Chapter 4.4, Problem 17E

To determine

The exact values of the sixtrigonometric functions of the angle if the point $(-5.4,7.2)$ is on the terminal side of an angle instandard position.

Answer to Problem 17E

The exact values of the sixtrigonometric functions are

  $\begin{array}{l}\text{sin}\theta \text{=}\frac{4}{5}\\ \text{cos}\theta \text{=}\frac{-3}{5}\\ \text{tan}\theta \text{=}\frac{-4}{3}\\ \text{csc}\theta \text{=}\frac{5}{4}\\ \text{sec}\theta \text{=}\frac{-5}{3}\\ \text{cot}\theta \text{=}\frac{-3}{4}\end{array}$

Explanation of Solution

Given info:

  $(x,y)=(-5.4,7.2)$

Formula used:

  

  $\begin{array}{l}r=\sqrt{x^2+y^2}\\ \text{sin}\theta \text{=}\frac{y}{r}\\ \text{cos}\theta \text{=}\frac{x}{r}\\ \text{tan}\theta \text{=}\frac{y}{x}\\ \text{csc}\theta \text{=}\frac{r}{y}\\ \text{sec}\theta \text{=}\frac{r}{x}\\ \text{cot}\theta \text{=}\frac{x}{y}\end{array}$

Calculation:

Given that $(x,y)=(-5.4,7.2)$

We have

  $\begin{array}{l}r=\sqrt{x^2+y^2}=\sqrt{{(-5.4)}^2+{7.2}^2}=\sqrt{81}=9\\ \text{sin}\theta \text{=}\frac{y}{r}\text{=}\frac{7.2}{9}\text{=}\frac{4}{5}\\ \text{cos}\theta \text{=}\frac{x}{r}\text{=}\frac{-5.4}{9}\text{=}\frac{-3}{5}\\ \text{tan}\theta \text{=}\frac{y}{x}\text{=}\frac{7.2}{-5.4}\text{=}\frac{-4}{3}\\ \text{csc}\theta \text{=}\frac{r}{y}\text{=}\frac{9}{7.2}\text{=}\frac{5}{4}\\ \text{sec}\theta \text{=}\frac{r}{x}\text{=}\frac{9}{-5.4}\text{=}\frac{-5}{3}\\ \text{cot}\theta \text{=}\frac{x}{y}\text{=}\frac{-5.4}{7.2}\text{=}\frac{-3}{4}\end{array}$

The exact values of the sixtrigonometric functions are

  $\begin{array}{l}\text{sin}\theta \text{=}\frac{4}{5}\\ \text{cos}\theta \text{=}\frac{-3}{5}\\ \text{tan}\theta \text{=}\frac{-4}{3}\\ \text{csc}\theta \text{=}\frac{5}{4}\\ \text{sec}\theta \text{=}\frac{-5}{3}\\ \text{cot}\theta \text{=}\frac{-3}{4}\end{array}$

Conclusion:

Thus,the exact values of the sixtrigonometric functions are

  $\begin{array}{l}\text{sin}\theta \text{=}\frac{4}{5}\\ \text{cos}\theta \text{=}\frac{-3}{5}\\ \text{tan}\theta \text{=}\frac{-4}{3}\\ \text{csc}\theta \text{=}\frac{5}{4}\\ \text{sec}\theta \text{=}\frac{-5}{3}\\ \text{cot}\theta \text{=}\frac{-3}{4}\end{array}$