Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 5 - Trigonometric Identities - Section 5.4 Sum and Difference Identities for Sine and Tangent - 5.4 Exercises - Page 220: 9

Answer

$$\sin\frac{5\pi}{12}=\frac{\sqrt2+\sqrt6}{4}$$

Work Step by Step

*STRATEGY: In this type of exercise, we would try to rewrite the angle in terms of already familiar angles, including $\frac{\pi}{6}$, $\frac{\pi}{4}$ and $\frac{\pi}{3}$ $$\sin\frac{5\pi}{12}$$ We try separating $5\pi$ into $2\pi$ and $3\pi$. In detail, $$\sin\frac{5\pi}{12}=\sin\Big(\frac{2\pi+3\pi}{12}\Big)=\sin\Big(\frac{2\pi}{12}+\frac{3\pi}{12}\Big)=\sin\Big(\frac{\pi}{6}+\frac{\pi}{4}\Big)$$ Now we use the sine sum identity: $$\sin(A+B)=\sin A\cos B+\cos A\sin B$$ Therefore, $$\sin\frac{5\pi}{12}=\sin\frac{\pi}{6}\cos\frac{\pi}{4}+\cos\frac{\pi}{6}\sin\frac{\pi}{4}$$ $$\sin\frac{5\pi}{12}=\frac{1}{2}\frac{\sqrt2}{2}+\frac{\sqrt3}{2}\frac{\sqrt2}{2}$$ $$\sin\frac{5\pi}{12}=\frac{\sqrt2}{4}+\frac{\sqrt6}{4}$$ $$\sin\frac{5\pi}{12}=\frac{\sqrt2+\sqrt6}{4}$$
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