Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 6 - Section 6.3 - Trigonometric Functions of Angles - 6.3 Exercises - Page 499: 54

Answer

$\sin\theta$ = $- \frac{4}{\sqrt {17}}$ $\cos\theta$ =$ -\frac{1}{\sqrt {17}}$ $\tan\theta$ =$-\frac{4}{1}$ = -4 (given) $\cot\theta$ =$ -\frac{1}{4}$ $\sec\theta$ = $-\sqrt {17}$ $\csc\theta$ =$- \frac{\sqrt {17}}{4}$

Work Step by Step

Considering $\theta$ as reference angle, we may draw a right triangle ABC to calculate other trigonometric functions- In right triangle ABC- Given $\tan\theta$ =$ \frac{y}{x}$ =-4 = $ \frac{4}{1}$ (Ignoring sign) Using Pythagorous theorem- $r =\sqrt {x^{2}+y^{2}}$ =$ \sqrt {1^{2}+4^{2}}$ = $\sqrt {17}$ i.e. x=1, y=4 and r=$\sqrt {17}$ Given tan is negative and sin is positive, therefore $\theta$ lies in quadrant II. As $\theta$ lies in quadrant II, therefore- $\sin\theta$ =$ \frac{y}{r}$ = $- \frac{4}{\sqrt {17}}$ $\cos\theta$ =$- \frac{x}{r}$ =$ -\frac{1}{\sqrt {17}}$ $\tan\theta$ =$ -\frac{y}{x}$ =$-\frac{4}{1}$ = -4 (given) $\cot\theta$ =$ -\frac{x}{y}$ =$ -\frac{1}{4}$ $\sec\theta$ =$ -\frac{r}{x}$ =$ -\frac{\sqrt {17}}{1}$ = $-\sqrt {17}$ $\csc\theta$ =$ \frac{r}{y}$ =$- \frac{\sqrt {17}}{4}$
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