Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 1 - Trigonometric Functions - Section 1.4 Using the Definitions of the Trigonometric Functions - 1.4 Exercises - Page 38: 73

Answer

$\sin\theta=\frac{y}{r}=\frac{15}{17}$ $\cos\theta=\frac{x}{r}=\frac{-8}{17}$ $\tan\theta=\frac{y}{x}=\frac{15}{-8}$ $\cot\theta=\frac{x}{y}=\frac{-8}{15}$ $\sec\theta=\frac{r}{x}=\frac{17}{-8}$ $\csc\theta=\frac{r}{y}=\frac{17}{15}$

Work Step by Step

1. The angle is in the Quadrant II, then x is negative and y is positive. $x=-8$ $y=15$ 2. Then calculate r, using x, y and distance formula $r=\sqrt {(-8)^{2}+(15)^{2}} =\sqrt {289}=17$ 3. Then when you have all values you need just insert them to find trig functions $\sin\theta=\frac{y}{r}=\frac{15}{17}$ $\cos\theta=\frac{x}{r}=\frac{-8}{17}$ $\tan\theta=\frac{y}{x}=\frac{15}{-8}$ $\cot\theta=\frac{x}{y}=\frac{-8}{15}$ $\sec\theta=\frac{r}{x}=\frac{17}{-8}$ $\csc\theta=\frac{r}{y}=\frac{17}{15}$
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