a, ar, ar²: GM = ∛(a·ar·ar²) = ar = middle term.
Geometric meaning
Altitude h drawn from right angle to hypotenuse: h = √(xy) where x,y are the two pieces of the hypotenuse. Proof via similar triangles: h/x = y/h ⇒ h² = xy.

Geometric Mean (n numbers)

WS4
$$ GM = (a_1 \cdot a_2 \cdots a_n)^{1/n} $$

In words. GM = ($a_1$ · $a_2$ · ... · $a_n$)^(1/n)

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Derivation
n-th root of the product.

Harmonic Mean (two numbers)

WS4
$$ HM = \dfrac{2ab}{a + b} = \dfrac{2}{\frac{1}{a} + \frac{1}{b}} $$

In words. HM = 2ab / (a + b)

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Derivation
Reciprocal of the average of the reciprocals. Shows up in rates, parallel resistance, crossing wires.
Derivation
HM = (((1/a) + (1/b))/2)^(-1). Common denominator: (b+a)/(2ab). Reciprocal: 2ab/(a+b).
Geometric meaning
Crossing wires (poles a, b): wires connect each top to opposite foot, meet at height h = ab/(a+b) = HM/2.

Harmonic Mean (n numbers)

WS4
$$ HM = \dfrac{n}{\frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_n}} $$

In words. HM = n / (1/$a_1$ + 1/$a_2$ + ... + 1/$a_n$)

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Derivation
n divided by the sum of reciprocals.

Quadratic Mean / RMS (two numbers)

WS4
$$ QM = \sqrt{\dfrac{a^2 + b^2}{2}} $$

In words. QM = √((a² + b²)/2)

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Derivation
Root-mean-square. Shows up in statistics, signal processing.

Quadratic Mean (n numbers)

WS4
$$ QM = \sqrt{\dfrac{a_1^2 + a_2^2 + \cdots + a_n^2}{n}} $$

In words. QM = √((Σ$a_i$²)/n)

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Derivation
Root mean square of n numbers.

AM-GM-HM-QM inequality

WS4
$$ HM \leq GM \leq AM \leq QM $$

In words. HM ≤ GM ≤ AM ≤ QM

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Derivation
For positive numbers, harmonic mean is smallest, then geometric, arithmetic, quadratic. Equality holds iff all numbers are equal.
Geometric proof
In a single circle diagram (WS4 Q9): four lengths visible — radius (AM), perpendicular to diameter at split point (GM), Pythagorean hypotenuse FD (QM), and altitude EG (HM). The geometric ordering of these segments proves the inequality.
Proofs
step: GM ≤ AM  ·  proof: (√a - √b)² ≥ 0 ⇒ a + b ≥ 2√(ab)
step: HM ≤ GM  ·  proof: Apply GM ≤ AM to 1/a and 1/b, then take reciprocals
step: AM ≤ QM  ·  proof: (a-b)² ≥ 0 ⇒ 2(a² + b²) ≥ (a+b)²

Insert k arithmetic means between A and B

WS4/WS10
$$ d = \dfrac{B - A}{k + 1} $$

In words. Common difference d = (B - A) / (k + 1)

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Derivation
To insert k arithmetic means between endpoints A and B, take k+1 equal steps. The arithmetic means are A + d, A + 2d, ..., A + kd.
Example
Insert 3 means between 5 and 25: d = 20/4 = 5; means are 10, 15, 20

Insert k geometric means between A and B

WS4/WS10
$$ r = \left(\dfrac{B}{A}\right)^{1/(k+1)} $$

In words. Common ratio r = (B/A)^(1/(k+1))

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Derivation
To insert k geometric means between A and B, find r so r^(k+1) = B/A. The k means are Ar, Ar², ..., A$r^k$.
Trap
For even k+1, r can be positive OR negative. e.g., r⁴ = 1/16 gives r = ±1/2 — two valid answers.
Example
Insert 2 geometric means between 27 and -125: r³ = -125/27, r = -5/3; sequence 27, -45, 75, -125

Insert k harmonic means between A and B

WS4/WS10
$$ \text{Flip endpoints} \to \text{insert arithmetic means} \to \text{flip back} $$

In words. Flip A and B, insert k arithmetic means in the reciprocal sequence, flip back

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Derivation
Use the universal harmonic trick: work in reciprocal space.
Example
Insert 2 harmonic means between 1/2 and 1/32: flip to 2 and 32, d = 10, sequence 2,12,22,32, flip: 1/2, 1/12, 1/22, 1/32

iteration

Iteration (general)

WS5
$$ x_n = f(x_{n-1}) $$

In words. Apply function f repeatedly to the previous output.

