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: 73

Answer

$$\cos195^\circ=\frac{-\sqrt2-\sqrt6}{4}$$

Work Step by Step

In Exercise 71, we have proved that $$\cos195^\circ=-\cos15^\circ$$ In Exercise 72, we found that $$-\cos15^\circ=\frac{-\sqrt2-\sqrt6}{4}$$ Therefore, $$\cos195^\circ=\frac{-\sqrt2-\sqrt6}{4}$$
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