Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 4 - Section 4.3 - Right Triangle Trigonometry - Exercise Set: 25

Answer

$\cot{(\frac{7\pi}{18})}$

Work Step by Step

RECALL: The cofunction Identities: (1) $\sin{\theta} = \cos{(\frac{\pi}{2}-\theta)}$ (2) $\cos{\theta} = \sin{(\frac{\pi}{2}-\theta)}$ (3) $\tan{\theta} = \cot{(\frac{\pi}{2}-\theta)}$ (4) $\cot{\theta} = \tan{(\frac{\pi}{2}-\theta)}$ (5) $\csc{\theta} = \sec{(\frac{\pi}{2}-\theta)}$ (6) $\sec{\theta} = \csc{(\frac{\pi}{2}-\theta)}$ Use identity (3) to obtain: $\tan{\frac{\pi}{9}} \\= \cot{(\frac{\pi}{2}-\frac{\pi}{9})} \\= \cot{(\frac{9\pi}{18}-\frac{2\pi}{18})} \\=\cot{(\frac{7\pi}{18})}$
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