Trigonometry is the largest single topic in IB Mathematics Analysis & Approaches HL by syllabus reference count, spanning SL 3.1–3.8 and AHL 3.9–3.11. It threads through almost every other topic — calculus, complex numbers, vectors and 3D geometry all assume fluency with the unit circle, the identities, and the compound-angle formulas. The trickiness is rarely conceptual; it is procedural. Students lose marks for forgetting that arc-length and sector formulas need radians, mixing up the range of $\arccos$, missing the second triangle in an ambiguous SSA case, or skipping the reasoning sentence "no solution since $|\sin\theta| \leq 1$".
This cheatsheet condenses every formula, identity, trick and trap from SL 3.1–3.8 and AHL 3.9–3.11 into one page you can revise from.
§1 — Right-Angle Trig & Exact Values SL 3.1–3.4
SOH CAH TOA & reciprocal trig
$\sin\theta = \dfrac{\text{Opp}}{\text{Hyp}}$ $\cos\theta = \dfrac{\text{Adj}}{\text{Hyp}}$ $\tan\theta = \dfrac{\text{Opp}}{\text{Adj}}$
$\csc\theta = \dfrac{1}{\sin\theta}$ $\sec\theta = \dfrac{1}{\cos\theta}$ $\cot\theta = \dfrac{\cos\theta}{\sin\theta}$
Exact values — must memorise for P1
| $\theta$ | $0$ | $\dfrac{\pi}{6}$ | $\dfrac{\pi}{4}$ | $\dfrac{\pi}{3}$ | $\dfrac{\pi}{2}$ |
|---|---|---|---|---|---|
| $\sin\theta$ | $0$ | $\dfrac{1}{2}$ | $\dfrac{\sqrt{2}}{2}$ | $\dfrac{\sqrt{3}}{2}$ | $1$ |
| $\cos\theta$ | $1$ | $\dfrac{\sqrt{3}}{2}$ | $\dfrac{\sqrt{2}}{2}$ | $\dfrac{1}{2}$ | $0$ |
| $\tan\theta$ | $0$ | $\dfrac{1}{\sqrt{3}}$ | $1$ | $\sqrt{3}$ | undef |
Radians & circle mensuration (SL 3.1)
$180° = \pi$ rad. Arc: $\ell = r\theta$. Sector: $A = \tfrac{1}{2}r^2\theta$. Segment: $\tfrac{1}{2}r^2(\theta - \sin\theta)$.
§2 — Unit Circle, ASTC & Reference Angles SL 3.3–3.5
ASTC: All in Q1 · Sin in Q2 · Tan in Q3 · Cos in Q4.ASTC rule — All Students Take Calculus
- Q1 ($0°$–$90°$): All positive
- Q2 ($90°$–$180°$): Sin positive
- Q3 ($180°$–$270°$): Tan positive
- Q4 ($270°$–$360°$): Cos positive
Reference angles: Q2 → $180° - \theta$; Q3 → $\theta - 180°$; Q4 → $360° - \theta$.
Symmetry identities (AHL 3.11)
§3 — Trigonometric Identities SL 3.6, AHL 3.9–3.10
Pythagorean identities (SL 3.6 / AHL 3.9)
$\sin^2\theta + \cos^2\theta = 1$ | $1 + \tan^2\theta = \sec^2\theta$ | $1 + \cot^2\theta = \csc^2\theta$.
Power-reduction: $\cos^2\theta = \dfrac{1 + \cos 2\theta}{2}$ and $\sin^2\theta = \dfrac{1 - \cos 2\theta}{2}$.
Double angle formulas (SL 3.6)
$\sin 2\theta = 2\sin\theta\cos\theta$ $\tan 2\theta = \dfrac{2\tan\theta}{1 - \tan^2\theta}$.
$\cos 2\theta = \cos^2\theta - \sin^2\theta = 2\cos^2\theta - 1 = 1 - 2\sin^2\theta$.
Compound angle formulas (AHL 3.10)
Solving $a\sin x + b\cos x = k$ (from AHL 3.10)
Derive using the compound angle formula (not a separate syllabus formula):
$$a\sin x + b\cos x = R\sin(x + \phi), \quad R\cos\phi = a,\;\; R\sin\phi = b$$
$$R = \sqrt{a^2 + b^2}, \quad \tan\phi = \frac{b}{a} \quad (\text{check quadrant of } \phi)$$
Then set $u = x + \phi$, solve $\sin u = k/R$, back-substitute tracking the range of $u$.