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Derivation
Build a sequence by repeatedly applying f. Need a seed $x_1$ (or $x_0$) to start.
Vocab
iteration: the process $x_n$ = f($x_{n-1}$)
seed: the starting value
orbit: the sequence $x_1$, $x_2$, $x_3$, ... generated
fixed_point: x* where f(x*) = x*
attracting: nearby seeds converge to x*
repelling: nearby seeds run away from x*
cycle: orbit eventually repeats with period k

Fixed point equation

WS5
$$ f(x^*) = x^* $$

In words. x* is a fixed point if plugging it into f returns x* itself

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Derivation
A fixed point is a value where iteration stops moving: plugging it in returns itself.
How to find
Set f(x) = x and solve algebraically.
Examples
f: f(x) = 3x - 1  ·  fixed_point: 3x - 1 = x ⇒ x = 1/2
f: f(x) = x/2 + 3  ·  fixed_point: x/2 + 3 = x ⇒ x = 6
f: f(x) = √x  ·  fixed_point: √x = x ⇒ x² = x ⇒ x = 0 or 1
f: f(x) = 1 + 1/x  ·  fixed_point: L² = L + 1 ⇒ L = φ = (1+√5)/2

Attracting/repelling test (linear)

WS5
$$ f(x) = mx + b: \text{ attracting iff } |m| < 1 $$

In words. For linear f(x) = mx + b: attracting if |m| < 1, repelling if |m| > 1

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Derivation
The slope determines stability. Smaller |m| = faster convergence.
Examples
f: 3x - 1, m = 3  ·  behavior: repelling (|m| > 1)
f: x/2 + 3, m = 1/2  ·  behavior: attracting (|m| < 1), globally
f: 2x/5 + 12, m = 2/5  ·  behavior: attracting

Attracting/repelling test (general)

WS5
$$ f \text{ general: attracting iff } |f'(x^*)| < 1 $$

In words. Attracting if |f'(x*)| < 1, repelling if |f'(x*)| > 1

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Derivation
Calculus version of the test. Evaluate the derivative at the fixed point.
Example: sqrt
f(x) = √x at x*=1: f'(1) = 1/(2√1) = 1/2 < 1, so 1 is attracting. At x*=0: f'(0) = ∞ > 1, so 0 is repelling.

Fixed-point shortcut for linear recurrence

WS5/WS9
$$ x_n = m \cdot x_{n-1} + b \implies L = \dfrac{b}{1 - m} $$

In words. For linear iteration $x_n$ = m·$x_{n-1}$ + b, fixed point L = b / (1 - m)

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Derivation
Quick way to find fixed point for linear recurrence.
Derivation
Set L = mL + b ⇒ L(1 - m) = b ⇒ L = b/(1-m)
Verification
After finding L, plug back: L should equal mL + b.
Examples
input: $x_n$ = $x_{n-1}$/2 + 3  ·  output: L = 3/(1 - 1/2) = 6
input: $x_n$ = 2$x_{n-1}$/5 + 12  ·  output: L = 12/(1 - 2/5) = 12/(3/5) = 20
input: $x_n$ = 3$x_{n-1}$ - 1  ·  output: L = -1/(1-3) = 1/2 (but |m|=3 > 1, REPELLING)

Iterative root-finding recipe

WS5
$$ g(x) = 0 \implies x = f(x) \implies x_n = f(x_{n-1}) $$

In words. To solve g(x) = 0: rewrite as x = f(x), iterate, watch for convergence

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Derivation
Method for approximating roots when algebra is hard.
Recipe
  • Rewrite g(x) = 0 as x = f(x) (multiple rearrangements possible)
  • Iterate $x_n$ = f($x_{n-1}$) from a seed near the suspected root
  • If it converges → you've found a root
  • If it diverges → try a different rearrangement
Convergence rule
Rearrangements where |f'(x*)| < 1 at the desired root will converge.
Common tricks
  • Solve for the linear x-term: x = (rest of equation)
  • Take a root: x = √(g(x)) or similar
  • Use x = x + g(x) as starting point

