Trigonometry · Units 1–9 · Quick reference
Trigonometry
Radians, right triangles, the unit circle, graphs, inverses, identities, equations, oblique triangles, vectors, polar coordinates and complex numbers.
1Angles, Radians & Circular Motionmeasure turning two ways
Coterminal angles
Same terminal side: add or subtract full turns, θ ± 360°k or θ ± 2πk.
EX840° − 2·360° = 120° · −45° + 360° = 315°
Degrees–minutes–seconds
1° = 60′, 1′ = 60″.
EX25°30′36″ = 25 + 3060 + 363600 = 25.51°
Radians in formulas
s = rθ, A = ½r²θ and v = rω need θ in radians. Convert degrees first.
Converting
Degrees → radians: × π180. Radians → degrees: × 180π.
π rad = 180°
EX150° = 5π6 · 7π4 = 315° · 1 rad ≈ 57.30°
Want π? Put π on top
Going to radians, multiply by π180 (π upstairs). Going to degrees, use 180π so the π cancels.
Arc length & sector area (θ in radians)
s = rθ · A = ½r²θ
EXr = 6, θ = 2π3: s = 4π ≈ 12.57, A = 12π ≈ 37.70
Angular & linear speed
- Angular speed ω = θt (rad per unit time)
- Linear speed v = st = rω
- 1 revolution = 2π rad
Bicycle wheel Worked
| Deg | Rad | Deg | Rad |
|---|---|---|---|
| 30° | π⁄6 | 120° | 2π⁄3 |
| 45° | π⁄4 | 180° | π |
| 60° | π⁄3 | 270° | 3π⁄2 |
| 90° | π⁄2 | 360° | 2π |
2Right-Triangle Trig & Applicationssix ratios, one triangle
The six ratios (acute θ)
sin θ = opphyp · cos θ = adjhyp · tan θ = oppadj
csc θ = hypopp · sec θ = hypadj · cot θ = adjopp
SOH-CAH-TOA and the special triangles: Geometry Part 2 §12 and Part 1 §3.
One "co" per pair
Each reciprocal pair has exactly one co: cosecant ↔ sine, secant ↔ cosine, cotangent ↔ tangent.
8-15-17 triangle, θ opposite 8 Worked
Solve the triangle Worked
Find an angle Worked
Elevation & depression
Both are measured from the horizontal. The angle of depression from A to B equals the angle of elevation from B to A (alternate interior angles).
Cliff to boat Worked
Bearings
- Compass bearing: N 35° E means 35° east of north.
- True bearing: clockwise from north (N 35° E = 035°).
EX12 km on N 35° E: east 12 sin 35° ≈ 6.88 km, north 12 cos 35° ≈ 9.83 km
Cofunctions
sin θ = cos(90° − θ), tan θ = cot(90° − θ), sec θ = csc(90° − θ): the two acute angles trade roles.
EXsin 28° = cos 62°
3The Unit Circleevery point is (cos θ, sin θ)
Definitions
For the point (x, y) where θ meets the circle x² + y² = 1:
cos θ = x, sin θ = y, tan θ = yx
sec θ = 1x, csc θ = 1y, cot θ = xy
For radius r: x = r cos θ, y = r sin θ.
Cos comes first
Alphabetical and on the axes: cos before sin, like x before y.
| θ | 0 | π⁄6 | π⁄4 | π⁄3 | π⁄2 |
|---|---|---|---|---|---|
| sin | 0 | 1/2 | √2/2 | √3/2 | 1 |
| cos | 1 | √3/2 | √2/2 | 1/2 | 0 |
| tan | 0 | √3/3 | 1 | √3 | — |
Count up the roots
sin of 0°, 30°, 45°, 60°, 90° = √02, √12, √22, √32, √42; cosine reads the list backward.
The left-hand trick
Palm up, fingers = 0°, 30°, 45°, 60°, 90° from the thumb. Fold one down: sin = √(fingers below)2, cos = √(fingers above)2. (Traditional.)