§4 — Solving Trigonometric Equations SL 3.8
General solutions
- $\sin\theta = k$: $\theta = \arcsin k + 2n\pi$ or $\pi - \arcsin k + 2n\pi$
- $\cos\theta = k$: $\theta = \pm\arccos k + 2n\pi$
- $\tan\theta = k$: $\theta = \arctan k + n\pi$
Step-by-step strategy
- Isolate the trig function (factorise if quadratic).
- Mixed $\sin$/$\cos$: substitute $\cos^2 = 1 - \sin^2$ or expand the double angle.
- Find reference angle $\alpha$; apply ASTC to find quadrants.
- Write both solutions; check the domain; reject $|\sin\theta| > 1$ with reason (R1).
§5 — Graphs of Trigonometric Functions SL 3.5
Key graph properties
| Fn | Period | Range | $x$-intercepts | Even / Odd |
|---|---|---|---|---|
| $\sin x$ | $2\pi$ | $[-1, 1]$ | $n\pi$ | Odd |
| $\cos x$ | $2\pi$ | $[-1, 1]$ | $\tfrac{\pi}{2} + n\pi$ | Even |
| $\tan x$ | $\pi$ | $\mathbb{R}$ | $n\pi$ | Odd |
| $\csc x$ | $2\pi$ | $|y| \geq 1$ | none | Odd |
| $\sec x$ | $2\pi$ | $|y| \geq 1$ | none | Even |
| $\cot x$ | $\pi$ | $\mathbb{R}$ | $\tfrac{\pi}{2} + n\pi$ | Odd |
General sinusoidal form
$f(x) = a\sin\!\big(b(x - h)\big) + k$:
- $|a|$ amplitude
- $T = \dfrac{2\pi}{|b|}$ period
- $h$ phase shift
- $k$ midline
$a = \dfrac{\max - \min}{2}$, $k = \dfrac{\max + \min}{2}$, Range: $[k - |a|, k + |a|]$.
§6 — Sine Rule, Cosine Rule & Area SL 3.7
Sine rule & ambiguous case
$\dfrac{\sin A}{a} = \dfrac{\sin B}{b} = \dfrac{\sin C}{c}$. Use: AAS or SSA.
Ambiguous case (SSA): $a < b\sin A$ → none; $a = b\sin A$ → one right-angled; $b\sin A < a < b$ → two!; $a \geq b$ → one.
Cosine rule & area
$a^2 = b^2 + c^2 - 2bc\cos A$ $\cos A = \dfrac{b^2 + c^2 - a^2}{2bc}$ Use: SAS or SSS.
Area $= \tfrac{1}{2}ab\sin C$.
§7 — 3D Trigonometry SL 3.2–3.3
- Draw and label the 3D diagram.
- Identify the 2D triangle containing the unknown.
- Apply SOH/CAH/TOA or sine/cosine rule.
- Work step by step.
Space diagonal: $d = \sqrt{l^2 + w^2 + h^2}$.
§8 — Inverse Trigonometric Functions AHL 3.9
Domains and ranges
| Function | Domain | Range |
|---|---|---|
| $\arcsin x$ | $[-1, 1]$ | $\left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]$ |
| $\arccos x$ | $[-1, 1]$ | $[0, \pi]$ |
| $\arctan x$ | $\mathbb{R}$ | $\left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)$ |
Right-triangle method & derivatives
Let $\theta = \arccos x$: adj $= x$, hyp $= 1$, opp $= \sqrt{1 - x^2}$ ⇒ $\sin(\arccos x) = \sqrt{1 - x^2}$.
$\sin\!\left(\arctan\dfrac{a}{b}\right) = \dfrac{a}{\sqrt{a^2 + b^2}}$ $\cos(\arctan x) = \dfrac{1}{\sqrt{1 + x^2}}$.
$(\arcsin x)' = \dfrac{1}{\sqrt{1 - x^2}}$ $(\arccos x)' = \dfrac{-1}{\sqrt{1 - x^2}}$ $(\arctan x)' = \dfrac{1}{1 + x^2}$.