Famous fixed points

WS5/WS10
$$ \text{Several iconic iterations and their fixed points} $$

In words. Iconic fixed points worth recognizing

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Note
For cos(x) use radians, not degrees
Fixed points
f: 1 + 1/x  ·  fixed_point: Golden ratio φ  ·  value: (1+√5)/2 ≈ 1.618
f: cos(x)  ·  fixed_point: Dottie number  ·  value: ≈ 0.7391
f: √x  ·  fixed_point: x = 0 (repelling), x = 1 (attracting)  ·  value: 0, 1
f: x/2 + 3  ·  fixed_point: Global attractor  ·  value: 6

Quadratic iteration cycles

WS5
$$ p(x) = \dfrac{x^2}{2} - \dfrac{3x}{2} - 1: \text{ period-3 cycle } \{1, -2, 4\} $$

In words. Quadratic iterations can have stable cycles of various periods

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Derivation
Verify cycle: p(1) = -2, p(-2) = 4, p(4) = 1. Some seeds join the cycle, others diverge.
Logistic map note
The logistic map $x_{n+1}$ = r·$x_n$(1-$x_n$) exhibits all these as r varies, including period-doubling cascade to chaos. 'Iteration Forever' is the MI4 subtitle named after this.
Iteration types
  • Fixed points (length-1 cycles)
  • Cycles of any length 2, 3, 4, ...
  • Chaotic behavior (bounded but no stable cycle)
  • Divergence to ±∞

Modular iteration (orbit cycles)

WS9
$$ a_n = (2 a_{n-1} + 3) \bmod 7 $$

In words. Modular iterations partition the modular group into orbits

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Derivation
Iteration on Z/7. Every element belongs to exactly one orbit (fixed point or cycle).
Example: orbits
orbit: {4}  ·  type: Fixed point
orbit: {0, 3, 2}  ·  type: Cycle of length 3
orbit: {1, 5, 6}  ·  type: Cycle of length 3

Fibonacci ratios from iteration

WS5
$$ x_n = 1 + \dfrac{1}{x_{n-1}}, x_1 = 1 \implies x_n \to \varphi = \dfrac{1 + \sqrt{5}}{2} $$

In words. The iteration x → 1 + 1/x generates ratios of consecutive Fibonacci numbers and converges to φ

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Derivation
Starting from $x_1$ = 1: 1, 2, 3/2, 5/3, 8/5, 13/8, 21/13, ... (Fibonacci ratios). Converges to golden ratio.
Fixed-point equation
L = 1 + 1/L ⇒ L² = L + 1 ⇒ L² - L - 1 = 0 ⇒ L = (1+√5)/2

series-notation

Sigma notation

WS6
$$ \sum_{k=1}^{n} a_k = a_1 + a_2 + \cdots + a_n $$

In words. Σ from k=1 to n of $a_k$ = $a_1$ + $a_2$ + ... + $a_n$

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Derivation
Compact notation for sums. The index variable (here k) runs over consecutive integers.
Trivial cases
empty_sum: n = 0 (or n < lower bound): sum = 0
single_term: n = 1: sum = $a_1$

Product notation (Π)

WS7
$$ \prod_{k=1}^{n} a_k = a_1 \cdot a_2 \cdots a_n $$

In words. Π from k=1 to n of $a_k$ = $a_1$ · $a_2$ · ... · $a_n$

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Derivation
Same indexed-family idea as Σ, but multiplication instead of addition.
Trivial cases
empty_product: n = 0 (or n < lower bound): product = 1

Indexed union

WS2/WS7
$$ \bigcup_{k=1}^{n} A_k = A_1 \cup A_2 \cup \cdots \cup A_n $$

In words. Union from k=1 to n of $A_k$

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Derivation
Every element in AT LEAST ONE of the indexed sets. Grows monotonically as n increases.
Interval rule
For closed intervals $A_k$ = [ℓ_k, $r_k$]: union has left endpoint min(ℓ_k), right endpoint max($r_k$) (provided min is achieved).

Indexed intersection

WS2/WS7
$$ \bigcap_{k=1}^{n} A_k = A_1 \cap A_2 \cap \cdots \cap A_n $$

In words. Intersection from k=1 to n of $A_k$

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Derivation
Every element in ALL of the indexed sets. Shrinks monotonically as n increases.
Interval rule
For closed intervals $A_k$ = [ℓ_k, $r_k$]: intersection has left endpoint max(ℓ_k), right endpoint min($r_k$). Empty if max(ℓ) > min(r).