4Reference Angles & Signsany angle, using quadrant I
| Quadrant | θ′ (degrees) | θ′ (radians) |
|---|---|---|
| I | θ | θ |
| II | 180° − θ | π − θ |
| III | θ − 180° | θ − π |
| IV | 360° − θ | 2π − θ |
All Students Take Calculus
From QI counterclockwise: All positive, Sine (and csc), Tangent (and cot), Cosine (and sec). (Traditional.)
Three-step evaluation
- Find a coterminal angle in [0°, 360°).
- Find θ′ and its quadrant-I value.
- Attach the sign from ASTC.
Evaluate Worked
From one ratio to all Worked
Quadrantal angles
At 0°, 90°, 180°, 270° there is no reference triangle: read (cos, sin) straight off the axes. tan 90° and sec 90° are undefined (x = 0).
5Graphs of the Six Functionsperiod, domain, range and asymptotes
| f | Period | Domain | Range | Zeros / asymptotes |
|---|---|---|---|---|
| sin | 2π | all reals | [−1, 1] | zeros at kπ |
| cos | 2π | all reals | [−1, 1] | zeros at π2 + kπ |
| tan | π | x ≠ π2 + kπ | all reals | zeros kπ; asymptotes π2 + kπ |
| csc | 2π | x ≠ kπ | (−∞, −1] ∪ [1, ∞) | asymptotes kπ |
| sec | 2π | x ≠ π2 + kπ | (−∞, −1] ∪ [1, ∞) | asymptotes π2 + kπ |
| cot | π | x ≠ kπ | all reals | zeros π2 + kπ; asymptotes kπ |
Asymptotes live where the denominator dies
tan and sec have asymptotes where cos = 0; cot and csc where sin = 0.
Symmetry
- cos and sec are even: graphs mirror across the y-axis.
- sin, csc, tan, cot are odd: symmetric about the origin.
Reciprocal sketch
Draw sin (or cos) lightly; put asymptotes at its zeros; the csc (sec) branches touch its peaks and valleys and open away.
6Sinusoids, Transformations & Harmonic Motiony = A sin(B(x − C)) + D
| Part | Meaning | Formula |
|---|---|---|
| |A| | amplitude (half the height) | max − min2 |
| B | cycles in 2π | period 2πB (tan, cot πB) |
| C | phase shift (right if +) | factor B out first |
| D | midline y = D | max + min2 |
Big B, busy wave
A bigger B squeezes more cycles in: the period shrinks to 2π/B.
Write the equation Worked
Read the graph Worked
Five key points
Split one period into quarters starting at the shift: midline, max, midline, min, midline (sine). Cosine starts at the max.
EXx = π4, π2, 3π4, π, 5π4 → y = 1, 4, 1, −2, 1
A-B-C-D
Amplitude, B sets the period, C shifts sideways, D moves the midline: read them in alphabetical order.
Simple harmonic motion
d = a sin ωt or d = a cos ωt: amplitude |a|, period 2πω, frequency f = ω2π (cycles per unit time).
EXd = 5 cos 4πt cm: amplitude 5 cm, period 0.5 s, frequency 2 Hz.
Explore
Add sine waves and you can draw almost any curve: Fourier: circles that draw (schoollaunch.org).
7Inverse Trig Functionsangles from ratios, on restricted ranges
| Function | Domain | Range (the answer) | Quadrants |
|---|---|---|---|
| sin⁻¹ x (arcsin) | [−1, 1] | [−π2, π2] | I, IV |
| cos⁻¹ x (arccos) | [−1, 1] | [0, π] | I, II |
| tan⁻¹ x (arctan) | all reals | (−π2, π2) | I, IV |
| cot⁻¹ x | all reals | (0, π) | I, II |
| sec⁻¹ x | |x| ≥ 1 | [0, π], ≠ π2 | I, II |
| csc⁻¹ x | |x| ≥ 1 | [−π2, π2], ≠ 0 | I, IV |
Sine & tangent answer right, cosine answers up
sin⁻¹ and tan⁻¹ give angles on the right half (QI or QIV); cos⁻¹ gives angles on the top half (QI or QII).
