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 194: 38

Answer

$$\sin\theta=\frac{\sqrt{15}}{4}$$ $$\sec\theta=-4$$ $$\csc\theta=\frac{4\sqrt{15}}{15}$$ $$\tan\theta=-\sqrt{15}$$ $$\cot\theta=-\frac{\sqrt{15}}{15}$$

Work Step by Step

$$\cos\theta=-\frac{1}{4}\hspace{1.5cm}\sin\theta\gt0$$ 1) Find $\sin\theta$ - Pythagorean Identities: $$\sin^2\theta=1-\cos^2\theta=1-(-\frac{1}{4})^2=1-\frac{1}{16}=\frac{15}{16}$$ $$\sin\theta=\pm\frac{\sqrt{15}}{4}$$ As $\sin\theta\gt0$, $$\sin\theta=\frac{\sqrt{15}}{4}$$ 2) Find $\sec\theta$ and $\csc\theta$ - Reciprocal Identities: $$\sec\theta=\frac{1}{\cos\theta}=\frac{1}{-\frac{1}{4}}=-4$$ $$\csc\theta=\frac{1}{\sin\theta}=\frac{1}{\frac{\sqrt{15}}{4}}=\frac{4}{\sqrt{15}}=\frac{4\sqrt{15}}{15}$$ 3) Find $\tan\theta$ and $\cot\theta$ - Quotient Identities: $$\cot\theta=\frac{\cos\theta}{\sin\theta}=\frac{-\frac{1}{4}}{\frac{\sqrt{15}}{4}}=-\frac{1}{\sqrt{15}}=-\frac{\sqrt{15}}{15}$$ - Reciprocal Identities: $$\tan\theta=\frac{1}{\cot\theta}=\frac{1}{-\frac{\sqrt{15}}{15}}=-\frac{15}{\sqrt{15}}=-\sqrt{15}$$
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