Trigonometry

Sketchbook fragment — graphs, identities, triangle laws, inverses, rotations.

Sine function

period · amplitude · phase · shift

$$y = a\sin\!\bigl(b(x - h)\bigr) + k$$

a (amplitude) 1.0 b (freq) 1.0

Period 2π/b; amplitude |a|; vertical shift k; horizontal shift h (right when h>0). Zeros every half-period at x=h+kπ/b.

Cosine function

even · cos(0)=1 · shift of sine

$$y = \cos(x - h), \qquad \cos x = \sin\!\bigl(x + \tfrac{\pi}{2}\bigr)$$

h (phase shift) 0.00

Cosine is sine shifted left by π/2. Even: cos(-x)=cos(x). Max at x=h+2πk, min at x=h+π+2πk. Dashed: reference cos(x).

Tangent function

period π · asymptotes at π/2+kπ

$$y = \tan x = \frac{\sin x}{\cos x}$$

Period π (half of sin/cos). Vertical asymptotes where cos x = 0: x = π/2 + kπ. Zeros at x = kπ. Always increasing on each branch.

Reciprocal functions

csc · sec · cot

$$\csc x=\tfrac{1}{\sin x},\quad \sec x=\tfrac{1}{\cos x},\quad \cot x=\tfrac{\cos x}{\sin x}$$

Each reciprocal blows up where its partner is zero. Dashed lines = vertical asymptotes; faint curve = the parent sin/cos for reference.

Unit circle (16 angles)

P(θ) = (cos θ, sin θ)

$$x^2 + y^2 = 1, \qquad P(\theta) = (\cos\theta,\ \sin\theta)$$

Every point on the circle is (cos θ, sin θ). Cosine reads off horizontally, sine vertically. tan θ = y/x = slope of the radius. Sign chart: Q1 all+; Q2 sin+; Q3 tan+; Q4 cos+ (A-S-T-C).

Reference angles & special triangles

30-60-90 · 45-45-90

$$|\text{trig}(\theta)| = |\text{trig}(\theta_{\text{ref}})|, \quad \theta_{\text{ref}}\in[0,\pi/2]$$

Drop a perpendicular from the terminal-side point to the x-axis — the acute angle made with the axis is θ_ref. Read magnitude from the special triangle; pull the sign from the quadrant.

Special-angle exact values

0, π/6, π/4, π/3, π/2

θ	sin θ	cos θ	tan θ
$0$	$0$	$1$	$0$
$\pi/6$	$\tfrac{1}{2}$	$\tfrac{\sqrt3}{2}$	$\tfrac{\sqrt3}{3}$
$\pi/4$	$\tfrac{\sqrt2}{2}$	$\tfrac{\sqrt2}{2}$	$1$
$\pi/3$	$\tfrac{\sqrt3}{2}$	$\tfrac{1}{2}$	$\sqrt3$
$\pi/2$	$1$	$0$	—

Sine row counts up: √0/2, √1/2, √2/2, √3/2, √4/2. Cosine row counts down (mirror). Tangent = sine ÷ cosine.

Inverse sine (arcsin)

domain [-1,1] · range [-π/2, π/2]

$$y = \sin^{-1}(x) \iff \sin y = x,\ y\in[-\tfrac{\pi}{2},\tfrac{\pi}{2}]$$

Restrict sin to its one-to-one piece [-π/2, π/2], then invert. Output is always in range. arcsin(sin x) snaps to range, NOT identity for all x.

Inverse cosine (arccos)

domain [-1,1] · range [0, π]

$$y = \cos^{-1}(x) \iff \cos y = x,\ y\in[0,\pi]$$

Restrict cos to [0, π] (monotone decreasing). Output is always non-negative; arccos(-1/2) = 2π/3, NOT -π/3.

Inverse tangent (arctan)

domain ℝ · range (-π/2, π/2)

$$y = \tan^{-1}(x) \iff \tan y = x,\ y\in(-\tfrac{\pi}{2},\tfrac{\pi}{2})$$

Tangent's periodic branch on (-π/2, π/2) is the entire reals. Inverting gives an S-shaped curve squeezed between horizontal asymptotes. Endpoints excluded (open interval).

Pythagorean identities

divide by cos² / sin²

$$\sin^2\theta + \cos^2\theta = 1$$ $$\tan^2\theta + 1 = \sec^2\theta$$ $$1 + \cot^2\theta = \csc^2\theta$$

Primary identity is Pythagoras applied to (cos θ, sin θ, 1). Divide both sides by cos²θ for the tan/sec form, by sin²θ for the cot/csc form.

Sum & difference identities

all six (sin, cos, tan)

$$\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta$$

$$\sin(\alpha - \beta) = \sin\alpha\cos\beta - \cos\alpha\sin\beta$$

$$\cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta$$

$$\cos(\alpha - \beta) = \cos\alpha\cos\beta + \sin\alpha\sin\beta$$

$$\tan(\alpha + \beta) = \dfrac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta}$$

$$\tan(\alpha - \beta) = \dfrac{\tan\alpha - \tan\beta}{1 + \tan\alpha\tan\beta}$$

Memory: sine MIXES (sin-cos + cos-sin); cosine STAYS PURE (cos-cos − sin-sin). Sign in the output matches sign in the input for sine, FLIPS for cosine. Derives from the rotation-matrix product R_α·R_β = R_α+β.

Double-angle identities

sub β = α into sum

$$\sin 2\alpha = 2\sin\alpha\cos\alpha$$ $$\cos 2\alpha = \cos^2\alpha - \sin^2\alpha$$ $$\phantom{\cos 2\alpha} = 2\cos^2\alpha - 1 = 1 - 2\sin^2\alpha$$ $$\tan 2\alpha = \dfrac{2\tan\alpha}{1 - \tan^2\alpha}$$

Cosine has THREE equivalent forms — pick whichever matches what you know. The 2cos²−1 form is the source of the cos half-angle; the 1−2sin² form is the source of the sin half-angle.

Half-angle (power-reduced)

rearranged double-angle

$$\cos^2\alpha = \dfrac{1 + \cos 2\alpha}{2}$$ $$\sin^2\alpha = \dfrac{1 - \cos 2\alpha}{2}$$ $$\tan^2\alpha = \dfrac{1 - \cos 2\alpha}{1 + \cos 2\alpha}$$

Solve the cos-double-angle forms for cos²/sin². "Power-reduction": a squared trig function becomes first-order in cos 2α — standard prep for integrating sin²/cos².

Law of Sines

AAS, ASA · SSA ambiguous

$$\dfrac{\sin A}{a} = \dfrac{\sin B}{b} = \dfrac{\sin C}{c}$$

Use when you have an angle-side pair plus one more piece (AAS/ASA/SSA). SSA caveat: if you're given two sides and a non-included angle A, compute h = c sin A; then 0, 1, or 2 triangles depending on whether a falls below, on, between, or beyond h and c.

Law of Cosines

SAS · SSS

$$c^2 = a^2 + b^2 - 2ab\cos C$$ $$\cos C = \dfrac{a^2 + b^2 - c^2}{2ab}$$

When C = π/2, cos C = 0 and the formula collapses to Pythagoras. Solved-for-angle form yields a unique angle in [0, π] — no SSA ambiguity. Use for SAS (find third side) or SSS (find any angle).

SAS area formula

K = ½ ab sin C

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

Drop altitude h = b sin C from the vertex of C. Area = ½ · base · height. The three permutations correspond to dropping the altitude from each of the three vertices.

