## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 5 - Trigonometric Identities - Section 5.1 Fundamental Identities - 5.1 Exercises - Page 200: 34

#### Answer

$$\sin\theta=-\frac{2}{5}$$ $$\cos\theta=-\frac{\sqrt{21}}{5}$$ $$\tan\theta=\frac{2\sqrt{21}}{21}$$ $$\cot\theta=\frac{\sqrt{21}}{2}$$ $$\sec\theta=-\frac{5\sqrt{21}}{21}$$

#### Work Step by Step

$$\csc\theta=-\frac{5}{2}\hspace{1.5cm}\theta\hspace{0.1cm}\text{in quadrant III}$$ 1) Reciprocal Identities: $$\csc\theta=\frac{1}{\sin\theta}$$ $$\sin\theta=\frac{1}{\csc\theta}=\frac{1}{-\frac{5}{2}}=-\frac{2}{5}$$ 2) Pythagorean Identities: $$\csc^2\theta=\cot^2\theta+1$$ $$\cot^2\theta=\csc^2\theta-1=(-\frac{5}{2})^2-1=\frac{25}{4}-1=\frac{21}{4}$$ $$\cot\theta=\pm\frac{\sqrt{21}}{2}$$ Also, $$\cos^2\theta=1-\sin^2\theta=1-(-\frac{2}{5})^2=1-\frac{4}{25}=\frac{21}{25}$$ $$\cos\theta=\pm\frac{\sqrt{21}}{5}$$ We know that $\theta$ is in quadrant IV, where both $\sin\theta\lt0$ and $\cos\theta\lt0$. Therefore, $$\cos\theta=-\frac{\sqrt{21}}{5}$$ 3) Quotient Identities: $$\cot\theta=\frac{\cos\theta}{\sin\theta}$$ We already have $\sin\theta\lt0$ and $\cos\theta\lt0$.That means, with the above identity, $\cot\theta\gt0$ $$\cot\theta=\frac{\sqrt{21}}{2}$$ Also, we have $$\tan\theta=\frac{\sin\theta}{\cos\theta}=\frac{-\frac{2}{5}}{-\frac{\sqrt{21}}{5}}=\frac{2}{\sqrt{21}}=\frac{2\sqrt{21}}{21}$$ 4) Reciprocal Identities: $$\sec\theta=\frac{1}{\cos\theta}=\frac{1}{-\frac{\sqrt{21}}{5}}=-\frac{5}{\sqrt{21}}=-\frac{5\sqrt{21}}{21}$$

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