Evaluate Worked
sin⁻¹ x ≠ 1/sin x
The −1 means inverse function. The reciprocal is (sin x)⁻¹ = csc x.
Compositions Worked
Triangle method
For cos(tan⁻¹ u): let θ = tan⁻¹ u, draw a right triangle with tan θ = u, find the third side, read the ratio. Signs follow the range.
8Identitiestrue for every angle where both sides are defined
Reciprocal & quotient
csc θ = 1sin θ, sec θ = 1cos θ, cot θ = 1tan θ
tan θ = sin θcos θ, cot θ = cos θsin θ
Pythagorean
÷ cos²θ: tan²θ + 1 = sec²θ · ÷ sin²θ: 1 + cot²θ = csc²θ
sin²θ + cos²θ = 1
One identity, three copies
Learn sin² + cos² = 1; divide by cos² or sin² to get the other two.
Even / odd & cofunction
- cos(−θ) = cos θ, sec(−θ) = sec θ (even)
- sin(−θ) = −sin θ; tan, csc, cot also odd
- sin(π2 − θ) = cos θ, tan(π2 − θ) = cot θ, sec(π2 − θ) = csc θ
Product-to-sum
- sin A cos B = ½[sin(A + B) + sin(A − B)]
- cos A cos B = ½[cos(A − B) + cos(A + B)]
- sin A sin B = ½[cos(A − B) − cos(A + B)]
Sum & difference
- sin(A ± B) = sin A cos B ± cos A sin B
- cos(A ± B) = cos A cos B ∓ sin A sin B
- tan(A ± B) = tan A ± tan B1 ∓ tan A tan B
EXsin 75° = sin(45° + 30°) = √6 + √24 ≈ 0.9659; tan 15° = tan(45° − 30°) = 2 − √3 ≈ 0.2679
Sine mixes, cosine matches
Sine mixes (sin·cos) and keeps the sign; cosine matches (cos·cos, sin·sin) and flips it.
Double angle
cos 2θ = cos²θ − sin²θ = 2cos²θ − 1 = 1 − 2sin²θ
sin 2θ = 2 sin θ cos θ
tan 2θ = 2 tan θ1 − tan²θ
EXsin θ = 35 in QII, so cos θ = −45: sin 2θ = −2425, cos 2θ = 1 − 2(925) = 725
Simplify Worked
Half angle (sign from the quadrant of θ/2)
tan θ2 = 1 − cos θsin θ = sin θ1 + cos θ
sin θ2 = ±√1 − cos θ2
cos θ2 = ±√1 + cos θ2
EXcos 22.5° = √1 + √2/22 = √(2 + √2)2 ≈ 0.9239
Power reducing
tan²x = 1 − cos 2x1 + cos 2x
sin²x = 1 − cos 2x2
cos²x = 1 + cos 2x2
Sum-to-product
- sin A + sin B = 2 sin A + B2 cos A − B2
- sin A − sin B = 2 cos A + B2 sin A − B2
- cos A + cos B = 2 cos A + B2 cos A − B2
- cos A − cos B = −2 sin A + B2 sin A − B2
9Proving Identities & Solving Equationsan identity is always true; an equation is true for some angles
Proof strategy
- Work on one side only (usually the messier one).
- Rewrite in sin and cos; combine fractions; factor.
- Use a Pythagorean identity to trade sin² and cos².
- Multiply by a conjugate (1 ± sin x) when you see 1 ± sin x.
Never cross the equals sign
In a proof, do not add, multiply or square both sides: transform one side until it matches the other.
Prove sec x − cos x = sin x tan x Worked
Prove (1 − cos²x)(1 + cot²x) = 1 Worked
Solving steps
- Isolate the trig function (or factor).
- Find solutions in one period, using reference angles and ASTC.
- General solution: add 2πk (sin, cos, sec, csc) or πk (tan, cot).
Linear Worked
Quadratic Worked
Use an identity Worked
Multiple angle Worked
Calculator angle Worked
Lost and fake solutions
- Do not divide by cos x: factor it out, or you lose cos x = 0.
