5.8 Addition and double-angle formulae — Practise with help

Write the angle as a sum or a difference of two special angles before you substitute. For a double angle, choose the form of the formula that uses the ratio you already have.

Warm-up: remember

Write down the exact values of $\sin 45^\circ$ , $\cos 45^\circ$ and $\sin 30^\circ$ .
Write down the exact value of $\cos 30^\circ$ .
Work out $1 - 2\left(\dfrac{1}{2}\right)^2$ .

Answers

$\sin 45^\circ = \dfrac{\sqrt{2}}{2}$ , $\cos 45^\circ = \dfrac{\sqrt{2}}{2}$ and $\sin 30^\circ = \dfrac{1}{2}$ .
$\cos 30^\circ = \dfrac{\sqrt{3}}{2}$ .
$1 - 2 \times \dfrac{1}{4} = 1 - \dfrac{1}{2} = \dfrac{1}{2}$ .

If question 2 was hard, revisit the exact values before you continue.

Practise with help

1. Use the addition formula to find the exact value of $\sin 75^\circ$ .

Hint

Write $75^\circ$ as $45^\circ + 30^\circ$ and use $\sin(A + B) = \sin A\cos B + \cos A\sin B$ .

Next step

Substitute the exact values: $\dfrac{\sqrt{2}}{2} \times \dfrac{\sqrt{3}}{2} + \dfrac{\sqrt{2}}{2} \times \dfrac{1}{2}$ .

Full working

\sin 75^\circ = \frac{\sqrt{2}}{2} \times \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \times \frac{1}{2} = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4}.

2. Use the addition formula to find the exact value of $\cos 15^\circ$ .

Hint

Write $15^\circ$ as $45^\circ - 30^\circ$ . The cosine formula has a plus sign for a difference.

Next step

$\cos 15^\circ = \cos 45^\circ\cos 30^\circ + \sin 45^\circ\sin 30^\circ$ .

Full working

\cos 15^\circ = \frac{\sqrt{2}}{2} \times \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \times \frac{1}{2} = \frac{\sqrt{6} + \sqrt{2}}{4}.

3. Use the addition formula to find the exact value of $\tan 75^\circ$ .

Hint

Write $75^\circ$ as $45^\circ + 30^\circ$ and use the tangent formula with $\tan 45^\circ = 1$ and $\tan 30^\circ = \dfrac{\sqrt{3}}{3}$ .

Next step

$\tan 75^\circ = \dfrac{1 + \frac{\sqrt{3}}{3}}{1 - 1 \times \frac{\sqrt{3}}{3}}$ . Simplify by multiplying top and bottom by $3$ .

Full working

\tan 75^\circ = \frac{1 + \frac{\sqrt{3}}{3}}{1 - \frac{\sqrt{3}}{3}} = \frac{3 + \sqrt{3}}{3 - \sqrt{3}}.

Rationalise the denominator:

= \frac{(3 + \sqrt{3})^2}{9 - 3} = \frac{9 + 6\sqrt{3} + 3}{6} = \frac{12 + 6\sqrt{3}}{6} = 2 + \sqrt{3}.

4. Given that $\sin A = \dfrac{5}{13}$ and $A$ is acute, find $\sin 2A$ .

Hint

Find $\cos A$ first, then use $\sin 2A = 2\sin A\cos A$ .

Next step

$\cos^2 A = 1 - \dfrac{25}{169} = \dfrac{144}{169}$ , so $\cos A = \dfrac{12}{13}$ . Then multiply.

Full working

\sin 2A = 2 \times \frac{5}{13} \times \frac{12}{13} = \frac{120}{169}.

5. For the same $A$ , find $\cos 2A$ using the sine form.

Hint

Use $\cos 2A = 1 - 2\sin^2 A$ , the form that needs only the sine.

Next step

Substitute $\sin A = \dfrac{5}{13}$ : $1 - 2 \times \dfrac{25}{169}$ .

Full working

\cos 2A = 1 - 2 \times \frac{25}{169} = 1 - \frac{50}{169} = \frac{119}{169}.

6. Given that $\tan A = \dfrac{1}{2}$ , find $\tan 2A$ .

Hint

Use $\tan 2A = \dfrac{2\tan A}{1 - \tan^2 A}$ .

Next step

$\tan 2A = \dfrac{1}{1 - \frac{1}{4}} = \dfrac{1}{\frac{3}{4}}$ .

Full working

\tan 2A = \frac{2 \times \frac{1}{2}}{1 - \frac{1}{4}} = \frac{1}{\frac{3}{4}} = \frac{4}{3}.