Infinite union endpoint trap

WS7
$$ \bigcup_{k=1}^{\infty} \left[1 + \dfrac{1}{k}, 4\right] = (1, 4] $$

In words. If left endpoint sequence converges to L but never reaches L, then L is NOT in the union

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Derivation
When endpoints approach a limit from above without reaching it, the limit is excluded from the union (open at L). This is the WS7 Q10 trap.
Reasoning
For any specific k, left endpoint 1+1/k > 1. The value 1 is the limit, not a left endpoint itself. The union takes OR — 1 has to be in at least one set, which it isn't. But every x > 1 eventually lands in some interval.

Sequence of partial sums

WS6
$$ S_n = \sum_{k=1}^{n} a_k; \quad \text{recursively: } S_1 = a_1, \; S_n = S_{n-1} + a_n $$

In words. The sequence $S_1$, $S_2$, $S_3$, ... where each $S_n$ is the partial sum up to term n

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Derivation
Add one more term at a time. Series convergence means the partial-sum sequence converges.
Recursive form
$S_1$ = $a_1$, $S_n$ = $S_{n-1}$ + $a_n$ for n ≥ 2

sigma-rules

Sigma linearity (constant multiple)

WS6
$$ \sum_{k=1}^{n} c \cdot a_k = c \sum_{k=1}^{n} a_k $$

In words. Σ c·$a_k$ = c · Σ $a_k$

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Derivation
Pull constants out of the sigma. Distributive/factor property.
Intuition
Expanding: c·$a_1$ + c·$a_2$ + ... + c·$a_n$ = c($a_1$ + $a_2$ + ... + $a_n$)

Sigma linearity (sum of sums)

WS6
$$ \sum_{k=1}^{n} (a_k + b_k) = \sum_{k=1}^{n} a_k + \sum_{k=1}^{n} b_k $$

In words. Σ ($a_k$ + $b_k$) = Σ $a_k$ + Σ $b_k$

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Derivation
Sigma distributes over addition. Combined with the constant-multiple rule, this says Σ is linear.
Intuition
By commutativity and associativity of addition, regroup the terms.

Index shift (subtract-the-prefix)

WS6
$$ \sum_{k=m}^{n} a_k = \sum_{k=1}^{n} a_k - \sum_{k=1}^{m-1} a_k $$

In words. Σ from m to n = Σ from 1 to n - Σ from 1 to m-1

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Derivation
Convert a non-standard index range to a difference of standard ones. Essential for using the special-sums formulas on Σ_{k=3}^{12} k³ type problems.

Counting terms in a sum

WS6
$$ \sum_{k=m}^{n} \text{ has } (n - m + 1) \text{ terms} $$

In words. Number of terms = n - m + 1 (NOT n - m, NOT n)

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Derivation
Off-by-one trap. The +1 catches both endpoints.
Trap
Σ_{k=4}^{17}(-3) has 14 terms, NOT 17. Sum = 14·(-3) = -42.
Examples
input: k=1 to 10  ·  output: 10 terms
input: k=5 to 20  ·  output: 20-5+1 = 16 terms

special-sums

Sum of a constant

WS6
$$ \sum_{k=1}^{n} c = nc $$

In words. Σ_{k=1}^{n} c = n · c

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Derivation
Adding the constant c to itself n times.
Index-shift trap
Σ_{k=4}^{17}(-3) has 14 terms (= 17 - 4 + 1), NOT 17. Sum = 14·(-3) = -42.
n = 1 check
n·c at n=1 gives c ✓
Degree
1 (linear in n)

Sum of first n positive integers

WS6
$$ \sum_{k=1}^{n} k = \dfrac{n(n+1)}{2} $$

In words. Σ_{k=1}^{n} k = n(n+1)/2

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Derivation
Gauss's schoolboy formula. The n-th triangular number $T_n$.
Derivation
Reverse-and-add (Gauss): write $S_n$ forward and reversed; each column sums to (n+1); n columns; 2$S_n$ = n(n+1).
Famous value
Σ_{k=1}^{100} k = 5050
n = 1 check
n(n+1)/2 at n=1 gives 1·2/2 = 1 ✓
Degree
2 (quadratic in n)

Sum of first n squares

WS8
$$ \sum_{k=1}^{n} k^2 = \dfrac{n(n+1)(2n+1)}{6} $$

In words. Σ_{k=1}^{n} k² = n(n+1)(2n+1)/6

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Derivation
Sum of consecutive squares.
Derivation
Polynomial-fit: 3rd differences constant (=2), so degree 3. Solve for An³+Bn²+Cn+D by plugging in n=0,1,2,3.
Famous value
Σ_{k=1}^{100} k² = 338,350
n = 1 check
n(n+1)(2n+1)/6 at n=1 gives 1·2·3/6 = 1 ✓
Degree
3 (cubic in n)

