A-level · Trigonometry

Trigonometric identitiesExam-style questions

The two AS identities — tan ≡ sin/cos and sin²+cos² ≡ 1 — where they come from, and how to use them to simplify expressions and prove "show that" results.

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8 questions · A-level level · mark scheme and worked solution on each · no sign-up to try

DifficultyAll Level 3 Level 4

1
In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.
(a) Show that

$(\tan\theta + \sin\theta)(1 - \cos\theta) \equiv \sin^2\theta\tan\theta$

(4 marks)
(b) Hence solve, for $0 < x < 2\pi$ , the equation

$(\tan x + \sin x)(1 - \cos x) = \tan x - 2\sin x\tan x$

giving your answers to 3 significant figures.

(4 marks)
Level 4 · 8 marksTry it
2
(a) Show that the equation

$6\cos^2\theta + \sin\theta - 5 = 0$

can be written as $6\sin^2\theta - \sin\theta - 1 = 0$ . (2 marks)
(b) Hence solve $6\cos^2\theta + \sin\theta - 5 = 0$ for $0^\circ \le \theta < 360^\circ$ , giving your answers to 1 decimal place where appropriate. (5 marks)
Level 4 · 7 marksTry it
3
Given that $90° < \theta < 180°$ and

$(\sin\theta + 3\cos\theta)^2 = 4 + 6\sin\theta\cos\theta$

find the exact value of $\sin\theta$

Fully justify your answer.

(6 marks)
Level 4 · 6 marksTry it
4
(a) Show that the equation $3\tan^{2}x - \dfrac{1}{\cos x} = 7$ can be expressed as $10\cos^{2}x + \cos x - 3 = 0$ .

(3 marks)
(b) Hence solve the equation $3\tan^{2}x - \dfrac{1}{\cos x} = 7$ for $0^{\circ} \leqslant x \leqslant 360^{\circ}$ .

(3 marks)
Level 4 · 6 marksTry it
5
(a) Prove the identity $\dfrac{\cos x}{1 - \sin x} - \tan x \equiv \dfrac{1}{\cos x}$ .

(3 marks)
(b) Hence solve the equation $\dfrac{\cos 2x}{1 - \sin 2x} - \tan 2x = 4$ for $0^{\circ} \leqslant x \leqslant 180^{\circ}$ .

(3 marks)
Level 4 · 6 marksTry it

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