Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 5 - Trigonometric Identities - Section 5.3 Sum and Difference Identities for Cosine - 5.3 Exercises - Page 219: 58


$$\cos(-24^\circ)=\cos16^\circ-\cos40^\circ$$ The statement is false.

Work Step by Step

$$\cos(-24^\circ)=\cos16^\circ-\cos40^\circ$$ We can rewrite $-24^\circ$ as follows, $$-24^\circ=16^\circ-40^\circ$$ That means $$\cos(-24^\circ)=\cos(16^\circ-40^\circ)$$ Using the cosine difference identity, which states $$\cos(A-B)=\cos A\cos B+\sin A\sin B$$ we can expand $\cos(16^\circ-40^\circ)$ $$\cos(-24^\circ)=\cos16^\circ\cos40^\circ+\sin16^\circ\sin40^\circ$$ And since $\cos16^\circ\cos40^\circ+\sin16^\circ\sin40^\circ\ne\cos16^\circ-\cos40^\circ$, that means $$\cos(-24^\circ)\ne\cos16^\circ-\cos40^\circ$$ The statement is false. Until now, it also should be clear to you that $\cos(16^\circ-40^\circ)\ne\cos16^\circ-\cos40^\circ$.
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