## Trigonometry (10th Edition)

Published by Pearson

# Chapter 1 - Trigonometric Functions - Section 1.4 Using the Definitions of the Trigonometric Functions - 1.4 Exercises: 74

#### Answer

$\sin\theta=\frac{y}{r}=\frac{1}{2}$ $\cos\theta=\frac{x}{r}=\frac{-\sqrt {3}}{2}$ $\tan\theta=\frac{y}{x}=\frac{1 }{-\sqrt {3}}=-\frac{\sqrt 3}{3}$ $\cot\theta=\frac{x}{y}=\frac{-\sqrt {3}}{1}=-\sqrt 3$ $\sec\theta=\frac{r}{x}=\frac{2}{-\sqrt {3}}=-\frac{2\sqrt 3}{3}$ $\csc\theta=\frac{r}{y}=\frac{2}{1}=2$

#### Work Step by Step

1.Quadrant II --&gt; x is negative and y is positive y=1 and r=2 2. FInd x by using distance formula $2=\sqrt {(x)^{2}+(1)^{2}}$ $4=1+x^{2}$ $x=-\sqrt 3$ 3. Plug the values to find trig functions $\sin\theta=\frac{y}{r}=\frac{1}{2}$ $\cos\theta=\frac{x}{r}=\frac{-\sqrt {3}}{2}$ $\tan\theta=\frac{y}{x}=\frac{1 }{-\sqrt {3}}=-\frac{\sqrt 3}{3}$ $\cot\theta=\frac{x}{y}=\frac{-\sqrt {3}}{1}=-\sqrt 3$ $\sec\theta=\frac{r}{x}=\frac{2}{-\sqrt {3}}=-\frac{2\sqrt 3}{3}$ $\csc\theta=\frac{r}{y}=\frac{2}{1}=2$

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