Trigonometry (10th Edition)

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

Chapter 1 - Review Exercises: 31

Answer

$\sin\theta=\frac{y}{r}=\frac{-6}{12}=-\frac{1}{2}$ $\cos\theta=\frac{x}{r}=\frac{6\sqrt 3}{12}=\frac{\sqrt 3}{2}$ $\csc\theta=\frac{r}{y}=\frac{12}{-6}=-2$ $\sec\theta=\frac{r}{x}=\frac{12}{6\sqrt 3}=\frac{2\sqrt 3}{3}$ $\tan\theta=\frac{y}{x}=\frac{-6}{6\sqrt 3}=-\frac{\sqrt 3}{3}$ $\cot\theta=\frac{x}{y}=\frac{6\sqrt 3}{-6}=-\sqrt 3$

Work Step by Step

$x=6\sqrt 3; y=-6$ $r=\sqrt {x^{2}+y^{2}}=\sqrt {(6\sqrt 3)^{2}+(-6)^{2}}=\sqrt {144}=12$
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