## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 5 - Trigonometric Identities - Section 5.1 Fundamental Identities - 5.1 Exercises - Page 200: 31

#### Answer

$$\cos\theta=-\frac{\sqrt 5}{3}$$ $$\tan\theta=-\frac{2\sqrt 5}{5}$$ $$\cot\theta=-\frac{\sqrt 5}{2}$$ $$\csc\theta=\frac{3}{2}$$ $$\sec\theta=\frac{-3\sqrt 5}{5}$$

#### Work Step by Step

$$\sin\theta=\frac{2}{3}\hspace{1.5cm}\theta\hspace{0.1cm}\text{in quadrant II}$$ 1) Reciprocal Identities: $$\csc\theta=\frac{1}{\sin\theta}$$ $$\csc\theta=\frac{1}{\frac{2}{3}}=\frac{3}{2}$$ 2) Pythagorean Identities: $$\cot^2\theta+1=\csc^2\theta$$ $$\cot^2\theta=\csc^2\theta-1$$ $$\cot^2\theta=(\frac{3}{2})^2-1=\frac{9}{4}-1=\frac{5}{4}$$ $$\cot\theta=\pm\frac{\sqrt5}{2}$$ 3) Quotient Identities: $$\cot\theta=\frac{\cos\theta}{\sin\theta}$$ We know that $\theta$ is in quadrant II, where $\sin\theta\gt0$ but $\cos\theta\lt0$. Therefore, with the above identity, we can conclude that $\cot\theta\lt0$. $$\cot\theta=-\frac{\sqrt 5}{2}$$ $$\frac{\cos\theta}{\sin\theta}=-\frac{\sqrt 5}{2}$$ $$\cos\theta=(-\frac{\sqrt 5}{2})\times\sin\theta=(-\frac{\sqrt 5}{2})\times\frac{2}{3}=-\frac{\sqrt 5}{3}$$ 4) Reciprocal Identities: $$\cot\theta=\frac{1}{\tan\theta}$$ $$\tan\theta=\frac{1}{\cot\theta}=\frac{1}{-\frac{\sqrt 5}{2}}=-\frac{2}{\sqrt 5}=-\frac{2\sqrt 5}{5}$$ Also, $$\sec\theta=\frac{1}{\cos\theta}=\frac{1}{-\frac{\sqrt 5}{3}}=-\frac{3}{\sqrt 5}=-\frac{3\sqrt 5}{5}$$

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