Sum of first n cubes

WS8
$$ \sum_{k=1}^{n} k^3 = \left[\dfrac{n(n+1)}{2}\right]^2 = \left(\sum_{k=1}^{n} k\right)^2 $$

In words. Σ_{k=1}^{n} k³ = [n(n+1)/2]² = (Σk)²

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Derivation
The remarkable identity: sum of cubes equals square of sum.
Discovery
$S_n$ values 1, 9, 36, 100, 225, 441 are squares of triangular numbers 1², 3², 6², 10², 15², 21².
Famous value
Σ_{k=1}^{10} k³ = 55² = 3025; Σ_{k=1}^{15} k³ = 120² = 14400
n = 1 check
1² = 1 ✓
Degree
4 (quartic in n)

Degree rule for polynomial sums

WS8
$$ \text{If } a_k \text{ is degree-}p \text{ in } k, \text{ then } S_n \text{ is degree-}(p+1) \text{ in } n $$

In words. Sum of degree-p polynomial in k → degree-(p+1) polynomial in n

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Derivation
Constants (degree 0) → linear sums. Arithmetic (degree 1 in k) → quadratic sums (degree 2). k² → cubic. k³ → quartic. Each Σ$k^p$ formula has one more factor than the last.

telescoping

Telescoping sum (finite)

WS6/WS7
$$ \sum_{k=1}^{n} (b_k - b_{k+1}) = b_1 - b_{n+1} $$

In words. Σ_{k=1}^{n} ($b_k$ - $b_{k+1}$) = $b_1$ - $b_{n+1}$

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Derivation
When each term is a difference of consecutive b-values, interior terms cancel pairwise; only the first and last survive.
Famous example
Σ_{k=1}^{n} (1/k - 1/(k+1)) = 1 - 1/(n+1)
Example
Σ_{m=1}^{20} (1/m - 1/(m+1)) = 1 - 1/21 = 20/21

Telescoping sum (infinite)

WS7
$$ \sum_{k=1}^{\infty} (b_k - b_{k+1}) = b_1 - \lim_{n \to \infty} b_{n+1} $$

In words. Σ_{k=1}^{∞} ($b_k$ - $b_{k+1}$) = $b_1$ - lim $b_{n+1}$

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Derivation
The infinite telescoping sum is finite iff the tail term $b_{n+1}$ has a finite limit.
Convergence criterion
lim $b_{n+1}$ exists and is finite
Famous examples
series: Σ (1/k - 1/(k+1))  ·  b_k: 1/k  ·  limit: 0  ·  sum: 1
series: Σ (6/(2k+1) - 6/(2k+3))  ·  b_k: 6/(2k+1)  ·  limit: 0  ·  sum: 2
series: Σ log₂((k+1)/(k+2))  ·  b_k: log₂(k+1)  ·  limit:  ·  sum: DIVERGES

Telescoping series can DIVERGE

WS7
$$ \text{If } \lim_{n\to\infty} b_{n+1} = \pm\infty, \text{ then } \sum (b_k - b_{k+1}) \text{ diverges} $$

In words. Tail term unbounded → telescoping series diverges

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Derivation
A telescoping series CAN diverge if the tail grows without bound. This is the WS7 Q9 lesson.
Important lesson
Terms going to zero is NECESSARY but NOT SUFFICIENT for convergence. Same lesson as the harmonic series in calculus.
Example
Σ log₂((k+1)/(k+2)) telescopes to log 2 - log(n+2) → -∞

Telescoping infinite product

WS7
$$ \prod_{k=1}^{n} \dfrac{k}{k+3} = \dfrac{6}{(n+1)(n+2)(n+3)} $$

In words. Factor into ratio of factorials, then cancel

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Derivation
WS7 Q11: infinite product k/(k+3). Numerators give n!; denominators give (n+3)!/3! = (n+3)!/6.
Derivation
$P_n$ = (1·2·3·...·n) / (4·5·6·...·(n+3)) = n! · 3! / (n+3)! = 6 / ((n+1)(n+2)(n+3))
Result
Infinite product converges to 0 (denominator → ∞)

series

Arithmetic series sum (using d)

WS6
$$ S_n = \dfrac{n}{2}\bigl(2a + (n-1)d\bigr) $$

In words. $S_n$ = (n/2)(2a + (n-1)d)