2D rotation matrix

columns = images of basis

$$R_\theta = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \phantom{-}\cos\theta \end{bmatrix}$$

θ (degrees) 30°

Column 1 = where (1,0) lands = (cos θ, sin θ). Column 2 = where (0,1) lands = (-sin θ, cos θ). Determinant = 1, so rotations preserve area and length. R_α·R_β = R_α+β — this is the engine of sum identities.

Periodicity templates

all solutions of trig = c

$$\sin\theta = c \;\Longleftrightarrow\; \theta = \arcsin c + 2\pi k \;\text{or}\; \pi - \arcsin c + 2\pi k$$ $$\cos\theta = c \;\Longleftrightarrow\; \theta = \pm\arccos c + 2\pi k$$ $$\tan\theta = c \;\Longleftrightarrow\; \theta = \arctan c + \pi k$$

Sine: two families (principal + supplement), period 2π. Cosine: two families (± principal), period 2π. Tangent: one family, period π. For sin α = sin β: either α = β + 2πk or α = π − β + 2πk.

Harmonic combination

A cos + B sin → R sin(x+φ)

$$A\cos x + B\sin x = R\sin(x + \phi)$$ $$R = \sqrt{A^2 + B^2}, \qquad \tan\phi = A/B$$

Two sinusoids of equal frequency add to one bigger sinusoid. Amplitude is the Pythagorean magnitude; phase is set by the ratio. Use to solve A cos x + B sin x = c: rewrite, then apply periodicity template.

Radian–degree conversion

180° = π rad

$$\text{rad} = \text{deg}\cdot\dfrac{\pi}{180°}, \qquad \text{deg} = \text{rad}\cdot\dfrac{180°}{\pi}$$

deg	rad	deg	rad
0°	$0$	180°	$\pi$
30°	$\pi/6$	210°	$7\pi/6$
45°	$\pi/4$	225°	$5\pi/4$
60°	$\pi/3$	240°	$4\pi/3$
90°	$\pi/2$	270°	$3\pi/2$
120°	$2\pi/3$	300°	$5\pi/3$
135°	$3\pi/4$	315°	$7\pi/4$
150°	$5\pi/6$	360°	$2\pi$

1 rad ≈ 57.30°. DMS: x° y' z'' = x + y/60 + z/3600 decimal degrees.

Even/odd & co-function

reflections of the unit circle

$$\sin(-\theta) = -\sin\theta,\quad \cos(-\theta) = \cos\theta,\quad \tan(-\theta) = -\tan\theta$$ $$\sin\!\bigl(\tfrac{\pi}{2}-\alpha\bigr) = \cos\alpha,\quad \cos\!\bigl(\tfrac{\pi}{2}-\alpha\bigr) = \sin\alpha$$

Sin/tan are odd (reflect across x-axis flips y); cos is even (x-coord untouched). Co-function: function at complement equals co-function — "cosine" literally means "sine of complement". Quick reductions: cos(π−α)=−cos α, sin(π+α)=−sin α, cos(x+3π/2)=sin x.

Slope & angle between lines

m = tan α

$$m = \tan\alpha,\quad \alpha\in[0,\pi)$$ $$\tan\theta = \left|\dfrac{m_1 - m_2}{1 + m_1 m_2}\right|$$ $$m_1 m_2 = -1 \iff \text{perpendicular}$$

Angle between two lines = difference of their inclinations = tan-difference identity applied to slopes. Denominator zero ⇒ perpendicular.

Angle bisector theorem

x/y = a/b

$$\dfrac{x}{y} = \dfrac{a}{b}$$

A bisector from a vertex splits the opposite side in the same ratio as the two adjacent sides. Same logic = Law of Sines applied to the two sub-triangles sharing the bisector.

Triangle inequality

a + b > c (all three)

$$a + b > c,\quad a + c > b,\quad b + c > a$$

Two shorter sides must reach across the longest. Violating any inequality = no triangle. Equality = degenerate (collinear). Quick check: sort the sides, then test whether the two smallest exceed the largest.

Bonus identities

triple-angle · tan+cot

$$\sin 3x = 3\sin x - 4\sin^3 x$$ $$\sin 2\theta = \dfrac{2\tan\theta}{1 + \tan^2\theta}$$ $$\tan\theta + \cot\theta = 2\csc 2\theta$$

Triple-angle: stack sin(2x+x) + double-angle + Pythagorean. Tangent-form sin(2θ): useful when only tan is known (Weierstrass substitution). tan+cot collapses to 1/(sin θ cos θ).

A vector is a displacement, not a point

WS1

The tail is not glued to the origin. A vector encodes how far and which way, not where. Slide the arrow around — same vector.

This is why we can add vectors tip-to-tail: an arrow's anchor is wherever you need it to be.

$\vec{v}$ = displacement, independent of starting point

Component form $\langle a, b\rangle$

WS1

Anchor the tail at the origin. The tip's $(x, y)$ coordinates are the components.

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

Three notations, one object. The basis vectors $\mathbf{i} = \langle 1, 0\rangle$ and $\mathbf{j} = \langle 0, 1\rangle$ point east and north with unit length, so $a\mathbf{i} + b\mathbf{j}$ literally says "go $a$ east, then $b$ north."

Magnitude — Pythagoras on the components

WS1

The components are the legs of a right triangle whose hypotenuse is the vector. Pythagoras, no thinking required.

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

Classic worked example: $\vec{v} = \langle 3, 4\rangle \Rightarrow |\vec{v}| = \sqrt{9 + 16} = 5$. The 3-4-5 triple shows up everywhere — memorize it.

Magnitude is always $\geq 0$, and $=0$ only for the zero vector.

Polar form $\langle r\cos\theta, r\sin\theta\rangle$ drag θ

WS1

θ 0°

Magnitude times direction. Scale the unit-circle point $\langle\cos\theta, \sin\theta\rangle$ by $r$ and you stretch it to the right length while preserving direction.

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

Worked: $r = 10,\ \theta = 45° \Rightarrow \vec{v} = \langle 10\cos 45°, 10\sin 45°\rangle = \langle 5\sqrt 2, 5\sqrt 2\rangle \approx \langle 7.07, 7.07\rangle$.

Bearings & compass directions — sin/cos swap

WS1

Bearings measure clockwise from North, while math angles measure counterclockwise from East. They're complementary: $\theta = 90° - B$.

$\vec{v} = \langle r\sin B,\ r\cos B\rangle$ (universal bearing form)

The trig swap: sine and cosine trade roles because the reference axis flipped. NαE means "start at North, rotate α toward East" — and the quadrant signs follow:

N α° E → $\langle +r\sin\alpha,\ +r\cos\alpha\rangle$ (Q1)
N α° W → $\langle -r\sin\alpha,\ +r\cos\alpha\rangle$ (Q2)
S α° E → $\langle +r\sin\alpha,\ -r\cos\alpha\rangle$ (Q4)
S α° W → $\langle -r\sin\alpha,\ -r\cos\alpha\rangle$ (Q3)

The arctan quadrant trap — sketch first

TRAP

Calculator's $\arctan(b/a)$ returns only $(-90°, 90°)$ — that's Q1 and Q4 only. If your vector lives in Q2 or Q3, the calculator is silently lying.

$\theta = \arctan\!\left(\dfrac{b}{a}\right) + \begin{cases}0° &\text{Q1, Q4}\\ 180° &\text{Q2, Q3}\end{cases}$

Example. $\vec{t} = \langle -1, 3\rangle$ is in Q2. $\arctan(3/-1) = \arctan(-3) \approx -71.57°$. Add 180°: $\theta \approx 108.43°$. ✓

Rule: sketch the quadrant before trusting the calculator.

