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: 14

Answer

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

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 $\frac{\pi}{3}$. $$\sin\frac{\pi}{12}$$ Now, we try writing $\pi$ as the difference of $4\pi$ and $3\pi$. $$\sin\frac{\pi}{12}=\sin \Big(\frac{4\pi-3\pi}{12}\Big)=\sin\Big(\frac{4\pi}{12}-\frac{3\pi}{12}\Big)=\sin\Big(\frac{\pi}{3}-\frac{\pi}{4}\Big)$$ We then use the sine difference identity: $$\sin(A-B)=\sin A\cos B-\cos A\sin B$$ Therefore, $$\sin\frac{\pi}{12}=\sin\frac{\pi}{3}\cos\frac{\pi}{4}-\cos\frac{\pi}{3}\sin\frac{\pi}{4}$$ $$\sin\frac{\pi}{12}=\frac{\sqrt3}{2}\frac{\sqrt2}{2}-\frac{1}{2}\frac{\sqrt2}{2}$$ $$\sin\frac{\pi}{12}=\frac{\sqrt6}{4}-\frac{\sqrt2}{4}$$ $$\sin\frac{\pi}{12}=\frac{\sqrt6-\sqrt2}{4}$$
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