## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 4 - Review Exercises - Page 645: 50

#### Answer

The trigonometric functions are: $\sin \theta =\frac{\sqrt{10}}{10}\text{; cos}\theta \text{=}-\frac{3\sqrt{10}}{10}\text{; csc}\theta \text{=}\sqrt{10};\text{ sec}\theta \text{=}-\frac{\sqrt{10}}{3};\text{ cot}\theta \text{=}-3$.

#### Work Step by Step

Since $\tan \theta <0\text{ and sin}\theta >0$, so $\theta$ lies in quadrant II. In quadrant II, x is negative and y is positive. Here, \begin{align} & \tan \theta =-\frac{1}{3} \\ & =\frac{y}{x} \end{align} So, $x=-3\text{ and }y=1$. Now, calculate r as: \begin{align} & {{r}^{2}}={{x}^{2}}+{{y}^{2}} \\ & r=\sqrt{{{\left( -3 \right)}^{2}}+{{1}^{2}}} \\ & r=\sqrt{10} \end{align} Then, evaluate the sine function as follows: \begin{align} & \sin \theta =\frac{y}{r} \\ & =\frac{1}{\sqrt{10}} \\ & =\frac{1\cdot \sqrt{10}}{\sqrt{10}\cdot \sqrt{10}} \\ & =\frac{\sqrt{10}}{10} \end{align} Now, evaluate the cosine function as follows: \begin{align} & \cos \theta =\frac{x}{r} \\ & =\frac{-3}{\sqrt{10}} \\ & =-\frac{3\cdot \sqrt{10}}{\sqrt{10}\cdot \sqrt{10}} \\ & =-\frac{3\sqrt{10}}{10} \end{align} Then, evaluate the cosecant function as follows: \begin{align} & \csc \theta =\frac{r}{y} \\ & =\frac{\sqrt{10}}{1} \\ & =\sqrt{10} \end{align} Then, evaluate the secant function as follows: \begin{align} & \sec \theta =\frac{r}{x} \\ & =\frac{\sqrt{10}}{-3} \\ & =-\frac{\sqrt{10}}{3} \end{align} Also, evaluate the cotangent function as follows: \begin{align} & \cot \theta =\frac{x}{y} \\ & =\frac{-3}{1} \\ & =-3 \end{align}

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