Trigonometry 7th Edition

Published by Cengage Learning
ISBN 10: 1111826854
ISBN 13: 978-1-11182-685-7

Chapter 1 - Section 1.3 - Definition I: Trigonometric Functions - 1.3 Problem Set - Page 31: 15

Answer

$\sin\theta$ = - $ \frac{4}{5}$ $\cos\theta$ = -$ \frac{3}{5}$ $\tan\theta$ = $ \frac{4}{3}$ $\cot\theta$ = $ \frac{3}{4}$ $\sec\theta$ = -$ \frac{5}{3}$ $\csc\theta$ = -$ \frac{5}{4}$

Work Step by Step

Given, point (-9a, -12a) is on the terminal side of $\theta$, we may apply Definition I to find all six trigonometric functions- We got $ x = -9a, y = -12a$ Therefore r= $\sqrt (x^{2} + y^{2})$ = $\sqrt ((-9a)^{2} + (-12a)^{2})$ = $\sqrt (81a^{2} + 144 a^{2})$ = $\sqrt (225 a^{2})$ = 15a i.e. $ x = -9a, y = -12a,$ and $ r= 15a$ Given that 'a' is a positive number Applying Definition I- $\sin\theta$ =$ \frac{y}{r}$ = $ \frac{-12a}{15a}$ = - $ \frac{4}{5}$ $\cos\theta$ =$ \frac{x}{r}$ =$ \frac{-9a}{15a}$ = -$ \frac{3}{5}$ $\tan\theta$ =$ \frac{y}{x}$ =$ \frac{-12a}{-9a}$ = $ \frac{4}{3}$ $\cot\theta$ =$ \frac{x}{y}$ =$ \frac{-9a}{-12a}$ = $ \frac{3}{4}$ $\sec\theta$ =$ \frac{r}{x}$ =$ \frac{15a}{-9a}$ = -$ \frac{5}{3}$ $\csc\theta$ =$ \frac{r}{y}$ =$ \frac{15a}{-12a}$= -$ \frac{5}{4}$
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