Addition — tip to tail, or parallelogram

WS2

Geometric: place $\vec w$'s tail at $\vec v$'s tip; the sum is the arrow from $\vec v$'s tail to $\vec w$'s tip. Equivalently, build the parallelogram and take the diagonal.

$\vec v + \vec w = \langle a_1 + a_2,\ b_1 + b_2\rangle$

Components just add. $\langle -5, 1\rangle + \langle 2, 6\rangle = \langle -3, 7\rangle$. Commutative and associative — order doesn't matter.

Subtraction — "from $\vec w$'s tip to $\vec v$'s tip"

WS2

w's tip to v's tip is v − w.

Two equivalent pictures: subtract components, or place tails together and draw from w's tip to v's tip.

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

The negative of $\vec w$ is just $\vec w$ flipped 180°. So subtraction is addition of the reversed vector.

$|\vec v - \vec w| \neq |\vec v| - |\vec w|$ in general — subtract first, then take magnitude.

Scalar multiplication — stretch, shrink, flip

WS3

Multiply each component by the scalar. Magnitude scales by $|c|$; sign of $c$ controls direction.

$c\vec v = \langle ca,\ cb\rangle,\qquad |c\vec v| = |c|\cdot|\vec v|$

$c > 1$: stretch in same direction
$c = 1$: identity
$0 < c < 1$: shrink in same direction
$c = 0$: collapses to $\vec 0$
$c < 0$: flip direction; magnitude scales by $|c|$

Linear combinations — two non-parallel vectors span the plane

WS3

Two non-parallel vectors $\vec x, \vec y$ form a basis: every vector in the plane is some unique $a\vec x + b\vec y$.

$\vec v = a\vec x + b\vec y$ ⇒ equate components → 2×2 system in $a, b$

The oblique lattice they generate is the plane — just with skewed grid lines.

Worked: $\vec v = \langle 5, -3\rangle,\ \vec x = \langle -4, 1\rangle,\ \vec y = \langle 3, -1\rangle$. System $-4a + 3b = 5,\ a - b = -3$ gives $a = 4, b = 7$. Verify: $4\langle -4,1\rangle + 7\langle 3,-1\rangle = \langle 5, -3\rangle$. ✓

Unique solution iff $x_1 y_2 - x_2 y_1 \neq 0$ (vectors not parallel).

Unit vectors — direction with magnitude stripped out

WS4

Divide a vector by its own length and you get a length-1 arrow pointing the same way. The "pure direction" of $\vec v$.

$\hat v = \dfrac{\vec v}{|\vec v|},\qquad \vec v = |\vec v|\,\hat v$

The second form is THE foundational identity for the entire unit. Every other formula builds on "magnitude × direction."

Standard basis $\mathbf i = \langle 1, 0\rangle$ and $\mathbf j = \langle 0, 1\rangle$ are the unit vectors along the axes. In 3D add $\mathbf k = \langle 0, 0, 1\rangle$.

Unit vector at angle θ — points on the unit circle

WS4

Every unit vector in 2D is a point on the unit circle. The bridge between trig (angles) and vectors (directions).

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

Magnitude check: $\sqrt{\cos^2\theta + \sin^2\theta} = 1$. ✓ At $\theta = 45°$: $\hat u = \langle\sqrt 2/2, \sqrt 2/2\rangle \approx \langle 0.707, 0.707\rangle$.

Multiplying by $r$ gives the polar form from card 04.

Dot product — algebraic: multiply matching components, sum

WS5

The dot product collapses two vectors into one number by pairing components.

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

$\vec v = \langle -2, 6\rangle$ and $\vec w = \langle 3, 4\rangle$: $\vec v \cdot \vec w = (-2)(3) + (6)(4) = -6 + 24 = 18$.

It's commutative, distributes over addition, and a scalar can move outside: $(c\vec v)\cdot\vec w = c(\vec v\cdot\vec w)$.

Dot product — geometric: $|\vec v||\vec w|\cos\theta$

WS5

Same scalar as before, but written geometrically: the product of the magnitudes scaled by the cosine of the angle between them.

$\vec v \cdot \vec w = |\vec v|\,|\vec w|\cos\theta$

Derived from the Law of Cosines on the triangle with sides $\vec v, \vec w, \vec v - \vec w$. The squared-magnitude terms cancel, leaving the dot product on one side and $|\vec v||\vec w|\cos\theta$ on the other.

Aligned vectors ($\theta = 0$): dot = $|\vec v||\vec w|$. Anti-aligned ($\theta = 180°$): dot = $-|\vec v||\vec w|$. Perpendicular ($\theta = 90°$): dot = 0.

Angle between vectors drag θ

WS5

θ 37°

Solve the geometric form for $\theta$, then arccos.

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

Sign trichotomy:

$\vec v \cdot \vec w > 0$ → acute ($\theta < 90°$)
$\vec v \cdot \vec w = 0$ → perpendicular ($\theta = 90°$)
$\vec v \cdot \vec w < 0$ → obtuse ($\theta > 90°$)

Classify without computing $\theta$. (Caveat: positive can also mean parallel; negative can mean anti-parallel. Check parallelism first if unsure.)

Projection — the shadow of $\vec v$ on $\vec w$ drag θ

WS6

θ 50°

Drop a perpendicular from $\vec v$'s tip to the line through $\vec w$. The "shadow" along $\vec w$ is the projection.

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

No cosines, no arccos, no magnitudes — just dot products. Works in any dimension.

Worked: $\vec v = \langle 3, 1\rangle,\ \vec w = \langle 2, 6\rangle$. $\vec v \cdot \vec w = 12$, $\vec w \cdot \vec w = 40$. $\text{proj}_{\vec w}\vec v = (12/40)\langle 2, 6\rangle = \langle 3/5, 9/5\rangle$.

Acute → projection along $\vec w$. Right → zero. Obtuse → projection opposite to $\vec w$. One formula, all cases.

Line equation as dot product: $\vec n \cdot \vec r = c$

WS5

The coefficients of any linear equation $ax + by = c$ form the normal vector $\vec n = \langle a, b\rangle$ perpendicular to the line. Varying $c$ slides the line parallel to itself.

$\vec n \cdot \vec r = c$ ⇔ $a x + b y = c$

Proof: pick two points $P_1, P_2$ on the line. $\vec n \cdot P_1 = c$ and $\vec n \cdot P_2 = c$. Subtract: $\vec n \cdot (P_2 - P_1) = 0$. Since $P_2 - P_1$ runs along the line, $\vec n$ is perpendicular to every direction on the line.

In 3D: $ax + by + cz = d$ is a plane with normal $\langle a, b, c\rangle$. Same trick, one dimension up.

2D rotation matrix $R(\theta)$

WS7

The $2\times 2$ matrix that rotates any vector CCW by $\theta$. Apply via matrix-vector multiplication.

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

Sanity check at $\theta = 90°$: $R(90°) = \begin{bmatrix}0 & -1\\ 1 & \phantom{-}0\end{bmatrix}$. Apply to $\mathbf i = \langle 1, 0\rangle$: get $\langle 0, 1\rangle = \mathbf j$. ✓ — $\mathbf i$ rotates to $\mathbf j$.

Properties: $\det R = 1$ (area-preserving), $|R\vec v| = |\vec v|$ (isometry), $R(\theta)^{-1} = R(-\theta) = R(\theta)^T$.

Composition: $R(\alpha)R(\beta) = R(\alpha + \beta)$ — the angle-sum identities, packaged.

