Chapter 4.4, Problem 24E

To determine

The exact values of the six trigonometric functions of $\theta$ with the given constraint.

Answer to Problem 24E

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

  $\begin{array}{l}\text{sin}\theta \text{=}\frac{-15}{17}\\ \text{cos}\theta \text{=}\frac{8}{17}\\ \tan \theta \text{=}\frac{-15}{8}\\ \text{csc}\theta \text{=}\frac{-17}{15}\\ \text{sec}\theta \text{=}\frac{17}{8}\\ \text{cot}\theta \text{=}\frac{-8}{15}\end{array}$

Explanation of Solution

Given info:

  $\text{cos}\theta \text{=}\frac{8}{17}\text{and tan}\theta <0$

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 $\text{cos}\theta >0$

Hence angle $\theta$ lies inquadrant I or quadrant IV

And $\text{tan}\theta <0$

Hence angle $\theta$ lies inquadrant II or quadrant IV

The common quadrant is quadrant IV.

So $\theta$ lies inquadrant IV.

Now we have

  $\begin{array}{l}\text{cos}\theta \text{=}\frac{8}{17}=\frac{x}{r}\\ r=17\\ x=8\\ r=\sqrt{x^2+y^2}\\ 17=\sqrt{8^2+y^2}\\ y=-15\\ \text{sin}\theta \text{=}\frac{y}{r}\text{=}\frac{-15}{17}\\ \text{tan}\theta \text{=}\frac{y}{x}\text{=}\frac{-15}{8}\\ \text{csc}\theta \text{=}\frac{r}{y}\text{=}\frac{-17}{15}\\ \text{sec}\theta \text{=}\frac{r}{x}\text{=}\frac{17}{8}\\ \text{cot}\theta \text{=}\frac{x}{y}\text{=}\frac{-8}{15}\end{array}$

Conclusion:

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

  $\begin{array}{l}\text{sin}\theta \text{=}\frac{-15}{17}\\ \text{cos}\theta \text{=}\frac{8}{17}\\ \tan \theta \text{=}\frac{-15}{8}\\ \text{csc}\theta \text{=}\frac{-17}{15}\\ \text{sec}\theta \text{=}\frac{17}{8}\\ \text{cot}\theta \text{=}\frac{-8}{15}\end{array}$ .