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Trigonometry Identities Cheat Sheet

CAPS / NSC & IEB · Grades 10–12 · Paper 2

△ Definitions

In a right triangle $\sin\theta = \dfrac{\text{opp}}{\text{hyp}}$

$\cos\theta = \dfrac{\text{adj}}{\text{hyp}}$

$\tan\theta = \dfrac{\text{opp}}{\text{adj}}$

Reciprocals $\operatorname{cosec}\theta = \dfrac{1}{\sin\theta}$

$\sec\theta = \dfrac{1}{\cos\theta}$

$\cot\theta = \dfrac{1}{\tan\theta}$

On the Cartesian plane $\sin\theta = \dfrac{y}{r},\ \cos\theta = \dfrac{x}{r},\ \tan\theta = \dfrac{y}{x}$

with $r = \sqrt{x^{2}+y^{2}}\,,\quad r > 0$

$r$ is a length and is never negative. The signs live in $x$ and $y$, which is the whole reason the CAST diagram works.

≡ Fundamental Identities

Quotient $\tan\theta = \dfrac{\sin\theta}{\cos\theta}$

Square (Pythagorean) $\sin^{2}\theta + \cos^{2}\theta = 1$

rearranged $\sin^{2}\theta = 1 - \cos^{2}\theta$

rearranged $\cos^{2}\theta = 1 - \sin^{2}\theta$

Derived $1 + \tan^{2}\theta = \sec^{2}\theta$

$1 + \cot^{2}\theta = \operatorname{cosec}^{2}\theta$

Only the first two are examinable as "the" identities in CAPS. The other two are worth knowing because they turn a hard proof into a one-line substitution.

↻ Reduction Formulae & the CAST Diagram

SQ2 · sin +

AQ1 · all +

TQ3 · tan +

CQ4 · cos +

	$180^\circ - \theta$	$180^\circ + \theta$	$360^\circ - \theta$	$-\theta$	$360^\circ + \theta$
$\sin$	$\sin\theta$	$-\sin\theta$	$-\sin\theta$	$-\sin\theta$	$\sin\theta$
$\cos$	$-\cos\theta$	$-\cos\theta$	$\cos\theta$	$\cos\theta$	$\cos\theta$
$\tan$	$-\tan\theta$	$\tan\theta$	$-\tan\theta$	$-\tan\theta$	$\tan\theta$

The ratio never changes for $180^\circ \pm \theta$ and $360^\circ \pm \theta$ — only the sign does, and the sign comes from the quadrant the angle lands in. Work out the quadrant first, then read the sign off CAST, then write the ratio down unchanged.

⇄ Co-functions (90°)

$90^\circ - \theta$ $\sin(90^\circ - \theta) = \cos\theta$

$\cos(90^\circ - \theta) = \sin\theta$

$90^\circ + \theta$ $\sin(90^\circ + \theta) = \cos\theta$

$\cos(90^\circ + \theta) = -\sin\theta$

Odd multiples of $90^\circ$ swap sine and cosine. Even multiples ($180^\circ$, $360^\circ$) leave the ratio alone. That single sentence replaces memorising the table above.

✦ Special Angles

$\theta$	$0^\circ$	$30^\circ$	$45^\circ$	$60^\circ$	$90^\circ$
$\sin\theta$	$0$	$\tfrac{1}{2}$	$\tfrac{\sqrt{2}}{2}$	$\tfrac{\sqrt{3}}{2}$	$1$
$\cos\theta$	$1$	$\tfrac{\sqrt{3}}{2}$	$\tfrac{\sqrt{2}}{2}$	$\tfrac{1}{2}$	$0$
$\tan\theta$	$0$	$\tfrac{\sqrt{3}}{3}$	$1$	$\sqrt{3}$	undefined

Read $\sin$ left to right as $\tfrac{\sqrt{0}}{2}, \tfrac{\sqrt{1}}{2}, \tfrac{\sqrt{2}}{2}, \tfrac{\sqrt{3}}{2}, \tfrac{\sqrt{4}}{2}$ and $\cos$ as the same row backwards. Then there is nothing to memorise.

± Compound & Double Angles

$\sin(A+B)$ $\sin A\cos B + \cos A\sin B$

$\sin(A-B)$ $\sin A\cos B - \cos A\sin B$

$\cos(A+B)$ $\cos A\cos B - \sin A\sin B$

$\cos(A-B)$ $\cos A\cos B + \sin A\sin B$

$\sin 2A$ $2\sin A\cos A$

$\cos 2A$ $\cos^{2}A - \sin^{2}A$

or $2\cos^{2}A - 1$

or $1 - 2\sin^{2}A$

The three forms of $\cos 2A$ exist so you can choose the one that matches what the rest of the expression already contains. If the question is full of $\sin$, take $1 - 2\sin^{2}A$ — picking the right one is usually the entire difficulty of the proof. Note the signs are the opposite way round for $\cos$ than for $\sin$.

◺ Sine, Cosine & Area Rules

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

Cosine rule $a^{2} = b^{2} + c^{2} - 2bc\cos A$

rearranged $\cos A = \dfrac{b^{2}+c^{2}-a^{2}}{2bc}$

Area rule $\text{Area} = \tfrac{1}{2}ab\sin C$

Use sine rule two angles and any side, or two sides and a non-included angle

Use cosine rule three sides, or two sides and the included angle

Use area rule two sides and the angle between them

Each rule pairs a side with the angle opposite it. If the angle you have is not opposite the side you want, you are in the cosine rule.

∀ General Solutions

$\sin\theta = k$ $\theta = \sin^{-1}k + n\cdot360^\circ$

or $\theta = 180^\circ - \sin^{-1}k + n\cdot360^\circ$

$\cos\theta = k$ $\theta = \pm\cos^{-1}k + n\cdot360^\circ$

$\tan\theta = k$ $\theta = \tan^{-1}k + n\cdot180^\circ$

in every case $n \in \mathbb{Z}$

$\tan$ repeats every $180^\circ$, not $360^\circ$, which is why it has one branch instead of two. Writing "$n \in \mathbb{Z}$" is worth a mark on its own — leave it off and the solution is incomplete.

∿ Graphs of $y = a\,f(bx + c) + q$

Graph	Amplitude	Period	Range (for $y=\sin x$ form)	Asymptotes
$y = a\sin bx + q$	$\|a\|$	$\dfrac{360^\circ}{\|b\|}$	$[\,q-\|a\|,\ q+\|a\|\,]$	none
$y = a\cos bx + q$	$\|a\|$	$\dfrac{360^\circ}{\|b\|}$	$[\,q-\|a\|,\ q+\|a\|\,]$	none
$y = a\tan bx + q$	none	$\dfrac{180^\circ}{\|b\|}$	$y \in \mathbb{R}$	every $\dfrac{180^\circ}{\|b\|}$

$a$ stretches vertically, $b$ compresses horizontally, $q$ shifts up or down, and $c$ shifts left or right. The tangent graph has no amplitude because it has no maximum — asking for "the amplitude of $\tan$" is a trick.

Most trig marks are lost to a sign, not to a forgotten identity.

The reduction formulae are easy to look up and easy to apply to the wrong quadrant. I tutor Grade 10–12 maths one-to-one in Gqeberha, at your own table, in English or Afrikaans. The first consultation is free.

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An independent revision aid aligned to the CAPS/NSC Mathematics syllabus. Not an official Department of Basic Education document.