Force on a ramp — decompose weight into $w_T$ and $w_N$

WS8

On a ramp at angle $\theta$, gravity splits into a tangential piece (tries to slide the block) and a normal piece (presses it into the surface).

Sanity at $\theta = 0$: $w_T = 0$ (flat ground, no sliding). At $\theta = 90°$: $w_N = 0$ (vertical wall, no normal force). Pythagorean check: $w_T^2 + w_N^2 = |w|^2$. ✓

If you only know the two components: $\tan\theta = |w_T|/|w_N|$ — the weight cancels.

Equilibrium means $\sum \vec F_i = \vec 0$ — every force adds to zero. In 2D this is two scalar equations (x and y components), so two unknowns are solvable.

3D vectors — add $\mathbf k$, Pythagoras stacks

WS9

Add one more axis. $\mathbf k = \langle 0, 0, 1\rangle$ points up; $\mathbf i, \mathbf j, \mathbf k$ are mutually perpendicular unit vectors forming a right-handed system.

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

Magnitude: Pythagoras applied twice. First in the xy-plane to get the floor diagonal $\sqrt{v_1^2 + v_2^2}$; then with $v_3$ to get the box diagonal.

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

Examples: $\langle 2, 2, 1\rangle$ has magnitude $\sqrt 9 = 3$. $\langle 3, 4, 12\rangle$ has magnitude $\sqrt{169} = 13$.

Dot product extends naturally: $\vec v \cdot \vec w = v_1 w_1 + v_2 w_2 + v_3 w_3$. Geometric form $|\vec v||\vec w|\cos\theta$ unchanged.

Cross product — determinant, area, right hand

WS10

3D-only operation that returns a vector perpendicular to both inputs, with magnitude equal to the parallelogram area.

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

Expand along the top row with alternating signs. Components:

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

Anti-commutative: $\vec w \times \vec v = -(\vec v \times \vec w)$. Swap → thumb flips.

Triangle area shortcut: a triangle with sides $\vec v, \vec w$ has area $\tfrac{1}{2}|\vec v \times \vec w|$.

Self-check: $(\vec v \times \vec w)\cdot \vec v = 0$ and $(\vec v \times \vec w)\cdot \vec w = 0$ always.

Always dot, never cross — for angles

PEAK 5 TRAP

Use the dot product for angles. Always.

$\theta = \arccos\!\left(\dfrac{\vec v \cdot \vec w}{|\vec v|\,|\vec w|}\right)$ — covers all of $[0°, 180°]$

Worked. $\vec m = \langle 1, 1, 1\rangle,\ \vec w = \langle -2, 3, -6\rangle$.

Dot: $\vec m \cdot \vec w = -5$; $\cos\theta = -5/(7\sqrt 3) \approx -0.412$; $\theta \approx 114.4°$ ✓
Cross: $|\vec m \times \vec w| = \sqrt{122}$; $\sin\theta \approx 0.911$; $\arcsin = 65.6°$ ✗ (supplementary)

Two angles, $65.6° + 114.4° = 180°$. $\arcsin$ picked the wrong one. The dot product always wins.

★

Dot vs Cross — the duality, side by side

SYNTHESIS

	Dot product	Cross product
Output	Scalar	Vector (3D only)
Trig form	$\|\vec v\|\|\vec w\|\cos\theta$	$\|\vec v\|\|\vec w\|\sin\theta$
Symmetry	Commutative	Anti-commutative
Zero iff	$\vec v \perp \vec w$	$\vec v \parallel \vec w$
Dimensions	Any	3D only
Use for	Angles, projections	Perpendiculars, areas

Dot extracts the cosine (parallel component). Cross extracts the sine (perpendicular direction). Together they characterize the entire geometric relationship between two vectors.

Unit IV · Polar Coordinates

Polar sketchbook

Every Cartesian point has infinite polar names — and every polar curve hides a wave that, when read counter-clockwise, traces its silhouette. This is a visual ledger of the shapes you'll need on sight: spin to the angle, walk r units. Negative r walks backward. The pole has no angle. Three pages: foundation, the canonical zoo, and the rectangular bridge.

01 / Foundation

corePolar coordinate system

Spin to angle θ, then walk |r| units — forward if r>0, backward if r<0. Cartesian uses two perpendicular distances; polar uses a direction plus a signed distance.

trade-off: circles get one symbol (r = 3), but every point gets infinite names.

02 / Conversion

P→RPolar → Rectangular

x = r cos θ y = r sin θ

Comes straight off the right triangle — cos θ = x/r, sin θ = y/r. The signs of cos and sin handle the quadrants automatically.

always works: no quadrant fix needed. This is the easy direction.

03 / Conversion

R→PRectangular → Polar (with quadrant fix)

r² = x² + y² tan θ = y / x

Calculator's arctan only returns angles in (−π/2, π/2) — Q1 and Q4 only. For Q2 and Q3, add π or compute π − α. The highest-yield Cartesian-to-polar trap, tested every time.

trap: (−2, 2) → tan θ = −1 → calc says −π/4, but real answer is 3π/4.

04 / Foundation — PEAK 1

peak 1Two families: every point has infinite polar names

Family A: (r, θ + 2πk) — same direction, full spins

Family B: (−r, θ + π + 2πk) — walk backward

(r, θ + 2πk) or (−r, θ + π + 2πk), k ∈ ℤ

Cartesian is unique. Polar is not. The same Cartesian point √3, 1 is (2, π/6), (2, 13π/6), (−2, 7π/6), (2, −11π/6), … all valid. The two families together generate every alias.

05 / Foundation

edgeThe pole — θ undefined, not zero

(0, θ) = pole, for any θ

When r = 0, the angle is undefined. Every θ works. This is also why tan θ = y/x blows up at the origin — both top and bottom are zero.

consequence: the pole has uncountably many representations, even within [0, 2π).

06 / Meta

notationConvention bridge: a, b vs D, A

video: r = a + b cos θ worksheet: r = D + A cos θ (constant)(coeff)

The YouTube video uses a + b cos θ with a = constant, b = coefficient. The worksheets use D + A cos θ with D = constant, A = coefficient. Same family — different letters. Don't memorize letters. Memorize the question.

THE QUESTION: Is the trig coefficient bigger, smaller, or equal to the constant?

classification (letter-free): coeff = const → cardioid · coeff > const → inner loop · coeff < const → dimpled or convex.

07 / Shapes

circler = A cos θ · off-axis circle (x-axis)

r = A cos θ ⇔ (x − A/2)² + y² = (A/2)²

Circle on the x-axis, diameter |A|, tangent to the y-axis at the pole. Center (A/2, 0), radius |A|/2. Multiply both sides by r, substitute, complete the square.

retrace warning: full circle is drawn in [0, π]. The second π retraces.

08 / Shapes

circler = A sin θ · off-axis circle (y-axis)

r = A sin θ ⇔ x² + (y − A/2)² = (A/2)²

Sine = cosine rotated 90° CCW. Circle on the y-axis, center (0, A/2), tangent to x-axis at pole. Sign of A flips above/below.

memory: sin = y-axis, cos = x-axis. The video says this at every shape transition.

09 / Shapes

rotatedr = A cos(θ − C) · phase-shifted circle

r = A cos(θ − C)

Replacing θ with θ − C rotates the entire circle by C radians CCW about the pole. Same shape, twisted frame. Center moves from (A/2, 0) to polar (A/2, C).

sine version: r = A sin(θ − C) has center at polar (A/2, C + π/2) — the base sine center is already 90° up.

