Chapter 4.4, Problem 28E

To determine

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

Answer to Problem 28E

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

  $\begin{array}{l}\text{sin}\theta \text{=}\frac{1}{4}\\ \text{cos}\theta \text{=}\frac{-\sqrt{15}}{4}\\ \text{tan}\theta \text{=}\frac{-1}{\sqrt{15}}\\ \text{csc}\theta \text{=}4\\ \text{sec}\theta \text{=}\frac{-4}{\sqrt{15}}\\ \text{cot}\theta \text{=}-\sqrt{15}\end{array}$

Explanation of Solution

Given info:

  $\text{csc}\theta =4\text{and cot}\theta <0$

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 $\text{csc}\theta >0$

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

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

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

The common quadrant is quadrant II.

So $\theta$ lies inquadrant II.

Now we have

  $\begin{array}{l}\text{csc}\theta \text{=}\frac{r}{y}=4\\ y=1\\ r=4\\ r=\sqrt{x^2+y^2}\\ 4=\sqrt{x^2+1^2}\\ x=-\sqrt{15}\\ \text{sin}\theta \text{=}\frac{y}{r}\text{=}\frac{1}{4}\\ \text{cos}\theta \text{=}\frac{x}{r}\text{=}\frac{-\sqrt{15}}{4}\\ \text{tan}\theta \text{=}\frac{y}{x}\text{=}\frac{1}{-\sqrt{15}}\text{=}\frac{-1}{\sqrt{15}}\\ \text{sec}\theta \text{=}\frac{r}{x}\text{=}\frac{4}{-\sqrt{15}}\text{=}\frac{-4}{\sqrt{15}}\\ \text{cot}\theta \text{=}\frac{x}{y}\text{=}\frac{-\sqrt{15}}{1}\text{=}-\sqrt{15}\end{array}$

Conclusion:

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

  $\begin{array}{l}\text{sin}\theta \text{=}\frac{1}{4}\\ \text{cos}\theta \text{=}\frac{-\sqrt{15}}{4}\\ \text{tan}\theta \text{=}\frac{-1}{\sqrt{15}}\\ \text{csc}\theta \text{=}4\\ \text{sec}\theta \text{=}\frac{-4}{\sqrt{15}}\\ \text{cot}\theta \text{=}-\sqrt{15}\end{array}$ .