5.8 Addition and double-angle formulae — Practise

Work through these on your own. Write each unfamiliar angle as a sum or a difference of special angles. Give exact values, and choose the form of $\cos 2A$ that fits the information.

Both the questions and the answers below come from a set that is checked by computer every time the book is built.

Find the exact value of $\sin 75^\circ$ .
Find the exact value of $\cos 15^\circ$ .
Find the exact value of $\tan 75^\circ$ .
Given that $\sin A = \dfrac{5}{13}$ and $A$ is acute, find $\sin 2A$ .
Given that $\sin A = \dfrac{5}{13}$ and $A$ is acute, find $\cos 2A$ .
Given that $\tan A = \dfrac{1}{2}$ , find $\tan 2A$ .
Find the exact value of $\sin 105^\circ$ .
Given that $\cos A = \dfrac{3}{5}$ , find $\cos 2A$ .
Show that $\sin(A + B) + \sin(A - B) \equiv 2\sin A\cos B$ .
Show that $\dfrac{\sin 2A}{2\sin A} \equiv \cos A$ for $\sin A \neq 0$ .

Answers and concise solutions

1. The value

Answer: Write $75^\circ = 45^\circ + 30^\circ$ . Then $\sin 75^\circ = \dfrac{\sqrt{2}}{2} \times \dfrac{\sqrt{3}}{2} + \dfrac{\sqrt{2}}{2} \times \dfrac{1}{2} = \dfrac{\sqrt{6} + \sqrt{2}}{4}$ .

2. The value

Answer: Write $15^\circ = 45^\circ - 30^\circ$ . Then $\cos 15^\circ = \dfrac{\sqrt{2}}{2} \times \dfrac{\sqrt{3}}{2} + \dfrac{\sqrt{2}}{2} \times \dfrac{1}{2} = \dfrac{\sqrt{6} + \sqrt{2}}{4}$ .

3. The value

Answer: Write $75^\circ = 45^\circ + 30^\circ$ . Then $\tan 75^\circ = \dfrac{1 + \frac{\sqrt{3}}{3}}{1 - \frac{\sqrt{3}}{3}} = \dfrac{3 + \sqrt{3}}{3 - \sqrt{3}}$ . Rationalising gives $2 + \sqrt{3}$ .

4. The value

Answer: From $\sin^2 A + \cos^2 A = 1$ and the acute condition, $\cos A = \dfrac{12}{13}$ . Then $\sin 2A = 2 \times \dfrac{5}{13} \times \dfrac{12}{13} = \dfrac{120}{169}$ .

5. The value

Answer: Use the sine form, which needs only $\sin A$ : $\cos 2A = 1 - 2\sin^2 A = 1 - \dfrac{50}{169} = \dfrac{119}{169}$ .

6. The value

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

7. The value

Answer: Write $105^\circ = 60^\circ + 45^\circ$ . Then $\sin 105^\circ = \dfrac{\sqrt{3}}{2} \times \dfrac{\sqrt{2}}{2} + \dfrac{1}{2} \times \dfrac{\sqrt{2}}{2} = \dfrac{\sqrt{6} + \sqrt{2}}{4}$ . This is the value the chapter problem needs for the angle at $L$ .

8. The value

Answer: Use the cosine form, which needs only $\cos A$ : $\cos 2A = 2\cos^2 A - 1 = 2 \times \dfrac{9}{25} - 1 = \dfrac{18}{25} - \dfrac{25}{25} = -\dfrac{7}{25}$ .

9. The identity

Answer: Add the two addition formulae: $\sin(A + B) + \sin(A - B) = (\sin A\cos B + \cos A\sin B) + (\sin A\cos B - \cos A\sin B)$ . The $\cos A\sin B$ terms cancel, leaving $2\sin A\cos B$ .

10. The identity

Answer: Replace $\sin 2A$ with $2\sin A\cos A$ : $\dfrac{2\sin A\cos A}{2\sin A} = \cos A$ , provided $\sin A \neq 0$ .