Chapter 4.4, Problem 34E

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 $y=\frac{1}{3}x$ and in thequadrant $III$ .

Answer to Problem 34E

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

  $\begin{array}{l}\text{sin}\theta =\frac{-1}{\sqrt{10}}\\ \text{cos}\theta =\frac{-3}{\sqrt{10}}\\ \text{tan}\theta =\frac{1}{3}\\ \text{csc}\theta \text{=}-\sqrt{10}\\ \text{sec}\theta \text{=}\frac{-\sqrt{10}}{3}\\ \text{cot}\theta =3\end{array}$

Explanation of Solution

Given info:

The terminal side of $\theta$ lies on the line $y=\frac{1}{3}x$ and in thequadrant $III$

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 $y=\frac{1}{3}x$ and in thequadrant $III$

We have

  $\begin{array}{l}y=\frac{1}{3}x\\ \frac{y}{x}=\frac{1}{3}\\ \text{For quadrant III x is -ve and y is -ve}\\ y=-1\\ x=-3\\ r=\sqrt{x^2+y^2}\\ r=\sqrt{{(-1)}^2+{(-3)}^2}=\sqrt{10}\\ \text{sin}\theta \text{=}\frac{y}{r}=\frac{-1}{\sqrt{10}}\\ \text{cos}\theta \text{=}\frac{x}{r}=\frac{-3}{\sqrt{10}}\\ \text{tan}\theta \text{=}\frac{y}{x}=\frac{-1}{-3}=\frac{1}{3}\\ \text{csc}\theta \text{=}\frac{r}{y}=\frac{\sqrt{10}}{-1}=-\sqrt{10}\\ \text{sec}\theta \text{=}\frac{r}{x}=\frac{\sqrt{10}}{-3}=\frac{-\sqrt{10}}{3}\\ \text{cot}\theta \text{=}\frac{x}{y}=\frac{-3}{-1}=3\end{array}$

Conclusion:

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

  $\begin{array}{l}\text{sin}\theta =\frac{-1}{\sqrt{10}}\\ \text{cos}\theta =\frac{-3}{\sqrt{10}}\\ \text{tan}\theta =\frac{1}{3}\\ \text{csc}\theta \text{=}-\sqrt{10}\\ \text{sec}\theta \text{=}\frac{-\sqrt{10}}{3}\\ \text{cot}\theta =3\end{array}$ .