AS & A Level · Practice questions · Exact values and the two identities

Exactly right.

Six original Cambridge-style questions on exact trigonometric values and the identities \(\tan\theta \equiv \dfrac{\sin\theta}{\cos\theta}\) and \(\sin^2\theta + \cos^2\theta \equiv 1\).

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Acute angle, quadrant sign, then the identities.

Exact values: rebuild them from the two special triangles.
Related angles: acute angle to the \(x\)-axis, then the sign of that quadrant.
One ratio to another: \(\sin^2\theta + \cos^2\theta = 1\), then choose the sign from the quadrant.
Proofs: start from one side, combine fractions, replace \(1 - \cos^2\theta\) by \(\sin^2\theta\) (or the reverse), finish on the other side.
Exact answers: rationalise surd denominators when asked.

Recall

[3 marks]

Without a calculator, find the exact values of (a) \(\sin 240^\circ\) (b) \(\cos\dfrac{7\pi}{4}\) (c) \(\tan\dfrac{5\pi}{6}\).

(a) \(240^\circ = 180^\circ + 60^\circ\), third quadrant: \(-\sin 60^\circ = -\dfrac{\sqrt{3}}{2}\) ✓
(b) \(2\pi - \dfrac{\pi}{4}\), fourth quadrant: \(\cos\dfrac{\pi}{4} = \dfrac{\sqrt{2}}{2}\) ✓
(c) \(\pi - \dfrac{\pi}{6}\), second quadrant: \(-\tan\dfrac{\pi}{6} = -\dfrac{1}{\sqrt{3}} = -\dfrac{\sqrt{3}}{3}\) ✓

Calculation

[2 marks]

Without a calculator, find the exact value of \(3\tan^2\dfrac{\pi}{6} + 4\sin\dfrac{3\pi}{4}\cos\dfrac{\pi}{4}\).

\(3 \times \left(\dfrac{1}{\sqrt{3}}\right)^2 = 3 \times \dfrac{1}{3} = 1\) ✓
\(\sin\dfrac{3\pi}{4} = \dfrac{\sqrt{2}}{2}\), so \(4 \times \dfrac{\sqrt{2}}{2} \times \dfrac{\sqrt{2}}{2} = 4 \times \dfrac{1}{2} = 2\) ✓
Total \(= 1 + 2 = 3\) ✓

Calculation

[3 marks]

Given that \(\cos\theta = -\dfrac{3}{4}\) and \(180^\circ < \theta < 270^\circ\), find the exact values of \(\sin\theta\) and \(\tan\theta\).

\(\sin^2\theta = 1 - \dfrac{9}{16} = \dfrac{7}{16}\), so \(\sin\theta = \pm\dfrac{\sqrt{7}}{4}\). ✓
Third quadrant: sine is negative, so \(\sin\theta = -\dfrac{\sqrt{7}}{4}\). ✓
\(\tan\theta = \dfrac{-\sqrt{7}/4}{-3/4} = \dfrac{\sqrt{7}}{3}\) (positive, as expected in the third quadrant). ✓

Proof

[4 marks]

Prove the identity \(\dfrac{1 + \sin\theta}{\cos\theta} + \dfrac{\cos\theta}{1 + \sin\theta} \equiv \dfrac{2}{\cos\theta}\).

LHS \(= \dfrac{(1 + \sin\theta)^2 + \cos^2\theta}{\cos\theta(1 + \sin\theta)}\) ✓
Numerator \(= 1 + 2\sin\theta + \sin^2\theta + \cos^2\theta = 2 + 2\sin\theta = 2(1 + \sin\theta)\) ✓
LHS \(= \dfrac{2(1 + \sin\theta)}{\cos\theta(1 + \sin\theta)} = \dfrac{2}{\cos\theta}\) = RHS ✓

Proof

[3 marks]

Prove the identity \(\tan\theta - \dfrac{1}{\tan\theta} \equiv \dfrac{2\sin^2\theta - 1}{\sin\theta\cos\theta}\).

LHS \(= \dfrac{\sin\theta}{\cos\theta} - \dfrac{\cos\theta}{\sin\theta} = \dfrac{\sin^2\theta - \cos^2\theta}{\sin\theta\cos\theta}\) ✓
Replace \(\cos^2\theta\) by \(1 - \sin^2\theta\): numerator \(= \sin^2\theta - 1 + \sin^2\theta = 2\sin^2\theta - 1\) ✓
So LHS \(= \dfrac{2\sin^2\theta - 1}{\sin\theta\cos\theta}\) = RHS ✓

Analysis

[6 marks]

A Ferris wheel has radius 20 m and its centre is 22 m above the ground. When a car has turned through an angle \(\theta\) anticlockwise from the horizontal line through the centre (on the right), its height is \(h = 22 + 20\sin\theta\) metres and its horizontal displacement to the right of the centre is \(x = 20\cos\theta\) metres.

(a) A car is 34 m above the ground on the left-hand side of the wheel. Find the exact values of \(\sin\theta\), \(\cos\theta\) and \(\tan\theta\), and the value of \(x\).

(b) Find the exact height of a car when \(\theta = \dfrac{7\pi}{6}\) and when \(\theta = \dfrac{5\pi}{4}\).

(a) \(22 + 20\sin\theta = 34\) gives \(\sin\theta = \dfrac{12}{20} = \dfrac{3}{5}\). ✓
Left-hand side means \(\theta\) is obtuse, so \(\cos\theta = -\sqrt{1 - \dfrac{9}{25}} = -\dfrac{4}{5}\) and \(\tan\theta = -\dfrac{3}{4}\). ✓
\(x = 20 \times \left(-\dfrac{4}{5}\right) = -16\): the car is 16 m to the left of the centre. ✓
(b) \(\sin\dfrac{7\pi}{6} = -\tfrac{1}{2}\), so \(h = 22 - 10 = 12\) m. ✓
\(\sin\dfrac{5\pi}{4} = -\dfrac{\sqrt{2}}{2}\), so \(h = 22 - 10\sqrt{2}\) m. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.