## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 1 - Trigonometric Functions - Section 1.4 Using the Definitions of the Trigonometric Functions - 1.4 Exercises - Page 38: 77

#### Answer

$\sin\theta=\frac{y}{r}=\frac{8}{\sqrt {67}}=\frac{8\sqrt {67}}{67}$ $\cos\theta=\frac{x}{r}=\frac{\sqrt {3}}{\sqrt{67}}=\frac{\sqrt {201}}{67}$ $\tan\theta=\frac{y}{x}=\frac{8 }{\sqrt {3}}=\frac{8\sqrt {3} }{3}$ $\cot\theta=\frac{x}{y}=\frac{\sqrt {3}}{8}$ $\sec\theta=\frac{r}{x}=\frac{\sqrt {67}}{\sqrt {3}}=\frac{\sqrt {201}}{3}$ $\csc\theta=\frac{r}{y}=\frac{\sqrt {67}}{8}$

#### Work Step by Step

1. I quadrant x and y are positive $x=\sqrt 3$ and $y=8$ 2. Calculate r using distnace formula $r=\sqrt {(\sqrt 3)^{2}+(8)^{2}} =\sqrt {67}$ 3. Plug the values to find trig function $\sin\theta=\frac{y}{r}=\frac{8}{\sqrt {67}}=\frac{8\sqrt {67}}{67}$ $\cos\theta=\frac{x}{r}=\frac{\sqrt {3}}{\sqrt{67}}=\frac{\sqrt {201}}{67}$ $\tan\theta=\frac{y}{x}=\frac{8 }{\sqrt {3}}=\frac{8\sqrt {3} }{3}$ $\cot\theta=\frac{x}{y}=\frac{\sqrt {3}}{8}$ $\sec\theta=\frac{r}{x}=\frac{\sqrt {67}}{\sqrt {3}}=\frac{\sqrt {201}}{3}$ $\csc\theta=\frac{r}{y}=\frac{\sqrt {67}}{8}$

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