Trigonometry Toolkit

A2 · P3 · 2 min

P3 trigonometry adds three reciprocal functions and a dozen identities to the two you learnt in P1. The difficulty is rarely the algebra; it is choosing which identity to use. This note collects everything in one place and gives a strategy for picking the right tool.

Every identity you need

Key result

Definitions

\sec\theta = \frac{1}{\cos\theta}, \qquad \operatorname{cosec}\theta = \frac{1}{\sin\theta}, \qquad \cot\theta = \frac{1}{\tan\theta} = \frac{\cos\theta}{\sin\theta}

Pythagorean

\sin^2\theta + \cos^2\theta \equiv 1, \qquad 1 + \tan^2\theta \equiv \sec^2\theta, \qquad 1 + \cot^2\theta \equiv \operatorname{cosec}^2\theta

Compound angles

\sin(A \pm B) \equiv \sin A\cos B \pm \cos A\sin B, \qquad \cos(A \pm B) \equiv \cos A\cos B \mp \sin A\sin B, \qquad \tan(A \pm B) \equiv \frac{\tan A \pm \tan B}{1 \mp \tan A\tan B}

Double angles

\sin 2A \equiv 2\sin A\cos A, \qquad \cos 2A \equiv \cos^2 A - \sin^2 A \equiv 2\cos^2 A - 1 \equiv 1 - 2\sin^2 A, \qquad \tan 2A \equiv \frac{2\tan A}{1 - \tan^2 A}

Harmonic form

a\sin\theta \pm b\cos\theta \equiv R\sin(\theta \pm \alpha), \qquad a\cos\theta \pm b\sin\theta \equiv R\cos(\theta \mp \alpha), \qquad R = \sqrt{a^2 + b^2},\ \tan\alpha = \frac{b}{a}

Choosing an identity

The expression contains	Reach for
$\sec$ , $\operatorname{cosec}$ , $\cot$	rewrite in terms of $\sin$ , $\cos$ , $\tan$ ; or the matching Pythagorean identity
$\tan^2$ and $\sec^2$ together, or $\sec$ and $\tan$	$1 + \tan^2 = \sec^2$
an angle and its double ( $\theta$ and $2\theta$ )	double-angle formulae; pick the $\cos 2A$ version that leaves one function
a sum or difference of angles, or $\theta \pm 30^\circ$ etc.	compound-angle formulae
$a\sin\theta + b\cos\theta$ with a constant on the other side	$R$ -form
$\sin^2$ or $\cos^2$ to integrate	$\cos 2A$ rearranged: $\sin^2 A = \tfrac{1}{2}(1 - \cos 2A)$

Simplifying a compound expression

Simplify $\cos(x - 30^\circ) - \sqrt{3}\sin(x - 60^\circ)$ .

Solution

$\cos(x - 30^\circ) = \cos x\cos 30^\circ + \sin x\sin 30^\circ = \tfrac{\sqrt{3}}{2}\cos x + \tfrac{1}{2}\sin x$ .

$\sqrt{3}\sin(x - 60^\circ) = \sqrt{3}\left(\tfrac{1}{2}\sin x - \tfrac{\sqrt{3}}{2}\cos x\right) = \tfrac{\sqrt{3}}{2}\sin x - \tfrac{3}{2}\cos x$ .

Subtracting: $\left(\tfrac{\sqrt{3}}{2} + \tfrac{3}{2}\right)\cos x + \left(\tfrac{1}{2} - \tfrac{\sqrt{3}}{2}\right)\sin x$ .

This is already simplified; it can be written as $\tfrac{\sqrt{3} + 3}{2}\cos x - \tfrac{\sqrt{3} - 1}{2}\sin x$ , or converted to $R$ -form if a single function is wanted.

Mixed functions in one equation

Solve $\tan\theta + \cot\theta = 4$ for $0^\circ < \theta < 180^\circ$ .

Solution

$\tan\theta + \dfrac{1}{\tan\theta} = 4 \Rightarrow \tan^2\theta - 4\tan\theta + 1 = 0 \Rightarrow \tan\theta = 2 \pm \sqrt{3}$ .

$\tan\theta = 3.732 \Rightarrow \theta = 75^\circ$ ; $\tan\theta = 0.268 \Rightarrow \theta = 15^\circ$ .

(Alternatively, $\tan\theta + \cot\theta = \dfrac{1}{\sin\theta\cos\theta} = \dfrac{2}{\sin 2\theta}$ , so $\sin 2\theta = \tfrac{1}{2}$ , giving $2\theta = 30^\circ, 150^\circ$ .)

Sec and tan together

Solve $\sec^2\theta = 2\tan\theta + 4$ for $0^\circ \leq \theta \leq 360^\circ$ .

Solution

$1 + \tan^2\theta = 2\tan\theta + 4 \Rightarrow \tan^2\theta - 2\tan\theta - 3 = 0 \Rightarrow (\tan\theta - 3)(\tan\theta + 1) = 0$ .

$\tan\theta = 3$ : $\theta = 71.6^\circ, 251.6^\circ$ . $\tan\theta = -1$ : $\theta = 135^\circ, 315^\circ$ .

Strategy for proving identities

Start from the more complicated side.
Convert everything to $\sin$ and $\cos$ if no other identity is obvious.
Combine fractions over a common denominator.
Look for $\sin^2 + \cos^2$ or a difference of two squares.
Do not cross-multiply as if solving; keep the sides separate until they match.

A proof through sin and cos

Prove that $\dfrac{\sec\theta - \cos\theta}{\tan\theta} \equiv \sin\theta$ .

Solution

\frac{\frac{1}{\cos\theta} - \cos\theta}{\frac{\sin\theta}{\cos\theta}} = \frac{\frac{1 - \cos^2\theta}{\cos\theta}}{\frac{\sin\theta}{\cos\theta}} = \frac{1 - \cos^2\theta}{\sin\theta} = \frac{\sin^2\theta}{\sin\theta} = \sin\theta.

Exam tip

Exam questions in this topic almost always come in two parts: "prove the identity" then "hence solve". The "hence" means the equation is the identity in disguise; substitute and you get a simple equation in one function.

Practice

Question

Simplify $\dfrac{\sin 2\theta}{1 + \cos 2\theta}$ .
Solve $\operatorname{cosec}^2\theta = 3\cot\theta - 1$ for $0^\circ < \theta < 360^\circ$ .
Express $\cos\theta + \sqrt{3}\sin\theta$ in the form $R\cos(\theta - \alpha)$ .
Prove that $\dfrac{1}{1 - \sin\theta} - \dfrac{1}{1 + \sin\theta} \equiv 2\tan\theta\sec\theta$ .

Answers

$\dfrac{2\sin\theta\cos\theta}{2\cos^2\theta} = \tan\theta$ .
$1 + \cot^2\theta = 3\cot\theta - 1 \Rightarrow \cot^2\theta - 3\cot\theta + 2 = 0 \Rightarrow \cot\theta = 1, 2$ : $\theta = 45^\circ, 225^\circ, 26.6^\circ, 206.6^\circ$ .
$2\cos(\theta - 60^\circ)$ .
LHS $= \dfrac{(1 + \sin\theta) - (1 - \sin\theta)}{1 - \sin^2\theta} = \dfrac{2\sin\theta}{\cos^2\theta} = 2\tan\theta\sec\theta$ .

Every identity you need

Choosing an identity

Strategy for proving identities

Practice

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