Trigonometry (10th Edition)

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

Chapter 1 - Review Exercises: 32

Answer

$\sin\theta=\frac{y}{r}=\frac{2\sqrt 2}{4}=\frac{\sqrt 2}{2}$ $\cos\theta=\frac{x}{r}=\frac{-2\sqrt 2}{4}=-\frac{\sqrt 2}{2}$ $\csc\theta=\frac{r}{y}=\frac{4}{2\sqrt 2}=\sqrt 2$ $\sec\theta=\frac{r}{x}=\frac{4}{-2\sqrt 2}=-\sqrt 2$ $\tan\theta=\frac{y}{x}=\frac{2\sqrt 2}{-2\sqrt 2}=-1$ $\cot\theta=\frac{x}{y}=\frac{-2\sqrt 2}{2\sqrt 2}=-1$

Work Step by Step

$x= -2\sqrt 2; y=2\sqrt 2$ $r=\sqrt {(x)^{2}+(y)^{2}}= \sqrt {(-2\sqrt 2)^{2}+(2\sqrt 2)^{2}}=\sqrt {16}=4$
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