Trigonometry (10th Edition)

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

Chapter 1 - Review Exercises: 43

Answer

$\sin\theta=\frac{y}{r}=\frac{2}{\sqrt 5}=\frac{2\sqrt 5}{5}$ $\cos\theta=\frac{x}{r}=\frac{-1}{\sqrt 5}=-\frac{\sqrt 5}{5}$ $\csc\theta=\frac{r}{y}=\frac{\sqrt 5}{2}$ $\sec\theta=\frac{r}{x}=\frac{\sqrt 5}{-1}=-\sqrt 5$ $\tan\theta=\frac{y}{x}=\frac{2}{-1}=-2$ $\cot\theta=\frac{x}{y}=\frac{-1}{2}$

Work Step by Step

From secant and II quadrant we know that $x=-1; r=\sqrt 5$, so lets find x: $x^{2}+y^{2}=r^{2}$ $(-1)^{2}+y^{2}=(\sqrt 5)^{2}$ $y^{2}+1=25$ $y^{2}=4$ (since in II quadrant y is positive) $y=2$
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