Trigonometry (10th Edition)

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

Chapter 5 - Trigonometric Identities - Section 5.3 Sum and Difference Identities for Cosine - 5.3 Exercises - Page 214: 74b

Answer

$$\cos\frac{11\pi}{12}=\frac{-\sqrt6-\sqrt2}{4}$$

Work Step by Step

$$\cos\frac{11\pi}{12}$$ 1) We can write $\frac{11\pi}{12}$ as the difference of $\pi$ and $\frac{\pi}{12}$, since $$\pi-\frac{\pi}{12}=\frac{12\pi}{12}-\frac{\pi}{12}=\frac{11\pi}{12}$$ So, $$\cos\frac{11\pi}{12}=\cos(\pi-\frac{\pi}{12})$$ Now we expand $\cos(\pi-\frac{\pi}{12})$ according to the cosine difference identity: $$\cos\frac{11\pi}{12}=\cos\pi\cos\frac{\pi}{12}+\sin\pi\sin\frac{\pi}{12}$$ $$\cos\frac{11\pi}{12}=(-1)\times\cos\frac{\pi}{12}+0\times\sin\frac{\pi}{12}$$ $$\cos\frac{11\pi}{12}=-\cos\frac{\pi}{12}$$ So now the job is to calculate $-\cos\frac{\pi}{12}$ 2) We can write $\frac{\pi}{12}$ as the difference of $\frac{\pi}{4}$ and $\frac{\pi}{6}$, since $$\frac{\pi}{4}-\frac{\pi}{6}=\frac{3\pi}{12}-\frac{2\pi}{12}=\frac{\pi}{12}$$ So, $$-\cos\frac{\pi}{12}=-\cos(\frac{\pi}{4}-\frac{\pi}{6})$$ Now we expand $\cos(\frac{\pi}{4}-\frac{\pi}{6})$ according to the cosine difference identity: $$-\cos\frac{\pi}{12}=-(\cos\frac{\pi}{4}\cos\frac{\pi}{6}+\sin\frac{\pi}{4}\sin\frac{\pi}{6})$$ $$-\cos\frac{\pi}{12}=-\Big[\Big(\frac{\sqrt2}{2}\Big)\Big(\frac{\sqrt3}{2}\Big)+\Big(\frac{\sqrt2}{2}\Big)\Big(\frac{1}{2}\Big)\Big]$$ $$-\cos\frac{\pi}{12}=-\Big(\frac{\sqrt6}{4}+\frac{\sqrt2}{4}\Big)$$ $$-\cos\frac{\pi}{12}=-\Big(\frac{\sqrt6+\sqrt2}{4}\Big)$$ $$-\cos\frac{\pi}{12}=\frac{-\sqrt6-\sqrt2}{4}$$ 3) Therefore, $$\cos\frac{11\pi}{12}=-\cos\frac{\pi}{12}=\frac{-\sqrt6-\sqrt2}{4}$$
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