Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 1 - Precalculus Review - 1.4 Trigonometric Functions - Exercises - Page 30: 6

Answer

$\sin{\theta} = \dfrac{8}{17}$ $\cos{\theta} =\dfrac{15}{17}$ $\tan{\theta}=\dfrac{8}{15}$ $\csc{\theta}=\dfrac{17}{8}$ $\sec{\theta} =\dfrac{17}{15}$ $\cot{\theta}=\dfrac{15}{8}$

Work Step by Step

Recall: $$\sin{\theta} = \dfrac{\text{opposite}}{\text{hypotenuse}}$$ $$\cos{\theta} = \dfrac{\text{adjacent}}{\text{hypotenuse}}$$ $\sin{\theta} = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac{8}{17}$ $\cos{\theta} = \dfrac{\text{adjacent}}{\text{hypotenuse}} = \dfrac{15}{17}$ $\tan{\theta}= \dfrac{\sin{\theta}}{\cos{\theta}} = \dfrac{8}{15}$ $\csc{\theta}= \dfrac{1}{\sin{\theta}} = \dfrac{17}{8}$ $\sec{\theta} = \dfrac{1}{\cos{\theta}} = \dfrac{17}{15}$ $\cot{\theta}=\dfrac{1}{\tan{\theta}}= \dfrac{15}{8}$
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