Trigonometry 7th Edition

Published by Cengage Learning
ISBN 10: 1111826854
ISBN 13: 978-1-11182-685-7

Chapters 1-3 - Cumulative Test - Page 176: 6


$\sin{\theta} = \frac{8}{10} = \frac{4}{5} \\\cos{\theta} = \frac{-6}{10} = -\frac{3}{5} \\\tan{\theta} = \frac{8}{-6} = -\frac{4}{3} \\\cot{\theta}=\frac{-6}{8} = -\frac{3}{4} \\\csc{\theta}=\frac{10}{8}=\frac{5}{4} \\\sec{\theta} = \frac{10}{-6} - -\frac{5}{3}$

Work Step by Step

RECALL: If P(x, y) is on the terminal side of an angle $\theta$ in standard position then: $\begin{array}{cc} &\sin{\theta} = \frac{y}{r} &\csc{\theta} = \frac{r}{y} \\&\cos{\theta} = \frac{x}{r} &\sec{\theta} = \frac{r}{x} \\&\tan{\theta} = \frac{y}{x} &\cot{\theta} = \frac{x}{y} \end{array}$ where $r= \sqrt{x^2+y^2}$ Solve for $r$ to obtain: $r=\sqrt{(-6)^2+8^2} = \sqrt{36+64}=\sqrt{100}=10$ Thus, using the ratios in the recall part above, and with P(-6, 8) and $r=10$, gives: $\sin{\theta} = \frac{8}{10} = \frac{4}{5} \\\cos{\theta} = \frac{-6}{10} = -\frac{3}{5} \\\tan{\theta} = \frac{8}{-6} = -\frac{4}{3} \\\cot{\theta}=\frac{-6}{8} = -\frac{3}{4} \\\csc{\theta}=\frac{10}{8}=\frac{5}{4} \\\sec{\theta} = \frac{10}{-6} - -\frac{5}{3}$
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