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 32: 22


$\cos{\theta} = \dfrac{12}{13} \\\sin{\theta} = \dfrac{5}{13}$

Work Step by Step

(Assumption: the angle is in standard position) If the point $(6.36, 2.65)$ is on the terminal side of $\theta$. then $r$ can be solved using the formula: $r=\sqrt{x^2 + y^2}$ The given point has $x=6.36$ and $y=2.65$. Substitute these values into the formula above to obtain: $r= \sqrt{6.36^2 + 2.65^2} \\r = 6.89$ RECALL: $\sin{\theta} = \dfrac{y}{r} \\\cos{\theta} = \dfrac{x}{r}$ Use the formulas above and the known values of x, y, and r to obtain: $\cos{\theta} = \dfrac{6.36}{6.89}=\dfrac{12}{13} \\\sin{\theta} = \dfrac{2.65}{6.89}=\dfrac{5}{13}$
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