10 / Limaçons

|A|=|D|Cardioid · meant for each other

r = D + A cos θ, |A| = |D| r = A(1 ± cos θ) or A(1 ± sin θ)

Heart with a cusp at the pole. Minimum r is exactly 0 — the curve touches the pole at exactly one θ (the cusp). Bulge tip sits at distance 2|A|. Cusp angle: where the trig term hits −1.

a = a meant for each other → forms a heart.

11 / Limaçons

1 < D/A < 2Dimpled limaçon · moved on

r = D + A cos θ, |A| < |D| < 2|A| r_max = D + A, r_min = D − A > 0

r stays strictly positive — no pole crossing, no loop. The modulation is strong enough to produce a visible dent on the side opposite the bulge, but not strong enough to flip sign.

moved on tip flattened into a dimple — partly recovered.

12 / Limaçons

D ≥ 2AConvex limaçon · fully moved on

r = D + A cos θ, |D| ≥ 2|A|

When D dominates A by at least a factor of 2, the modulation is too small to dent. The curve looks like an off-center oval — almost a circle, but slightly egg-shaped on the bulge side.

fully moved on no trace of the old shape.

13 / Limaçons

|A| > |D|Inner-loop limaçon · hole in the heart

r = D + A cos θ, |A| > |D| outer tip = A + D loop length = A − D pole crossings: cos θ = −D/A

When the trig coefficient exceeds the constant, r goes negative for some θ. Those negative-r portions flip through the pole, drawing a small inner loop on the same side as the bulge.

hole in heart sad — not compatible.

14 / Mnemonic

memoryThe heart mnemonic · three side-by-side

|A| = |D| · cardioid · meant for each other

|A| > |D| · inner loop · hole in the heart

|A| < |D| < 2|A| · dimpled · moved on

The video's mnemonic. Meant for each other when A and D are equal (cardioid). Hole in the heart when the trig is too strong, sad and not compatible (inner loop). Moved on when the constant dominates, just a dimple left (dimpled or convex). Memorize all three at once.

15 / Roses — PEAK 3

peak 3Rose curves · B odd → B petals · B even → 2B petals

r = 3 cos(3θ) · B=3 odd · 3 petals · spacing 120°

r = 3 cos(2θ) · B=2 even · 4 petals · spacing 90°

r = A cos(Bθ) or r = A sin(Bθ) B odd → B petals (negative-r petals overlap) B even → 2B petals (negative-r petals land in new positions)

Why odd is half: for odd B, the negative-r petals trace the same positions as the positive-r petals — they overlap perfectly, so only B distinct petals appear. For even B, the negative-r petals fall at new angles, so all 2B appear.

most-tested polar fact on the final.

16 / Roses

workflowAnchor petal + even spacing

Δθ = 2π / (petal count) = 360°/count ANCHOR: set Bθ = π/2 → θ_anc = π/(2B) (if leading sign negative, add π)

Pretty roses: petals are equally spaced around the pole. Find one petal by making the trig argument equal π/2 (so trig = +1, r at max), then step around evenly.

cosine vs sine orientation: cosine roses have a petal on the +x-axis · sine roses are rotated by π/(2B).

17 / Roses — TRAP

trapEven-B sine rose · NO petals on axes

r = A sin(Bθ), B even petal tips at θ = π/(2B), 3π/(2B), 5π/(2B), … none lands on a coordinate axis

All 2B petals fall on the diagonals, offset from the axes by π/(2B). Cosine roses with even B sit on the axes; sine roses with even B sit between them.

flagged by reviewer as a high-yield trap.

18 / Reference

refLemniscate · figure-8

r² = a² cos(2θ) horizontal r² = a² sin(2θ) diagonal

Infinity symbol. Squared r² means r can take either sign (two branches per loop). Mentioned in the YT review video for reference, not deep-tested on the worksheets.

19 / Intersections — PEAK 4

peak 4 · trapThe pole trap · algebra misses pole intersections

r₁ = 2 − 2 sin θ (cardioid) r₂ = 2 sin θ (circle) Algebra: 2 − 2 sin θ = 2 sin θ sin θ = 1/2 θ = π/6, 5π/6 → (1, π/6), (1, 5π/6) But: pole is ALSO an intersection, because BOTH curves reach r=0 (cardioid at θ=π/2; circle at θ=0). Different θ values → algebra misses it.

Setting r₁ = r₂ requires the same θ. But polar curves hit the pole at different θ values. Always check the pole separately as Step 2.

procedure: 1) algebraic r₁=r₂ · 2) check pole · 3) include pole if both curves have any θ with r=0.

20 / Bridge

bridgeRectangular ↔ polar · y = 2 − 2 sin x becomes r = 2 − 2 sin θ

Cartesian: y = 2 − 2 sin x

Polar: r = 2 − 2 sin θ (cardioid, cusp opens up)

y = f(x) ↔ r = f(θ) (replace y → r, x → θ) y = 0 crossing ↔ pole crossing y < 0 region ↔ inner loop max y ↔ max distance from pole

Every rectangular sinusoid has a polar twin. If the wave dips below zero, the polar version has an inner loop (negative-r region). If it just touches zero, you get a cardioid. If it stays above zero, dimpled or convex.

21 / Bridge

conditionInner loop ↔ wave below axis

y = D + A sin x dips below zero ⇔ r = D + A sin θ has inner loop condition: |A| > |D|

The polar inner loop is created exactly by negative-r values — which correspond to the rectangular wave dipping below the x-axis. The shaded region is where r becomes negative.

22 / Edge

domainNon-integer B · extended θ-domain

r = f(p/q · θ), gcd(p,q)=1 ⇒ θ ∈ [0, 2πq] for full graph

For rational non-integer B = p/q in lowest terms, the curve does not close at θ = 2π. Extend to 2πq. Calculator-discipline rule: set Tmax = 2πq, or you'll see only a partial petal pattern.

example: r = cos(2θ/3): q = 3, full domain [0, 6π].

23 / Interactive

live Limaçon morpher

D 2.0

A 2.0

r = 2 + 2 cos θ
D/A = 1.00 → cardioid
max r = 4, min r = 0

Slide D and A to watch the curve transition cardioid → dimpled → convex → inner-loop. The classification updates live. Try D=0 (pure cosine circle) and D=A (cardioid cusp at pole).

24 / Interactive

live Rose petal counter

B 3

trig cos

r = 3 cos(3θ)
B = 3 (odd) → 3 petals
spacing = 120° · petal length = 3

Slide B from 1 to 8. Watch petals halve when B is odd and double when B is even. Toggle cos ↔ sin to see the rotation.

25 / Interactive

live Phase-shift rotator · r = A cos(θ − C)

A 3.0

C 0

r = 3 cos(θ − 0)
circle, diameter 3
center at polar (1.5, 0) = Cartesian (1.5, 0)

Slide C from 0 to 2π. The whole circle rotates CCW about the pole. The center traces its own circle of radius A/2. Note: phase shift inside the trig is a rotation, not a translation.

Complex Numbers

UNIT 07 · ARGAND PLANE · CIS FORM · DEMOIVRE

The reals weren't enough — once you allow i ² = −1, the number line becomes a plane, multiplication becomes rotation, and the trig identities you memorized turn out to be the same fact written in different ink.

01 / 18

The defining identity: i² = −1

Postulate · extends ℝ to ℂ

i² = −1 · equivalently i = √(−1)

Introduce a symbol i with i² = −1. Extend the reals to ℂ = {a + bi : a, b ∈ ℝ} with the usual ring operations. Multiplication by i is later revealed to be a 90° rotation of the plane.

