IGCSE · Practice questions · Exact trigonometric values

No calculator, no problem.

Extended. Six original Cambridge-style non-calculator questions on the exact values of sine, cosine and tangent.

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

The method

Sketch the special triangle, then read off the ratio.

\(45^\circ\): half a square, sides \(1, 1, \sqrt{2}\).
\(30^\circ\) and \(60^\circ\): half an equilateral triangle, sides \(1, \sqrt{3}, 2\).
\(0^\circ\) and \(90^\circ\): sin goes \(0 \to 1\), cos goes \(1 \to 0\).
Squares: \(\sin^2 x = (\sin x)^2\); square top and bottom.
Surds: \(\sqrt{3} \times \sqrt{3} = 3\); rationalise if asked.

Recall

[3 marks]

Write down the exact value of (a) \(\sin 45^\circ\) (b) \(\cos 60^\circ\) (c) \(\tan 30^\circ\).

(a) \(\dfrac{1}{\sqrt{2}} = \dfrac{\sqrt{2}}{2}\) ✓
(b) \(\dfrac{1}{2}\) ✓
(c) \(\dfrac{1}{\sqrt{3}} = \dfrac{\sqrt{3}}{3}\) ✓

Proof

[2 marks]

Without a calculator, show that \(\sin^2 60^\circ + \cos^2 60^\circ = 1\).

\(\sin^2 60^\circ = \left(\dfrac{\sqrt{3}}{2}\right)^2 = \dfrac{3}{4}\) and \(\cos^2 60^\circ = \left(\dfrac{1}{2}\right)^2 = \dfrac{1}{4}\). ✓
\(\dfrac{3}{4} + \dfrac{1}{4} = 1\), as required. ✓

Calculation

[2 marks]

Without a calculator, find the exact value of \(4\sin 30^\circ + \sqrt{3}\tan 60^\circ\).

\(4 \times \dfrac{1}{2} = 2\) and \(\sqrt{3} \times \sqrt{3} = 3\). ✓
\(2 + 3 = 5\) ✓

Calculation

[4 marks]

In triangle \(ABC\), angle \(B = 90^\circ\), angle \(A = 30^\circ\) and \(AC = 12\) cm.

Without a calculator, find (a) \(BC\) (b) \(AB\), giving your answer in surd form (c) the exact area of the triangle.

(a) \(BC = 12\sin 30^\circ = 6\) cm ✓
(b) \(AB = 12\cos 30^\circ = 12 \times \dfrac{\sqrt{3}}{2} = 6\sqrt{3}\) cm ✓
(c) Area \(= \tfrac{1}{2} \times 6 \times 6\sqrt{3} = 18\sqrt{3}\) cm² ✓

Calculation

[2 marks]

Without a calculator, find the acute angle \(x\) in each case. (a) \(2\sin x = 1\) (b) \(\sqrt{3}\tan x = 3\)

(a) \(\sin x = \dfrac{1}{2}\), so \(x = 30^\circ\). ✓
(b) \(\tan x = \dfrac{3}{\sqrt{3}} = \sqrt{3}\), so \(x = 60^\circ\). ✓

Analysis

[4 marks]

A regular hexagon is made of six equilateral triangles, each of side 10 cm.

Without a calculator, find (a) the exact height of one triangle (b) the exact area of the hexagon.

(a) Each angle of an equilateral triangle is \(60^\circ\), so height \(= 10\sin 60^\circ = 10 \times \dfrac{\sqrt{3}}{2} = 5\sqrt{3}\) cm. ✓
(b) Area of one triangle \(= \tfrac{1}{2} \times 10 \times 5\sqrt{3} = 25\sqrt{3}\) cm². ✓
Hexagon \(= 6 \times 25\sqrt{3} = 150\sqrt{3}\) cm². ✓

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.