## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 5 - Trigonometric Identities - Section 5.3 Sum and Difference Identities for Cosine - 5.3 Exercises - Page 219: 66

#### Answer

$$\sin\Big(x-\frac{\pi}{2}\Big)=\cos x$$ The statement is false.

#### Work Step by Step

$$\sin\Big(x-\frac{\pi}{2}\Big)=\cos x$$ Now what we've already known from the cofunction identities is $$\sin\Big(\frac{\pi}{2}-x\Big)=\cos x$$ Yet here, the statement involves $\sin\Big(x-\frac{\pi}{2}\Big)$, not $\sin\Big(\frac{\pi}{2}-x\Big)$. Unfortunately, 2 expressions are not the same, so we need to find a way to transform $\sin\Big(x-\frac{\pi}{2}\Big)$ to the already known. And we can do like this: $$\sin\Big(x-\frac{\pi}{2}\Big)=\sin\Big[-\Big(\frac{\pi}{2}-x\Big)\Big]$$ And we already know from the negative-angle identities that $$\sin(-\theta)=-\sin\theta$$ Therefore, $$\sin\Big(x-\frac{\pi}{2}\Big)=-\sin\Big(\frac{\pi}{2}-x\Big)$$ $$\sin\Big(x-\frac{\pi}{2}\Big)=-\cos x$$ Since $\cos x\ne-\cos x$, $$\sin\Big(x-\frac{\pi}{2}\Big)\ne\cos x$$ The statement thus is false.

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