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Derivation
Sum of first n terms of arithmetic sequence with first term a and common difference d.
Derivation
Gauss's reverse-and-add. Each column of $S_n$ forward + $S_n$ reversed sums to 2a + (n-1)d. n columns. 2$S_n$ = n(2a + (n-1)d).
Variables
S_n: sum of first n terms
a: first term (sometimes $a_1$)
d: common difference
n: number of terms
n = 1 check
$S_1$ = (1/2)(2a + 0) = a ✓

Arithmetic series sum (using last term)

WS6
$$ S_n = \dfrac{n}{2}(a + \ell) $$

In words. $S_n$ = (n/2)(a + ℓ) where ℓ is the last term

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Derivation
Equivalent to d-form, using the last term ℓ = a + (n-1)d.
Use case
When you know first and last terms, this form skips computing d.
n = 1 check
At n=1, ℓ = a, so $S_1$ = (1/2)(a+a) = a ✓
Examples
input: Σ_{k=1}^{50} k  ·  output: $S_5$0 = (50/2)(1 + 50) = 25 · 51 = 1275
input: 3 + 7 + 11 + ... + 99 (n=25)  ·  output: $S_2$5 = (25/2)(3 + 99) = 1275

Geometric series sum (finite, r ≠ 1)

WS6
$$ S_n = \dfrac{a(1 - r^n)}{1 - r}, \quad r \neq 1 $$

In words. $S_n$ = a(1 - $r^n$) / (1 - r), provided r ≠ 1

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Derivation
Sum of first n terms of geometric sequence with first term a and ratio r.
Derivation
Multiply-by-r-and-subtract: $S_n$ = a + ar + ar² + ... + ar^(n-1). r$S_n$ = ar + ar² + ... + a$r^n$. Subtract: $S_n$ - r$S_n$ = a - a$r^n$. Factor: $S_n$(1-r) = a(1-$r^n$).
First-term warning
For Σ_{k=1}^{∞} 9(2/3)^k, the first term (k=1) is 9·2/3 = 6, NOT 9. Always check by plugging k=1.
Variables
S_n: sum of first n terms
a: first term of the actual series (NOT always the coefficient — multiply out and check)
r: common ratio
n: number of terms
n = 1 check
$S_1$ = a(1-r)/(1-r) = a ✓

Geometric series sum (r = 1 edge case)

WS6
$$ S_n = na, \quad r = 1 $$

In words. $S_n$ = n · a when r = 1

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Derivation
If r = 1, denominator 1-r is zero — formula breaks. Fall back to Σa = na from the constant-sum special formula.
Removable singularity
The formula a(1-$r^n$)/(1-r) has a removable singularity at r = 1. L'Hôpital or direct algebra recovers na.

series-infinite

Arithmetic series (infinite) — always diverges

WS7
$$ \sum_{k=1}^{\infty} (a + (k-1)d) \text{ diverges (except trivially)} $$

In words. Infinite arithmetic series ALWAYS diverges, except when a = 0 AND d = 0

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Derivation
Terms eventually grow without bound in one direction (if d ≠ 0), or the series is a + a + a + ... (if d = 0, a ≠ 0). Only convergent case: trivial zero series.

Infinite geometric series sum

WS7
$$ \sum_{k=1}^{\infty} ar^{k-1} = \dfrac{a}{1 - r}, \quad |r| < 1 $$

In words. S = a / (1 - r), provided |r| < 1

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Derivation
The most important infinite-sum formula in the unit.
Derivation
Start from $S_n$ = a(1-$r^n$)/(1-r). As n → ∞ with |r| < 1, $r^n$ → 0, so $S_n$ → a/(1-r).
Convergence condition
|r| < 1 (strictly less than 1)
First-term warning
a is the first term of the ACTUAL series, not always the leading coefficient. For Σ_{k=1}^{∞} 9(2/3)^k, a = 6 (first term at k=1).

Convergence cases for geometric $r^n$

WS7
$$ \lim_{n\to\infty} r^n = \begin{cases} 0 & |r| < 1 \\ 1 & r = 1 \\ \text{DNE/oscillates} & r = -1 \\ \pm\infty & |r| > 1 \end{cases} $$

In words. $r^n$ behavior as n → ∞

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Derivation
What $r^n$ does determines what the geometric series does.
Cases
case: |r| < 1  ·  rn_behavior: → 0  ·  series: Converges to a/(1-r)
case: |r| > 1  ·  rn_behavior: → ±∞  ·  series: Diverges
case: r = 1  ·  rn_behavior: = 1 always  ·  series: $S_n$ = na, diverges (unless a=0)
case: r = -1  ·  rn_behavior: oscillates ±1  ·  series: Partial sums oscillate a, 0, a, 0; no limit

geometric-applications

Trapezoid parallel slices form arithmetic sequence

WS4/WS9
$$ \text{Inserting } k \text{ parallels equally spaced: arithmetic with } d = \dfrac{B - A}{k+1} $$