Powers of i cycle with period 4:

i¹ = i

i² = −1

i³ = −i

i⁴ = 1

i²⁰²⁴ = i^4·506 = 1 · i²⁰²⁵ = i

02 / 18

z = a + bi on the Argand plane

Definition · rectangular form

z = a + bi, a, b ∈ ℝ → point (a, b)

Horizontal axis = real, vertical axis = imaginary. Every complex number is a 2D point — or equivalently, the position vector from the origin to that point. The reals live on the x-axis; i is one unit straight up.

z = −3 + 3√3 i sits at (−3, 3√3) (quadrant II)

03 / 18

Modulus |z| — distance from origin

Definition · Pythagorean theorem

|z| = r = √(a² + b²)

Right triangle with legs a and b in the Argand plane: the hypotenuse is r. Equivalently, |z|² = z · z̄ (a number times its conjugate).

z = −4 − 4√3 i → |z| = √(16 + 48) = √64 = 8

04 / 18

Argument arg(z) = θ

Angle from positive real axis · CCW

θ = arctan(b/a) + quadrant correction

Sweep counterclockwise from the positive real axis to the position vector of z. Defined mod 360°.

Quadrant-fix table (α = arctan(|b|/|a|))

QIθ = α QIIθ = 180° − α QIIIθ = 180° + α QIVθ = 360° − α

#1 burn. arctan(b/a) only sees the ratio — it can't tell QII from QIV. Plot first, then choose θ.

INTERACTIVE 05 / 18

Cis form z = r · cis(θ)

Polar representation · cis(θ) := cosθ + i sinθ

z = r · cis(θ) = r(cosθ + i sinθ) · r = |z|, θ = arg(z)

Same point in the plane — described by length r from the origin and angle θ from the positive real axis. The natural language for rotations, because multiplication of cis-form numbers just adds angles.

r 6.0 θ 60°

z = 6.0 cis(60°)
= 3.000 + 5.196 i
(a, b) = (3.000, 5.196)

z (vector) a = r cosθ b = r sinθ θ arc

06 / 18

Rect ↔ Cis conversion

Two languages · same point

a + bi

(rectangular)

↔

r cis(θ)

(polar)

Cis → Rect: a = r cosθ, b = r sinθ

Rect → Cis: r = √(a²+b²), θ = arctan(b/a) + quadrant fix

15 cis(90°) = 15(0 + i) = 15i

4 cis(30°) = 4(√3/2 + i/2) = 2√3 + 2i

z = −3 + 3√3 i → r=6, θ=120° → 6 cis(120°)

07 / 18

Complex conjugate z̄ = a − bi

Reflection across the real axis

z̄ = a − bi · in cis: r cis(θ)̄ = r cis(−θ)

Flip the sign of the imaginary part. Geometrically: mirror across the horizontal axis. The magic identity:

z · z̄ = (a+bi)(a−bi) = a² + b² = |z|²

(1 + √3 i)(1 − √3 i) = 1 − 3i² = 4 = |1+√3 i|²

The conjugate trick rationalizes complex denominators — multiply top and bottom by the conjugate.

08 / 18

Multiplication by i = 90° rotation

Geometric reveal · i = cis(90°)

z · i = r cis(θ + 90°)

Since i = 0 + 1·i sits at (0, 1) on the unit circle, i = cis(90°). Multiplying by it rotates z by 90° CCW with no change in length. Apply it four times — you're back where you started.

W1 P7: z = 4 + 4√3 i = 8 cis(60°) → zi = 8 cis(150°)

09 / 18

Multiplication: multiply radii, add angles PEAK 1

The whole unit, in one line

r₁ cis(θ₁) · r₂ cis(θ₂) = (r₁r₂) cis(θ₁ + θ₂)

Two arrows from the origin. The product is one arrow: length = r₁·r₂, angle = θ₁ + θ₂. Multiplication in ℂ is rotation + scaling — and the proof is just FOIL plus the trig angle-addition identities.

W1 P1: 6 cis(120°) · 4 cis(30°) = 24 cis(150°)

W1 P4: 3 cis(20°) · 5 cis(70°) = 15 cis(90°) = 15i

INTERACTIVE 10 / 18

Multiplication visualizer

Drag both factors · watch the product rotate & scale

r₁ 4.0 θ₁ 30° r₂ 3.0 θ₂ 70°

z₁ = 4.0 cis(30°)
z₂ = 3.0 cis(70°)
z₁ · z₂ = 12.0 cis(100°)

Notice: the product's radius is the product of radii, the angle is the sum of angles. Try r₂ < 1 — the product shrinks. Try θ₂ = 90° — pure rotation.

11 / 18

Division: divide radii, subtract angles

Inverse of multiplication · proved by the conjugate trick

r₁ cis(θ₁) / [r₂ cis(θ₂)] = (r₁/r₂) cis(θ₁ − θ₂)

Proof technique: multiply numerator and denominator by the conjugate of the denominator. The denominator becomes r₂²(cos²β + sin²β) = r₂² by the Pythagorean identity. The numerator FOILs into the angle-subtraction identities — so cos(α−β) and sin(α−β) drop out.

W1 P5: 12 cis(90°) / 2 cis(60°) = 6 cis(30°)

W1 P6: 15 cis(140°) / 5 cis(35°) = 3 cis(105°)

12 / 18

DeMoivre's theorem PEAK 3

The multiplication rule applied n times

[r cis(θ)]ⁿ = rⁿ cis(nθ)

Each step scales by r and rotates by θ. Angles climb linearly with n; radii scale exponentially with n. Trace gives a log spiral when |z| ≠ 1.

Proof by induction. Base case z¹ trivial. Inductive step: z^k+1 = z^k · z = r^kcis(kθ) · r cis(θ) = r^k+1cis((k+1)θ). Negative n via division rule.

W2 P3a: (2 cis(12°))⁶ = 64 cis(72°)

W2 P4: (2 cis(45°))⁵ = 32 cis(225°)

13 / 18

Powers trajectory: spiral or circle

|z| decides everything · log spiral family

|z| > 1 → spiral out · |z| = 1 → march on unit circle · |z| < 1 → spiral in

|z| > 1 — outward

|z| = 1 — unit circle

|z| < 1 — inward

Same family as nautilus shells, hurricane bands, galaxy arms. W2 P4: z = 2 cis(45°) doubles in length and rotates 45° per step — outward spiral.

14 / 18

Cycle period of cis(θ)

When do powers come home?

n = 360° / gcd(θ, 360°)

For z on the unit circle, zⁿ = 1 when nθ ≡ 0 (mod 360°). The smallest such n is the cycle length — and the powers trace a regular n-gon.

θ = 90° → gcd(90, 360) = 90 → n = 4 (powers of i)

θ = 72° → gcd(72, 360) = 72 → n = 5 (pentagon)

θ = 15° → gcd(15, 360) = 15 → n = 24 (W2 P5 reveal)

Modular reduction: z^k = z^{k mod n}. So z³⁰ = z⁶ when n = 24.

15 / 18

n-th roots of a complex number

n distinct roots on a circle of radius ⁿ√R

z = ⁿ√R · cis((α + 360°k)/n), k = 0, 1, …, n−1

To solve zⁿ = R cis(α): match moduli (r = R^1/n) and match angles mod 360°. The n choices of k give n distinct roots spaced 360°/n apart around a circle.