- After squaring both sides, check every answer in the original.
10Oblique Triangles & Areaany triangle, no right angle needed
Law of Sines (AAS, ASA, SSA)
asin A = bsin B = csin C
Law of Cosines (SAS, SSS)
cos C = a² + b² − c²2ab
c² = a² + b² − 2ab cos C
Parallelogram law
c² + d² = 2(a² + b²)
Law of Cosines on c (angle θ) and d (180° − θ); cos(180° − θ) = −cos θ, so the 2ab cos θ terms cancel. EX5, 3, 60°: 19 + 49 = 2(25 + 9) ✓
Sines need a pair
Sines need a matched pair: a side and its opposite angle. No pair (SAS, SSS)? Use cosines.
ASA: A = 40°, B = 75°, a = 10 Worked
SSS: 7, 9, 12 (largest angle first) Worked
SAS Worked
| A acute, h = b sin A | Triangles |
|---|---|
| a < h | 0 |
| a = h | 1 (right) |
| h < a < b | 2 |
| a ≥ b | 1 |
Two triangles Worked
Area
ASA: K = a² sin B sin C2 sin A
K = ½ab sin C
Heron's formula (SSS)
K = √(s(s − a)(s − b)(s − c))
s = a + b + c2
EX7, 9, 12: s = 14, K = √(14·7·5·2) = 14√5 ≈ 31.30 (= ½·7·9·sin C ✓)
11Vectorsmagnitude and direction
Component form
- v = ⟨a, b⟩ = ai + bj
- |v| = √(a² + b²); direction tan θ = ba (check the quadrant)
- v = ⟨|v| cos θ, |v| sin θ⟩; unit vector v|v|
Operations
- Add/subtract componentwise; kv scales the length by |k|.
- Dot product: u · v = u₁v₁ + u₂v₂ = |u||v| cos θ
- u ⊥ v exactly when u · v = 0
Angle between Worked
Plane and wind Worked
12Polar Coordinates & Graphsdistance and angle instead of x and y
Conversions
x = r cos θ, y = r sin θ
r² = x² + y², tan θ = yx
Convert Worked
Many names, one point
(r, θ) = (r, θ + 2πk) = (−r, θ + π). A negative r points the opposite way.
EXr = 4 cos θ → r² = 4r cos θ → x² + y² = 4x: circle, center (2, 0), radius 2
Odd roses keep, even roses double
r = a cos nθ: odd n gives n petals, even n gives 2n.
| Equation | Graph |
|---|---|
| r = a; θ = α | circle at the pole; line through the pole |
| r = a cos θ, r = a sin θ | circle of diameter |a| through the pole |
| r = a cos nθ | rose: n petals (n odd), 2n (n even) |
| r = a(1 ± cos θ) | cardioid |
| r = a + b cos θ | limaçon; inner loop if |a| < |b| |
| r² = a² cos 2θ | lemniscate |
| r = aθ | spiral |
13Complex Numbers & De Moivrerotate and scale in the complex plane
Trigonometric (polar) form
r = |z| = √(a² + b²), tan θ = ba (check the quadrant)
z = a + bi
= r(cos θ + i sin θ) = r cis θ
Multiply & divide
- z₁z₂ = r₁r₂ cis(θ₁ + θ₂)
- z₁z₂ = r₁r₂ cis(θ₁ − θ₂)
EX2 cis 30° · 3 cis 60° = 6 cis 90° = 6i
Multiply lengths, add angles
Multiplying complex numbers multiplies their lengths and adds their angles.
De Moivre's theorem
[r cis θ]ⁿ = rⁿ cis nθ
EX(1 + i)⁸ = (√2 cis 45°)⁸ = 16 cis 360° = 16
nth roots
The n roots of r cis θ: ⁿ√r · cis(θ + 360°kn), k = 0, 1, …, n − 1. They sit evenly around a circle.
EX8 = 8 cis 0° → 2 cis 0°, 2 cis 120°, 2 cis 240° = 2, −1 ± √3 i