## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 4 - Section 4.4 - Trigonometric Functions of Any Angle - Exercise Set - Page 575: 30

#### Answer

$\sin \theta =\dfrac{y}{r}=\dfrac{\sqrt {10}}{10} \\ \cos \theta =\dfrac{x}{r}=\dfrac{- 3 \sqrt{10}}{10} \\ \tan \theta =\dfrac{y}{x}=\dfrac{-1}{3}$ and $\csc \theta =\dfrac{r}{y}=\sqrt {10} \\ \sec \theta =\dfrac{r}{x}=-\dfrac{\sqrt {10}}{3} \\ \cot \theta =\dfrac{x}{y}=-3$

#### Work Step by Step

Here, $x=-3; y=1$ and $r=\sqrt {x^2+y^2}$ This gives: $r=\sqrt {(-3)^2+(1)^2}=\sqrt {10}$; Because $\theta$ lies in Quadrant-II. The trigonometric ratios are as follows: $\sin \theta =\dfrac{y}{r}=\dfrac{\sqrt {10}}{10} \\ \cos \theta =\dfrac{x}{r}=\dfrac{- 3 \sqrt{10}}{10} \\ \tan \theta =\dfrac{y}{x}=\dfrac{-1}{3}$ and $\csc \theta =\dfrac{r}{y}=\sqrt {10} \\ \sec \theta =\dfrac{r}{x}=-\dfrac{\sqrt {10}}{3} \\ \cot \theta =\dfrac{x}{y}=-3$

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