5.6 Secant, cosecant, cotangent and the inverse trigonometric functions — Practise with help

Write each new function as a fraction before you calculate. Then the reciprocal and the domain are both visible.

Warm-up: remember

Work out the reciprocal of $\dfrac{4}{5}$ and of $\dfrac{3}{4}$ .
Use $\sin^2\theta + \cos^2\theta = 1$ to find $\sin\theta$ when $\cos\theta = \dfrac{5}{13}$ and $\theta$ is acute.
Write down the value of $\cos\dfrac{\pi}{3}$ .

Answers

The reciprocals are $\dfrac{5}{4}$ and $\dfrac{4}{3}$ .
$\sin^2\theta = 1 - \dfrac{25}{169} = \dfrac{144}{169}$ , so $\sin\theta = \dfrac{12}{13}$ .
$\cos\dfrac{\pi}{3} = \dfrac{1}{2}$ .

If question 1 was hard, revise reciprocals before you continue.

1. Write down the value of $\sec\dfrac{\pi}{3}$ .

Hint

Secant is the reciprocal of cosine.

Next step

$\cos\dfrac{\pi}{3} = \dfrac{1}{2}$ , so take its reciprocal.

Full working

\sec\frac{\pi}{3} = \frac{1}{\cos\frac{\pi}{3}} = \frac{1}{\frac{1}{2}} = 2.

2. Write down the value of $\operatorname{cosec}\dfrac{\pi}{6}$ .

Hint

Cosecant is the reciprocal of sine.

Next step

$\sin\dfrac{\pi}{6} = \dfrac{1}{2}$ , so take its reciprocal.

Full working

\operatorname{cosec}\frac{\pi}{6} = \frac{1}{\sin\frac{\pi}{6}} = \frac{1}{\frac{1}{2}} = 2.

3. Write down the value of $\cot\dfrac{\pi}{4}$ .

Hint

$\cot\theta = \dfrac{\cos\theta}{\sin\theta}$ .

Next step

Both $\cos\dfrac{\pi}{4}$ and $\sin\dfrac{\pi}{4}$ equal $\dfrac{\sqrt{2}}{2}$ . Divide them.

Full working

\cot\frac{\pi}{4} = \frac{\cos\frac{\pi}{4}}{\sin\frac{\pi}{4}} = \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = 1.

4. Simplify $\operatorname{cosec}^2\theta - \cot^2\theta$ .

Hint

Use the identity $1 + \cot^2\theta = \operatorname{cosec}^2\theta$ .

Next step

Replace $\operatorname{cosec}^2\theta$ with $1 + \cot^2\theta$ , then subtract $\cot^2\theta$ .

Full working

\operatorname{cosec}^2\theta - \cot^2\theta = (1 + \cot^2\theta) - \cot^2\theta = 1.

5. Given that $\cos\theta = \dfrac{5}{13}$ and $\theta$ is acute, find $\sec\theta$ and $\operatorname{cosec}\theta$ .

Hint

Secant is immediate. For cosecant, find sine first with $\sin^2\theta + \cos^2\theta = 1$ .

Next step

$\sec\theta = \dfrac{13}{5}$ . Also $\sin^2\theta = 1 - \dfrac{25}{169} = \dfrac{144}{169}$ , so $\sin\theta = \dfrac{12}{13}$ .

Full working

\sec\theta = \frac{1}{\cos\theta} = \frac{13}{5}.

\operatorname{cosec}\theta = \frac{1}{\sin\theta} = \frac{13}{12}.

6. Work out $\arcsin\dfrac{\sqrt{3}}{2}$ in radians, giving the principal value.

Hint

Ask which angle in the principal range has a sine of $\dfrac{\sqrt{3}}{2}$ .

Next step

The principal range of arcsine is $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$ . Find the angle in that range whose sine is $\dfrac{\sqrt{3}}{2}$ .

Full working

\arcsin\frac{\sqrt{3}}{2} = \frac{\pi}{3}.

This is $60^\circ$ , and it lies inside the principal range.