Cube roots of 8 cis(60°) = 2 cis(20°), 2 cis(140°), 2 cis(260°)

INTERACTIVE 16 / 18

Roots of unity PEAK 4

Solutions to zⁿ = 1 · a regular n-gon on the unit circle

ζₖ = cis(360°k/n), k = 0, 1, …, n−1

n 6

Step angle: 60°
Roots: 6 · spaced 60° apart
One vertex always at z = 1

Try n = 4 (powers of i), n = 6 (hexagon), n = 24 (the W2 P5 24-gon). Each root is a rotation of the previous by 360°/n.

Group structure: the n roots form a cyclic group under multiplication — they're closed (any product is another root) and the n powers of ζ₁ = cis(360°/n) generate them all.

17 / 18

Trig identities ARE complex multiplication

The grand unification · two languages, one fact

(cosα + i sinα)(cosβ + i sinβ) = cos(α+β) + i sin(α+β)

FOIL the left side. Use i² = −1. Group real and imaginary parts:

REAL PART

        cosα cosβ − sinα sinβ

        = cos(α+β)

IMAGINARY PART

        sinα cosβ + cosα sinβ

        = sin(α+β)

The angle-addition identities you memorized in trig class — they are literally the real and imaginary parts of cis(α) · cis(β). Two languages for the same equation. The proof of the multiplication rule is just this expansion run backward.

18 / 18

Fundamental Theorem of Algebra

Why ℂ is the right place for polynomials

A degree-n polynomial over ℂ has exactly n roots in ℂ (counted with multiplicity).

Stated without proof at this level (proofs use complex analysis / topology). Justifies the warm-up claim that xⁿ = 1 has exactly n complex solutions, no more no fewer.

x⁴ − 1 = (x−1)(x+1)(x−i)(x+i) → 4 roots: 1, −1, i, −i

Bonus: roots of unity menagerie

n = 3

n = 4

n = 6

n = 8

Sequences & Series

Unit 01 — MI4 — list vs. sum, five flavors, eight formulas, three techniques

Thesis. A sequence is a structured ordered list (arithmetic, geometric, harmonic, iterated, or pattern); a series is the sum of that list. Every problem reduces to identify the type → pick the formula → plug in. Recognition speed and arithmetic carefulness are the whole game. The n=1 verify is a 5-second insurance policy on every closed form.

ΔSequence vs. Series

A sequence is an ordered list. A series is the sum of that list. They are not interchangeable.

notation Sequence: {a_n}_n=1^N vs. Series: Σ_n=1^N a_n

Trap: "limit of the sequence" ≠ "sum of the series". Constant sequence 1,1,1,… converges to 1; its series diverges to ∞.

notationThe Indexed-Family Atlas

One index variable, one family, one combiner. Σ adds, Π multiplies, ⋃ unions, ⋂ intersects. Same parsing logic.

Op	Acts on	Combines via	Empty
Σ	numbers a_k	addition	= 0
Π	numbers a_k	multiplication	= 1
⋃	sets A_k	OR (in ≥1)	= ∅
⋂	sets A_k	AND (in all)	= universe

Explicit form a_n = f(n) plugs n directly. Recursive form a_n = g(a_n-1) needs a seed.

Set ≠ sequence. Curly-set {2,5,7} discards order. Use indexed {a_k}_k=1ⁿ when order matters.

arithArithmetic Sequence — Add Fixed d

Each term differs from the previous by a fixed common difference d. Plot of (n, a_n) is a straight line of slope d.

explicit form a_n = a₁ + (n − 1) · d

Off-by-one trap (7× on WS3): writing (n) or (n−2) where it should be (n−1).

geomGeometric Sequence — Multiply by Fixed r

Each term is the previous times common ratio r. Plot is exponential.

explicit form a_n = a₁ · rⁿ⁻¹

Exponent trap: rⁿ at n=1 gives a₁·r, not a₁. The exponent is always (n−1).

harmHarmonic — Flip → Arithmetic → Flip

A sequence is harmonic iff its reciprocals form an arithmetic sequence. There is no separate formula — just flip.

the flip trick 1/a_n = 1/a₁ + (n − 1) · d ⇒ a_n = 1 / [1/a₁ + (n−1)d]

recognitionThe Five Flavors of Sequence

Given a list, run differences and ratios on the first 4–5 terms. Whichever returns a constant tells you the type.

Test on first diffs / ratios	Type	Form
Δ constant	Arithmetic	a + (n−1)d
Δ² constant = c	Quadratic	lead = c/2
Δ^p constant	Polynomial deg p	lead = Δ^p/p!
Ratios constant = r	Geometric	a · rⁿ⁻¹
Reciprocal Δ constant	Harmonic	flip → arith → flip
a_n = a_n−1+a_n−2	Fibonacci-style	recursive

Example. 3, 8, 15, 24, 35 → Δ = 5, 7, 9, 11 → Δ² = 2 (constant). Quadratic with lead = 1. Match constants: a_n = n² + 2n.

habitThe n=1 Diagnostic — 5-Second Insurance

After deriving any closed form, plug n=1. If you don't get a₁, the formula is wrong. This habit catches the most common slip in the unit.

three checks every formula must pass n(n+1)/2 at n=1 → 1 · 2/2 = 1 ✓
(n/2)(2a + (n−1)d) at n=1 → (1/2)(2a) = a ✓
a · rⁿ⁻¹ at n=1 → a · r⁰ = a ✓

If it works at n=1, you've eliminated the off-by-one family of errors at almost zero cost. 7 errors on the WS3 submission would have been caught here.

Anti-pattern. Skipping the check because "it looks right." It does, until you compute a₁₀ and the answer is off by exactly d or by a factor of r.

seriesArithmetic Series — Reverse-and-Add (Gauss)

Pair the first and last term, then the second and second-to-last; each pair sums to (a₁ + a_n). There are n/2 pairs.

two equivalent forms S_n = (n/2)(a₁ + ℓ) = (n/2)(2a₁ + (n−1)d)

derivation — reverse and add S = a + (a+d) + (a+2d) + … + ℓ S = ℓ + (ℓ-d) + (ℓ-2d) + … + a ──────────────────────────────────── 2S = (a+ℓ) + (a+ℓ) + … ← n copies S = (n/2)(a + ℓ)

Gauss-as-a-kid: Σ_k=1¹⁰⁰ k = (100/2)(1 + 100) = 50·101 = 5050.

seriesGeometric Series (finite) — Multiply-and-Subtract

finite sum, r ≠ 1 S_n = a (1 − rⁿ) / (1 − r)

derivation — multiply by r, subtract S = a + ar + ar² + … + ar^(n-1) rS = ar + ar² + … + ar^(n-1) + ar^n ───────────────────────────────────────── S − rS = a − ar^n ⇒ S(1−r) = a(1−r^n) ⇒ S = a(1 − r^n)/(1 − r)

Edge case r = 1. The formula's denominator is zero. Use the constant-sum fallback S_n = n · a.

First-term warning. For Σ_k=1^∞ 9·(2/3)^k, the first term at k=1 is 6, not 9. Always plug k=1 to find a.

10 · interactive

∞ seriesInfinite Geometric — Convergence Slider

converges iff |r| < 1 S_∞ = a / (1 − r)

r = 0.50 a = 1.0

CONVERGES S∞ = 2.000

Drag r past ±1 to see partial sums explode or oscillate. The dashed green line is the limit when |r|<1.

specialThe Special-Sums Library

Σ_k=1ⁿ c = n · c    (deg 1)
Σ_k=1ⁿ k = n(n+1) / 2    (deg 2)
Σ_k=1ⁿ k² = n(n+1)(2n+1) / 6    (deg 3)
Σ_k=1ⁿ k³ = [n(n+1)/2]² = (Σk)²    (deg 4)

Degree rule. Summing a degree-p polynomial in k gives a degree-(p+1) polynomial in n. Each formula has one more factor than the last.