§9 — GDC Skills (Paper 2) SL 3.5–3.8
TI-Nspire: mode, syntax, solving
Mode check: type sin(90) → 1 = degrees; 0.894 = radians (IB default).
| Wrong | Right |
|---|---|
xsinx | x*sin(x) |
2sinx | 2*sin(x) |
sin2x | sin(2*x) |
Use Zoom-Trig for trig graphs. Menu → Analyse → Intersection to solve graphically. solve(eq,x) | 0<=x<=6.2832 — always restrict the domain.
solve command returns one solution. Always graph first to count intersections, then find each one separately.§10 — Identity Proofs AHL 3.10
- Start one side only (the more complex side).
- Convert to $\sin$ / $\cos$.
- Use $1 - \cos 2x = 2\sin^2 x$ or $1 + \cos 2x = 2\cos^2 x$.
- Factorise / expand to reach the other side.
- Write $\equiv$ throughout — never move terms across it.
§11 — Exam Attack Plan All sections
| If you see… | First move |
|---|---|
| $a\cos^2 x \pm b\sin x + c = 0$ | Substitute $\cos^2 x = 1 - \sin^2 x$ |
| $\sin 2x$ in the equation | Expand $2\sin x \cos x$ |
| $a\sin x + b\cos x = k$ | Compound angle: find $R$, $\phi$ |
| "Show that" identity | Start LHS; convert to $\sin/\cos$ |
| Arc / sector / segment | Check $\theta$ in radians first |
| SSA triangle | Check ambiguous case |
| 3D elevation problem | Identify the 2D right triangle |
| $\arcsin / \arccos +$ trig | Draw a right triangle |
| Sinusoidal model (P2) | Read $a, k$ from max/min; $b$ from period; $h$ last |
| Trig equation (P2) | Graph both sides; count intersections |
Top mark-losing errors
- Not stating why $\sin\theta = k$ is impossible when $|k| > 1$ (R1 mark).
- Missing the second triangle in the ambiguous SSA case.
- Wrong range for $\arccos$: must be $[0, \pi]$, not $[-\pi/2, \pi/2]$.
- Rounding intermediate values — keep them exact until the final answer.
- No domain restriction in the GDC
solvecommand. - Moving terms across $\equiv$ in a proof.
- Not tracking the range of $u$ when solving $\sin(x + \phi) = k$.
- Wrong angle mode on the GDC (always check with $\sin(90)$ first).
Worked Example — IB-Style Compound Angle Equation
Question (HL Paper 1 style — 7 marks)
Solve the equation $\sqrt{3}\sin x + \cos x = 1$ for $x \in [0, 2\pi]$, giving exact answers.
Solution
- Identify the compound-angle form: $a\sin x + b\cos x$ with $a = \sqrt{3}$, $b = 1$. Rewrite as $R\sin(x + \phi)$ where $R\cos\phi = a$, $R\sin\phi = b$. (M1)
- Find $R$ and $\phi$: $R = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{4} = 2$. From $R\cos\phi = \sqrt{3}$ and $R\sin\phi = 1$ both positive, $\phi$ is in Q1: $\tan\phi = \dfrac{1}{\sqrt{3}} \Rightarrow \phi = \dfrac{\pi}{6}$. (A1)(A1)
- Rewrite the equation: $2\sin\!\left(x + \dfrac{\pi}{6}\right) = 1 \Rightarrow \sin\!\left(x + \dfrac{\pi}{6}\right) = \dfrac{1}{2}$. (M1)
- Solve for $u = x + \dfrac{\pi}{6}$ over the shifted interval: since $x \in [0, 2\pi]$, $u \in \left[\dfrac{\pi}{6}, \dfrac{13\pi}{6}\right]$. Solutions of $\sin u = \dfrac{1}{2}$ in this range: $u = \dfrac{\pi}{6}, \dfrac{5\pi}{6}, \dfrac{13\pi}{6}$. (R1)
- Back-substitute $x = u - \dfrac{\pi}{6}$: $x = 0,\; \dfrac{2\pi}{3},\; 2\pi$. (A1)(A1)
Examiner's note: the R1 in step 4 is awarded for explicitly stating the shifted range of $u$; missing it costs the mark even if the final answers are correct. Students who use the R-formula without showing the derivation via the compound-angle formula also lose marks — the IB syllabus does not list the R-form as a given identity (AHL 3.10).