Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 1 - Review Exercises: 24

Answer

$\sin\theta=\frac{y}{r}=\frac{-3}{3\sqrt 2}=-\frac{\sqrt 2}{2}$ $\cos\theta=\frac{x}{r}=\frac{-3}{3\sqrt 2}=-\frac{\sqrt 2}{2}$ $\csc\theta=\frac{r}{y}=\frac{3\sqrt 2}{-3}=-\sqrt 2$ $\sec\theta=\frac{r}{x}=\frac{3\sqrt 2}{-3}=-\sqrt 2$ $\tan\theta=\frac{y}{x}=\frac{-3}{-3}=1$ $\\theta=\frac{x}{y}=\frac{-3}{-3}=1$

Work Step by Step

x=-3; y=-3 $r=\sqrt {x^{2}+y^{2}}=\sqrt {(-3)^{2}+(-3)^{2}}=\sqrt {18}=3\sqrt 2$
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