The remarkable identity. Sum of cubes equals square of sum: 1³+2³+3³+…+n³ = (1+2+3+…+n)². Visual proof: each cube k³ unfolds into a square shell of side T_k.

techniqueTelescoping — Adjacent Terms Cancel

If each term is a difference b_k − b_k+1, interior terms cancel pairwise. Only the head and tail survive.

Σ_k=1ⁿ (b_k − b_k+1) = b₁ − b_n+1
Σ_k=1^∞ (b_k − b_k+1) = b₁ − lim b_n+1

famous example — Σ (1/k − 1/(k+1)) (1 − ½) + (½ − ⅓) + (⅓ − ¼) + … ↑ each ½, ⅓, ¼, … appears with + then − = 1 − 1/(n+1) → 1 as n → ∞

Telescoping can DIVERGE. Σ log₂((k+1)/(k+2)) telescopes to log 2 − log(n+2) → −∞. Terms-to-zero is necessary but NOT sufficient for convergence.

13 · interactive

iterationFixed-Point Iteration — Cobweb

Apply x ↦ m·x + b repeatedly. Fixed point: L = b/(1−m). Attracting iff |m| < 1.

m = 0.50 b = 3.0 x₀ = 0.0

ATTRACTING · L = 6.00

14 · interactive

meansAM ≥ GM ≥ HM (Semicircle Proof)

On a semicircle of diameter a+b, the radius is the AM, the perpendicular at the split is the GM, the foot of the perpendicular onto the radius is the HM. The picture proves the inequality.

a = 3.0 b = 7.0

numerical readout AM = (a+b)/2 = 5.00 GM = √(ab) = 4.58 HM = 2ab/(a+b) = 4.20

The five flavors collapse to three moves. Reverse-and-add for arithmetic. Multiply-and-subtract for geometric. Flip the reciprocals for harmonic. Iteration is a 4th flavor that doesn't sum; pattern is a meta-flavor that hides one of the above.

applicationRepeating Decimal = Geometric Series

A pure repeating decimal with p-digit block is a geometric series with r = 10^−p.

0.d₁d₂…d_p = (block) / (p nines)

0.999… = 1 (the classic) 0.999… = 9/10 + 9/100 + 9/1000 + … = (9/10) / (1 − 1/10) = (9/10) / (9/10) = 1 ← the same number, not "almost" more 0.7̄ = 7/9 0.21̄ = 21/99 = 7/33 0.723̄ = 723/999 = 241/333

Mixed (prefix + repeat). Scale by 10^q and 10^q+p, subtract to eliminate the repeating tail. 0.7̄24̄: 1000x − 10x = 717 → x = 717/990.

wordBouncing Ball & HM Word Templates

bouncing ball total distance D = h · (1 + ρ) / (1 − ρ)

Why (1+ρ)/(1−ρ). Initial drop counts once. Every rebound height counts twice (up + down). Total = h + 2(hρ + hρ² + …) = h + 2hρ/(1−ρ) = h(1+ρ)/(1−ρ).

HM-family word problems Crossing wires: h = ab/(a+b)
Work-rate (simultaneous): t = t_At_B/(t_A+t_B)
Both equal HM(a,b)/2 — the same formula in disguise.

sigma rulesSigma Manipulation Toolkit

linearity (constants out) Σ_k c · a_k = c · Σ_k a_k

linearity (sums split) Σ_k (a_k + b_k) = Σ_k a_k + Σ_k b_k

subtract-the-prefix Σ_k=mⁿ a_k = Σ_k=1ⁿ a_k − Σ_k=1^m−1 a_k

counting terms # terms in Σ_k=mⁿ = n − m + 1

Off-by-one (WS6 trap). Σ_k=4¹⁷(−3) has 14 terms, not 17. Sum = 14·(−3) = −42.

Recognition table — final reference card.

Pattern	Formula	Pattern	Formula
Σ k	n(n+1)/2	Σ ar^k−1 finite	a(1−rⁿ)/(1−r)
Σ k²	n(n+1)(2n+1)/6	Σ ar^k−1 infinite	a/(1−r), \|r\|<1
Σ k³	[n(n+1)/2]²	Σ (b_k−b_k+1)	b₁ − lim b_n+1
Σ c	n·c	0.d p-block	block / (p nines)
arith series	(n/2)(2a+(n−1)d)	linear x_n=mx_n−1+b	L=b/(1−m), attract iff \|m\|<1
bouncing ball	h(1+ρ)/(1−ρ)	crossing wires	ab/(a+b)

Vectors

corePolar coordinate system

P→RPolar → Rectangular

R→PRectangular → Polar (with quadrant fix)

peak 1Two families: every point has infinite polar names

edgeThe pole — θ undefined, not zero

notationConvention bridge: a, b vs D, A

circler = A cos θ · off-axis circle (x-axis)

circler = A sin θ · off-axis circle (y-axis)

rotatedr = A cos(θ − C) · phase-shifted circle

|A|=|D|Cardioid · meant for each other

1 < D/A < 2Dimpled limaçon · moved on

D ≥ 2AConvex limaçon · fully moved on

|A| > |D|Inner-loop limaçon · hole in the heart

memoryThe heart mnemonic · three side-by-side

peak 3Rose curves · B odd → B petals · B even → 2B petals

workflowAnchor petal + even spacing

trapEven-B sine rose · NO petals on axes

refLemniscate · figure-8

peak 4 · trapThe pole trap · algebra misses pole intersections

bridgeRectangular ↔ polar · y = 2 − 2 sin x becomes r = 2 − 2 sin θ

conditionInner loop ↔ wave below axis

domainNon-integer B · extended θ-domain

live Limaçon morpher

live Rose petal counter

live Phase-shift rotator · r = A cos(θ − C)

Complex Numbers

The defining identity: i² = −1

z = a + bi on the Argand plane

Modulus |z| — distance from origin

Argument arg(z) = θ

Cis form z = r · cis(θ)

Rect ↔ Cis conversion

Complex conjugate z̄ = a − bi

Multiplication by i = 90° rotation

Multiplication: multiply radii, add angles PEAK 1

Multiplication visualizer

Division: divide radii, subtract angles

DeMoivre's theorem PEAK 3

Powers trajectory: spiral or circle

Cycle period of cis(θ)

n-th roots of a complex number

Roots of unity PEAK 4

Trig identities ARE complex multiplication

Fundamental Theorem of Algebra

Sequences & Series

ΔSequence vs. Series

notationThe Indexed-Family Atlas

arithArithmetic Sequence — Add Fixed d

geomGeometric Sequence — Multiply by Fixed r

harmHarmonic — Flip → Arithmetic → Flip

recognitionThe Five Flavors of Sequence

habitThe n=1 Diagnostic — 5-Second Insurance

seriesArithmetic Series — Reverse-and-Add (Gauss)

seriesGeometric Series (finite) — Multiply-and-Subtract

∞ seriesInfinite Geometric — Convergence Slider

specialThe Special-Sums Library

techniqueTelescoping — Adjacent Terms Cancel

iterationFixed-Point Iteration — Cobweb

meansAM ≥ GM ≥ HM (Semicircle Proof)

applicationRepeating Decimal = Geometric Series

wordBouncing Ball & HM Word Templates

sigma rulesSigma Manipulation Toolkit