In words. Equally-spaced parallel slices of a trapezoid (bases A, B) form arithmetic sequence

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Derivation
In a trapezoid with parallel bases A and B, equally-spaced parallel slices have lengths that vary linearly with vertical position.
Reason
The slanted sides are straight lines; parallel slice length is a linear function of height. Linear function at evenly-spaced positions → arithmetic sequence.
Example
Bases 8 and 23, insert 2 parallels: d = 15/3 = 5; lengths 8, 13, 18, 23

Annulus area sequence

WS9
$$ A_n = \pi(n+1)^2 - \pi n^2 = \pi(2n + 1) $$

In words. Area of annulus with outer radius n+1, inner radius n is π(2n+1)

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Derivation
Difference of squares: (n+1)² - n² = 2n + 1.
Sum formula
Σ_{n=1}^{N} $A_n$ = π·Σ(2n+1) = π[2·N(N+1)/2 + N] = π·N(N+2)
Example
Σ_{n=1}^{100} $A_n$ = π(2·5050 + 100) = 10200π

applications

Pure repeating decimal as fraction

WS7
$$ 0.\overline{d_1 d_2 \cdots d_p} = \dfrac{d_1 d_2 \cdots d_p}{\underbrace{99\cdots 9}_{p \text{ nines}}} $$

In words. Length-p repeating block: block digits over p nines

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Derivation
Convert pure repeating decimals using geometric series formula with r = 10^(-p).
Derivation
0.dddd... = d/10 + d/100 + d/1000 + ... is geometric with a = d/10, r = 1/10 (for single digit). Sum = (d/10)/(1 - 1/10) = d/9. Generalize for p-digit blocks.
Examples
input: 0.7̄  ·  output: 7/9 (a = 0.7, r = 0.1)
input: 0.21̄  ·  output: 21/99 = 7/33 (a = 0.21, r = 0.01)
input: 0.723̄  ·  output: 723/999 = 241/333

Mixed repeating decimal as fraction

WS9
$$ 0.a_1\ldots a_q \overline{b_1\ldots b_p}: \quad 10^{q+p}x - 10^q x = (\text{integer}) $$

In words. Scale by 1$0^a$ and 10^(a+p), subtract to eliminate the repeating part

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Derivation
For decimals with non-repeating prefix followed by repeating block. Set up two equations, subtract.
Example: 724
0.7̄24̄: let x = 0.7̄24̄; 10x = 7.̄24̄; 1000x = 724.̄24̄; subtract: 990x = 717; x = 717/990 = 239/330

Sum of (2k)^2 pattern

WS9
$$ \sum_{k=1}^{N}(2k)^2 = 4 \sum_{k=1}^{N} k^2 $$

In words. Σ(2k)² factors out: 4·Σk²

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Derivation
For sums of even squares, factor out the 4.
Example
2² + 4² + 6² + ... + 60² = Σ_{k=1}^{30}(2k)² = 4·Σ_{k=1}^{30}k² = 4·9455 = 37820

word-problems

Crossing wires height

WS4/WS10
$$ h = \dfrac{ab}{a+b} = \dfrac{HM(a,b)}{2} $$

In words. h = ab/(a+b) = HM/2

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Derivation
Two poles of heights a, b; wires from each top to opposite foot. Wires cross at height h independent of the distance between poles.
Derivation
By similar triangles: h/a = x/(x+y) and h/b = y/(x+y). Add: h(1/a + 1/b) = 1, so h = 1/(1/a + 1/b) = ab/(a+b).
Example
Poles 8 and 12: h = 96/20 = 4.8 ft

Work-rate combined time

WS4/WS10
$$ t_{\text{combined}} = \dfrac{t_A \cdot t_B}{t_A + t_B} = \dfrac{HM(t_A, t_B)}{2} $$

In words. $t_c$ombined = ($t_A$ · $t_B$) / ($t_A$ + $t_B$)

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Derivation
Person A does the job in $t_A$, person B in $t_B$. Working simultaneously, they finish in HM/2.
Derivation
Rates add: 1/$t_c$ombined = 1/$t_A$ + 1/$t_B$. Flip both sides.
Rule of thumb
Simultaneous work → HM. Sequential → AM.
Example
George 40 min, Sue 30 min together: 120/7 ≈ 17.14 min

