Trigonometry (10th Edition)

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

Chapter 1 - Review Exercises: 42

Answer

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

Work Step by Step

We know that angle is in the III quadrant, therefore x and y are negative, then using tangent of the angle and knowledge about the signs (of x and y) we will find r. $x= -1; y=-2$ $r=\sqrt {(x)^{2}+(y)^{2}}= \sqrt {(-1)^{2}+(-2)^{2}}=\sqrt 5=\sqrt 5$
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