Chapter 4.4, Problem 33E

To determine

The values of the six trigonometric functions of $\theta$ by finding a point on the line, the terminal side of $\theta$ lies on the line $y=-x$ and in the quadrant $II$ .

Answer to Problem 33E

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

  $\begin{array}{l}\text{sin}\theta =\frac{1}{\sqrt{2}}\\ \text{cos}\theta =\frac{-1}{\sqrt{2}}\\ \text{tan}\theta =-1\\ \text{csc}\theta =\sqrt{2}\\ \text{sec}\theta =-\sqrt{2}\\ \text{cot}\theta =-1\end{array}$

Explanation of Solution

Given info:

The terminal side of $\theta$ lies on the line $y=-x$ and in thequadrant $II$

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 $y=-x$ and in thequadrant $II$

We have

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

Conclusion:

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

  $\begin{array}{l}\text{sin}\theta =\frac{1}{\sqrt{2}}\\ \text{cos}\theta =\frac{-1}{\sqrt{2}}\\ \text{tan}\theta =-1\\ \text{csc}\theta =\sqrt{2}\\ \text{sec}\theta =-\sqrt{2}\\ \text{cot}\theta =-1\end{array}$ .