Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 1 - Precalculus Review - 1.4 Trigonometric Functions - Exercises - Page 30: 22

Answer

$\sin\theta=\frac{1}{17}$, $\cos \theta=\frac{4}{\sqrt {17}}$ and $\sec\theta=\frac{\sqrt {17}}{4}$

Work Step by Step

$\tan\theta=\frac{1}{\cot\theta}=\frac{1}{4}$ $\sec^{2}\theta =1+\tan^{2}\theta=1+\frac{1}{16}=\frac{17}{16}$ $\implies\,\sec\theta=\frac{\sqrt {17}}{4}$ ($\sec \theta$ is positive as $0\leq\theta\lt\frac{\pi}{2}$) $\cos \theta=\frac{1}{\sec \theta}=\frac{4}{\sqrt {17}}$ $\sin\theta= 1-\cos^{2}\theta=1-\frac{16}{17}=\frac{1}{17}$
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