Bouncing ball total distance

WS7/WS10
$$ \text{Total} = h + 2h\rho + 2h\rho^2 + \cdots = h + \dfrac{2h\rho}{1 - \rho} = \dfrac{h(1 + \rho)}{1 - \rho} $$

In words. Total = h(1 + ρ) / (1 - ρ)

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Derivation
Ball dropped from height h, each rebound is ρ × previous height. Initial drop counts once; each rebound height counts twice (up + down).
Example
h = 9 m, ρ = 7/8: Total = 9 + 2(9·7/8)/(1-7/8) = 9 + 126 = 135 m

Compound interest (geometric growth)

WS3/WS10
$$ a_n = a_0 \cdot r^n $$

In words. Compound interest: r = 1 + rate (growth) or r = 1 - rate (decay)

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Derivation
Anything compounding by a fixed percentage rate is geometric.
Rule of thumb
Percentage rate (6% per year) → geometric. Absolute amount ($50/month) → arithmetic.
Examples
input: $5000 at 6% APR, compounded monthly  ·  output: $a_n$ = 5000(1.005)^n where r = 1 + 0.06/12 = 1.005

meta

Type recognition for rapid recall

WS10
$$ \text{Decision tree for picking the right formula} $$

In words. Identify type → pick formula → plug in

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Derivation
The whole unit reduces to recognition.
Decision tree
pattern: Σ k  ·  use: n(n+1)/2
pattern: Σ k²  ·  use: n(n+1)(2n+1)/6
pattern: Σ k³  ·  use: [n(n+1)/2]²
pattern: Σ c (constant)  ·  use: nc
pattern: Σ_{k=1}^{n}(arithmetic in k)  ·  use: (n/2)(2a + (n-1)d) or (n/2)(a+ℓ)
pattern: Σ_{k=1}^{n} ar^(k-1) finite  ·  use: a(1-$r^n$)/(1-r), r≠1
pattern: Σ_{k=1}^{∞} ar^(k-1)  ·  use: a/(1-r) iff |r|<1, else diverges
pattern: Σ infinite arithmetic  ·  use: Always diverges (except trivial)
pattern: Σ telescoping ($b_k$ - $b_{k+1}$)  ·  use: $b_1$ - lim $b_{n+1}$
pattern: Repeating decimal 0.d̄  ·  use: block/(p nines), geometric with r = 10^(-p)
pattern: $x_n$ = f($x_{n-1}$)  ·  use: Solve f(x)=x for fixed point; |f'(x*)|<1 for attracting
pattern: Bouncing ball  ·  use: h(1+ρ)/(1-ρ)
pattern: Crossing wires (a,b)  ·  use: h = ab/(a+b) = HM/2
pattern: Work-rate ($t_A$, $t_B$ together)  ·  use: t = $t_A$·$t_B$/($t_A$+$t_B$)

Degree rule for partial sums of polynomials

WS8
$$ \deg(a_k) = p \implies \deg(S_n) = p + 1 $$

In words. If $a_k$ is degree-p polynomial in k, then $S_n$ is degree-(p+1) polynomial in n

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Derivation
Sum of constants (deg 0) → linear sum (deg 1). Sum of linear (arithmetic) → quadratic. Σk² → cubic. Σk³ → quartic.
Method
Use finite differences on first p+2 values of $S_n$ to identify the polynomial; solve linear system for coefficients.

The five named sequence types

WS1/WS10
$$ \text{Arithmetic, Geometric, Harmonic, Iterated, Pattern} $$

In words. The five sequence types

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Types
name: Arithmetic  ·  form: $a_n$ = a + (n-1)d  ·  test: Add fixed d  ·  example: 2, 5, 8, 11, ... (d=3)
name: Geometric  ·  form: $a_n$ = a · r^(n-1)  ·  test: Multiply by fixed r  ·  example: 3, 6, 12, 24, ... (r=2)
name: Harmonic  ·  form: $a_n$ = 1/(a + (n-1)d)  ·  test: Reciprocals are arithmetic  ·  example: 1, 1/2, 1/3, 1/4, ...
name: Iterated  ·  form: $a_n$ = f($a_{n-1}$)  ·  test: Apply f repeatedly  ·  example: 1 + 1/x → φ ≈ 1.618
name: Pattern  ·  form: Find via differences, ratios, or recognition  ·  test: No closed form unless found  ·  example: 1, 4, 9, 16, 25, ... (squares)
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