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
Show that $\tan^2 x - \sin^2 x \equiv \sin^2 x \tan^2 x$ .
Level 3 · 3 marksTry it
2
Show that $\dfrac{\sin\theta}{1 + \cos\theta} + \dfrac{1 + \cos\theta}{\sin\theta} \equiv \dfrac{2}{\sin\theta}$ .
Level 3 · 3 marksTry it
3
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
4
(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
5
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

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