Trigonometry 7th Edition

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

Chapter 1 - Section 1.3 - Definition I: Trigonometric Functions - 1.3 Problem Set - Page 33: 55

Answer

$\sin\theta=\frac{12}{13}$ $\cos\theta=\frac{5}{13}$ $\tan\theta=\frac{12}{5}$

Work Step by Step

$\sin\theta=\frac{y}{r}$ $\sin\theta=\frac{12}{13}$ Substitute into $x^2+y^2=r$ $x=\pm5$ So x=5 because $\theta$ terminates in the first quadrant and can only be positive. Knowing x, we can now substitute it into the rest of the trigonometric functions. $\cos\theta=\frac{5}{13}$ $\tan\theta=\frac{12}{5}$
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