Trigonometry (10th Edition)

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

Chapter 1 - Review Exercises - Page 42: 44

Answer

$\sin\theta=\frac{y}{r}=\frac{-2}{5}$ $\cos\theta=\frac{x}{r}=\frac{-\sqrt {21}}{5}$ $\csc\theta=\frac{r}{y}=\frac{5}{-2}$ $\sec\theta=\frac{r}{x}=\frac{5}{-\sqrt {21}}=-\frac{5\sqrt {21}}{21}$ $\tan\theta=\frac{y}{x}=\frac{-2}{-\sqrt {21}}=\frac{2\sqrt {21}}{21}$ $\cot\theta=\frac{x}{y}=\frac{-\sqrt {21}}{-2}=\frac{\sqrt {21}}{2}$

Work Step by Step

From sine we know quadrant that $y=-2; r=5$, so lets find x: $x^{2}+y^{2}=r^{2}$ $x^{2}+(-2)^{2}=5^{2}$ $x^{2}+4=25$ $x^{2}=21$ (since $x\lt0$ in second quadrant) $x=-\sqrt {21}$
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