Chapter 4.4, Problem 18E

To determine

The exact values of the six trigonometric functions of the angle if the point $\left(3\frac{1}{2},-7\frac{3}{4}\right)$ is on the terminal side of an angle in standard position.

Answer to Problem 18E

The exact values of the six trigonometric functions are

  $\begin{array}{c}\text{sin}\theta =\frac{-31\sqrt{1157}}{1157}\\ \text{cos}\theta \text{=}\frac{14\sqrt{1157}}{1157}\\ \text{tan}\theta \text{=}\frac{-31}{14}\\ \text{csc}\theta \text{=}\frac{\sqrt{1157}}{-31}\\ \text{sec}\theta \text{=}\frac{\sqrt{1157}}{14}\\ \text{cot}\theta \text{=}-\frac{14}{31}\end{array}$

Explanation of Solution

Given information:

  $\left(x,y\right)=\left(3\frac{1}{2},-7\frac{3}{4}\right)$

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 $\left(x,y\right)=\left(3\frac{1}{2},-7\frac{3}{4}\right)$

Now, calculate the value of r as shown:

  $\begin{array}{c}r=\sqrt{x^2+y^2}\\ =\sqrt{{\left(\frac{7}{2}\right)}^2+{\left(-\frac{31}{4}\right)}^2}\\ =\sqrt{\frac{49}{4}+\frac{961}{16}}\\ =\frac{\sqrt{1157}}{4}\end{array}$

The value of trigonometric function can be calculated as

  $\begin{array}{c}\text{sin}\theta \text{=}\frac{y}{r}\text{=}\frac{-\frac{31}{4}}{\frac{\sqrt{1157}}{4}}\text{=}\frac{-31\sqrt{1157}}{1157}\\ \text{cos}\theta \text{=}\frac{x}{r}\text{=}\frac{\frac{7}{2}}{\frac{\sqrt{1157}}{4}}\text{=}\frac{14\sqrt{1157}}{1157}\\ \text{tan}\theta \text{=}\frac{y}{x}\text{=}\frac{-\frac{31}{4}}{\frac{7}{2}}\text{=}\frac{-31}{14}\\ \text{csc}\theta \text{=}\frac{r}{y}\text{=}\frac{\frac{\sqrt{1157}}{4}}{-\frac{31}{4}}\text{=}\frac{\sqrt{1157}}{-31}\\ \text{sec}\theta \text{=}\frac{r}{x}\text{=}\frac{\frac{\sqrt{1157}}{4}}{\frac{7}{2}}\text{=}\frac{\sqrt{1157}}{14}\\ \text{cot}\theta \text{=}\frac{x}{y}\text{=}\frac{\frac{7}{2}}{-\frac{31}{4}}\text{=}-\frac{14}{31}\end{array}$

The exact values of the six trigonometric functions are

  $\begin{array}{c}\text{sin}\theta =\frac{-31\sqrt{1157}}{1157}\\ \text{cos}\theta \text{=}\frac{14\sqrt{1157}}{1157}\\ \text{tan}\theta \text{=}\frac{-31}{14}\\ \text{csc}\theta \text{=}\frac{\sqrt{1157}}{-31}\\ \text{sec}\theta \text{=}\frac{\sqrt{1157}}{14}\\ \text{cot}\theta \text{=}-\frac{14}{31}\end{array}$