Chapter 4.4, Problem 36E

To determine

The values of the six trigonometric functions of $\theta$ by finding a point onthe line, the terminal side of $\theta$ lies on the line $4x+3y=0$ and in thequadrant $IV$ .

Answer to Problem 36E

The values of the six trigonometric functions of $\theta$ are

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

Explanation of Solution

Given info:

The terminal side of $\theta$ lies on the line $4x+3y=0$ and in thequadrant $IV$

Formula used:

In quadrant I all trigonometric functions are positive.

In quadrant IIsine functions are positive.

In quadrant IIItangent functions are positive.

In quadrant IVcosine functions are positive.

  

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

Calculation:

Given that the terminal side of $\theta$ lies on the line $4x+3y=0$ and in thequadrant $IV$

We have

  $\begin{array}{l}4x+3y=0\\ \frac{y}{x}=\frac{-4}{3}\\ \text{For quadrant IV x is +ve and y is -ve}\\ y=-4\\ x=3\\ r=\sqrt{x^2+y^2}\\ r=\sqrt{3^2+{(-4)}^2}=5\\ \text{sin}\theta \text{=}\frac{y}{r}=\frac{-4}{5}\\ \text{cos}\theta \text{=}\frac{x}{r}=\frac{3}{5}\\ \text{tan}\theta \text{=}\frac{y}{x}=\frac{-4}{3}\\ \text{csc}\theta \text{=}\frac{r}{y}=\frac{5}{-4}=\frac{-5}{4}\\ \text{sec}\theta \text{=}\frac{r}{x}=\frac{5}{3}\\ \text{cot}\theta \text{=}\frac{x}{y}=\frac{3}{-4}=\frac{-3}{4}\end{array}$

Conclusion:

Thus,the values of the six trigonometric functions of $\theta$ are

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