Chapter 4.4, Problem 16E

To determine

The exact values of the sixtrigonometric functions of the angle if the point $(-4,10)$ is on the terminal side of an angle instandard position.

Answer to Problem 16E

The exact values of the sixtrigonometric functions are

  $\begin{array}{l}\text{sin}\theta \text{=}\frac{5}{\sqrt{29}}\\ \text{cos}\theta \text{=}\frac{-2}{\sqrt{29}}\\ \text{tan}\theta \text{=}\frac{-5}{2}\\ \text{csc}\theta \text{=}\frac{\sqrt{29}}{5}\\ \text{sec}\theta \text{=}\frac{-\sqrt{29}}{2}\\ \text{cot}\theta \text{=}\frac{-2}{5}\end{array}$

Explanation of Solution

Given info:

  $(x,y)=(-4,10)$

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 $(x,y)=(-4,10)$

We have

  $\begin{array}{l}r=\sqrt{x^2+y^2}=\sqrt{{(-4)}^2+{10}^2}=\sqrt{116}=2\sqrt{29}\\ \text{sin}\theta \text{=}\frac{y}{r}\text{=}\frac{10}{2\sqrt{29}}\text{=}\frac{5}{\sqrt{29}}\\ \text{cos}\theta \text{=}\frac{x}{r}\text{=}\frac{-4}{2\sqrt{29}}\text{=}\frac{-2}{\sqrt{29}}\\ \text{tan}\theta \text{=}\frac{y}{x}\text{=}\frac{10}{-4}\text{=}\frac{-5}{2}\\ \text{csc}\theta \text{=}\frac{r}{y}\text{=}\frac{2\sqrt{29}}{10}\text{=}\frac{\sqrt{29}}{5}\\ \text{sec}\theta \text{=}\frac{r}{x}\text{=}\frac{2\sqrt{29}}{-4}\text{=}\frac{-\sqrt{29}}{2}\\ \text{cot}\theta \text{=}\frac{x}{y}\text{=}\frac{-4}{10}\text{=}\frac{-2}{5}\end{array}$

The exact values of the sixtrigonometric functions are

  $\begin{array}{l}\text{sin}\theta \text{=}\frac{5}{\sqrt{29}}\\ \text{cos}\theta \text{=}\frac{-2}{\sqrt{29}}\\ \text{tan}\theta \text{=}\frac{-5}{2}\\ \text{csc}\theta \text{=}\frac{\sqrt{29}}{5}\\ \text{sec}\theta \text{=}\frac{-\sqrt{29}}{2}\\ \text{cot}\theta \text{=}\frac{-2}{5}\end{array}$

Conclusion:

Thus,the exact values of the sixtrigonometric functions are

  $\begin{array}{l}\text{sin}\theta \text{=}\frac{5}{\sqrt{29}}\\ \text{cos}\theta \text{=}\frac{-2}{\sqrt{29}}\\ \text{tan}\theta \text{=}\frac{-5}{2}\\ \text{csc}\theta \text{=}\frac{\sqrt{29}}{5}\\ \text{sec}\theta \text{=}\frac{-\sqrt{29}}{2}\\ \text{cot}\theta \text{=}\frac{-2}{5}\end{array}$