Trigonometry (10th Edition)

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

Chapter 5 - Trigonometric Identities - Section 5.1 Fundamental Identities - 5.1 Exercises - Page 193: 8

Answer

$$\sin\theta=\frac{\sqrt{11}}{6}$$

Work Step by Step

$$\cos\theta=\frac{5}{6}$$ We can transform the Pythagorean Identities $$\sin^2 x+\cos^2 x=1$$ into $$\sin^2 x=1-\cos^2 x$$ That means $$\sin^2\theta=1-\cos^2\theta=1-\Big(\frac{5}{6}\Big)^2=1-\frac{25}{36}=\frac{11}{36}$$ So, $$\sin\theta=\pm\frac{\sqrt{11}}{6}$$ Another given information shows that $\theta$ is in quadrant I, which means $\sin\theta\gt0$. Therefore, $$\sin\theta=\frac{\sqrt{11}}{6}$$
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