University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 1 - Section 1.3 - Trigonometric Functions - Exercises - Page 28: 44

Answer

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

Work Step by Step

As suggested by the exercise, we would rewrite: $$\frac{11\pi}{12}=\frac{\pi}{4}+\frac{2\pi}{3}$$ Therefore, $$\cos\frac{11\pi}{12}=\cos\Big(\frac{\pi}{4}+\frac{2\pi}{3}\Big)$$ Apply the addition formula for cosine here: $$\cos\frac{11\pi}{12}=\cos\frac{\pi}{4}\cos\frac{2\pi}{3}-\sin\frac{\pi}{4}\sin\frac{2\pi}{3}$$ Remember that $\sin\frac{\pi}{4}=\cos\frac{\pi}{4}=\frac{\sqrt2}{2}$ and $\sin\frac{2\pi}{3}=\frac{\sqrt3}{2}$ and $\cos\frac{2\pi}{3}=-\frac{1}{2}$ So, $$\cos\frac{11\pi}{12}=\frac{\sqrt2}{2}\times\Big(-\frac{1}{2}\Big)-\frac{\sqrt2}{2}\times\frac{\sqrt3}{2}$$ $$\cos\frac{11\pi}{12}=-\frac{\sqrt2}{4}-\frac{\sqrt6}{4}$$ $$\cos\frac{11\pi}{12}=\frac{-\sqrt2-\sqrt6}{4}$$
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