Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 5 - Trigonometric Identities - Section 5.6 Half-Angle Identities - 5.6 Exercises - Page 236: 8

Answer

$$\tan\Big(-\frac{\pi}{8}\Big)=1-\sqrt2$$ 8 should be matched with E.

Work Step by Step

$$\tan\Big(-\frac{\pi}{8}\Big)$$ - From negative-angle Identities: $\tan(-\theta)=-\tan\theta$. Therefore, $$\tan\Big(-\frac{\pi}{8}\Big)=-\tan\frac{\pi}{8}$$ Examine $\frac{\pi}{8}$: $$\frac{\pi}{8}=\frac{\pi}{4\times2}=\frac{1}{2}\times\frac{\pi}{4}$$ Therefore, $$\tan\Big(-\frac{\pi}{8}\Big)=-\tan\Big(\frac{1}{2}\times\frac{\pi}{4}\Big)$$ - Recall the half-angle identities: $$\tan\Big(\frac{A}{2}\Big)=\frac{1-\cos A}{\sin A}$$ So, if we replace $A=\frac{\pi}{4}$, we would have $$\tan\Big(\frac{1}{2}\times\frac{\pi}{4}\Big)=\frac{1-\cos\frac{\pi}{4}}{\sin\frac{\pi}{4}}$$ $$\tan\Big(\frac{1}{2}\times\frac{\pi}{4}\Big)=\frac{1-\frac{\sqrt2}{2}}{\frac{\sqrt2}{2}}$$ $$\tan\Big(\frac{1}{2}\times\frac{\pi}{4}\Big)=\frac{\frac{2-\sqrt2}{2}}{\frac{\sqrt2}{2}}$$ $$\tan\Big(\frac{1}{2}\times\frac{\pi}{4}\Big)=\frac{2-\sqrt2}{\sqrt2}$$ $$\tan\Big(\frac{1}{2}\times\frac{\pi}{4}\Big)=\sqrt2-1$$ Thus, $$\tan\Big(-\frac{\pi}{8}\Big)=-\tan\Big(\frac{1}{2}\times\frac{\pi}{4}\Big)=-(\sqrt2-1)=1-\sqrt2$$ 8 should be matched with E.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.