IGCSE · Practice questions · The six trigonometric functions

Size from the triangle, sign from CAST.

Six original Cambridge-style questions on the six trigonometric functions: sec, cosec and cot at the special angles, angles of any size with the reference angle and CAST, and finding every ratio from one given ratio, all without a calculator.

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

Quadrant, reference angle, sign, flip.

Reciprocals: \(\sec = \dfrac{1}{\cos}\), \(\operatorname{cosec} = \dfrac{1}{\sin}\), \(\cot = \dfrac{1}{\tan}\) (third letter tells the partner).
Bring the angle into range: add or subtract \(360^{\circ}\) or \(2\pi\).
Reference angle: the acute angle to the \(x\)-axis; its exact value gives the size.
Sign: CAST; each reciprocal has its partner's sign.
One ratio given: positive right-angled triangle, Pythagoras, then signs from the quadrant.

Recall

[3 marks]

Find the exact values of (a) \(\operatorname{cosec} 60^{\circ}\) (b) \(\sec\dfrac{\pi}{4}\) (c) \(\cot\dfrac{\pi}{6}\). Rationalise any denominators.

(a) \(\sin 60^{\circ} = \dfrac{\sqrt{3}}{2}\), so \(\operatorname{cosec} 60^{\circ} = \dfrac{2}{\sqrt{3}} = \dfrac{2\sqrt{3}}{3}\). ✓
(b) \(\cos\dfrac{\pi}{4} = \dfrac{1}{\sqrt{2}}\), so \(\sec\dfrac{\pi}{4} = \sqrt{2}\). ✓
(c) \(\tan\dfrac{\pi}{6} = \dfrac{1}{\sqrt{3}}\), so \(\cot\dfrac{\pi}{6} = \sqrt{3}\). ✓

Calculation

[4 marks]

Without a calculator, find the exact values of (a) \(\sin 330^{\circ}\) (b) \(\sec 150^{\circ}\) (c) \(\cot(-120^{\circ})\) (d) \(\operatorname{cosec}\dfrac{9\pi}{4}\).

(a) Fourth quadrant, reference \(30^{\circ}\), sin negative: \(-\dfrac{1}{2}\). ✓
(b) Second quadrant, reference \(30^{\circ}\), cos negative: \(\cos 150^{\circ} = -\dfrac{\sqrt{3}}{2}\), so \(\sec 150^{\circ} = -\dfrac{2\sqrt{3}}{3}\). ✓
(c) \(-120^{\circ} + 360^{\circ} = 240^{\circ}\), third quadrant, reference \(60^{\circ}\), tan positive: \(\cot(-120^{\circ}) = \dfrac{1}{\sqrt{3}} = \dfrac{\sqrt{3}}{3}\). ✓
(d) \(\dfrac{9\pi}{4} - 2\pi = \dfrac{\pi}{4}\), first quadrant: \(\operatorname{cosec}\dfrac{9\pi}{4} = \sqrt{2}\). ✓

Calculation

[4 marks]

Given that \(\cos\theta = -\dfrac{2}{3}\) and \(180^{\circ} < \theta < 270^{\circ}\), find the exact values of \(\tan\theta\) and \(\operatorname{cosec}\theta\).

Triangle: adjacent 2, hypotenuse 3, so the opposite side is \(\sqrt{9 - 4} = \sqrt{5}\). ✓
Third quadrant: tan positive, sin negative. ✓
\(\tan\theta = \dfrac{\sqrt{5}}{2}\). ✓
\(\sin\theta = -\dfrac{\sqrt{5}}{3}\), so \(\operatorname{cosec}\theta = -\dfrac{3}{\sqrt{5}} = -\dfrac{3\sqrt{5}}{5}\). ✓

Analysis

[5 marks]

The angle \(x\) satisfies \(\cot x = 2\) and \(\sin x < 0\), with \(0 < x < 2\pi\).

(a) State the quadrant in which \(x\) lies, giving a reason.

(b) Find the exact values of \(\sec x\) and \(\operatorname{cosec} x\).

(a) \(\cot x > 0\) so \(\tan x > 0\): first or third quadrant. \(\sin x < 0\) rules out the first, so \(x\) is in the third quadrant. ✓
(b) \(\tan x = \dfrac{1}{2}\): triangle opposite 1, adjacent 2, hypotenuse \(\sqrt{5}\). ✓
Third quadrant: \(\cos x = -\dfrac{2}{\sqrt{5}}\), so \(\sec x = -\dfrac{\sqrt{5}}{2}\); \(\sin x = -\dfrac{1}{\sqrt{5}}\), so \(\operatorname{cosec} x = -\sqrt{5}\). ✓
(c) \(\sec^{2}x = \dfrac{5}{4}\) and \(1 + \tan^{2}x = 1 + \dfrac{1}{4} = \dfrac{5}{4}\), so they are equal. ✓

Proof

[4 marks]

Show, without a calculator, that

\[\sec^{2}\frac{\pi}{6} + \operatorname{cosec}^{2}\frac{\pi}{6} = \sec^{2}\frac{\pi}{6}\,\operatorname{cosec}^{2}\frac{\pi}{6}.\]

\(\cos\dfrac{\pi}{6} = \dfrac{\sqrt{3}}{2}\), so \(\sec^{2}\dfrac{\pi}{6} = \dfrac{4}{3}\). ✓
\(\sin\dfrac{\pi}{6} = \dfrac{1}{2}\), so \(\operatorname{cosec}^{2}\dfrac{\pi}{6} = 4\). ✓
Left side: \(\dfrac{4}{3} + 4 = \dfrac{16}{3}\). Right side: \(\dfrac{4}{3} \times 4 = \dfrac{16}{3}\). ✓
Both sides equal \(\dfrac{16}{3}\), as required. ✓

Analysis

[6 marks]

The point \(P(-5, 12)\) is joined to the origin \(O\). The angle \(\theta\), with \(0^{\circ} < \theta < 360^{\circ}\), is measured anticlockwise from the positive \(x\)-axis to \(OP\).

(a) Find \(OP\) and state the quadrant of \(\theta\).

(b) Find the exact values of \(\sec\theta\), \(\operatorname{cosec}\theta\) and \(\cot\theta\).

(c) The angle \(\phi = \theta + 180^{\circ}\). Find \(\tan\phi\) and \(\sec\phi\), and explain why \(\tan\phi = \tan\theta\).

(a) \(OP = \sqrt{25 + 144} = 13\). \(P\) has \(x < 0\), \(y > 0\), so \(\theta\) is in the second quadrant. ✓
(b) \(\cos\theta = -\dfrac{5}{13}\), \(\sin\theta = \dfrac{12}{13}\), \(\tan\theta = -\dfrac{12}{5}\). ✓
\(\sec\theta = -\dfrac{13}{5}\), \(\operatorname{cosec}\theta = \dfrac{13}{12}\), \(\cot\theta = -\dfrac{5}{12}\). ✓
(c) Adding \(180^{\circ}\) takes \(P\) to \((5, -12)\), in the fourth quadrant: \(\cos\phi = \dfrac{5}{13}\), \(\sin\phi = -\dfrac{12}{13}\). ✓
\(\tan\phi = -\dfrac{12}{5}\) and \(\sec\phi = \dfrac{13}{5}\). ✓
Both sine and cosine change sign, so their quotient \(\tan\) is unchanged: \(\tan\phi = \tan\theta\). ✓

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.