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 213: 65

Answer

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

Work Step by Step

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