Chapter 4.4, Problem 38E

To determine

To find: The values of the six trigonometric functions of $\theta$ if $y=\frac{-4}{3}x$ and $\theta$ lies in the fourth quadrant.

Answer to Problem 38E

The values of the six trigonometric functions of $\theta$ are as follows:

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

Explanation of Solution

Given information:

  $y=\frac{-4}{3}x$ where $\theta$ lies in fourth quadrant.

Calculation:

Since, the given equation of the line is $y=\frac{-4}{3}x$ . Compare the given equation with the standard equation of the line $y=mx+c$ .

  $m=\frac{-4}{3}$ and $c=0$

The slope of the line can be calculated by,

  $\begin{array}{c}m=\tan \theta \\ m=\frac{-4}{3}\end{array}$

The slope of the equation is $-\frac{4}{3}$ .

  $\begin{array}{c}\tan \theta =\frac{-4}{3}\\ \tan \theta =\frac{-y}{x}\end{array}$

On caparison, we will get

  $x=3$ and $y=-4$

Applying Pythagoras theorem to calculate the hypotenuse $r$ ,

  $\begin{array}{c}r=\sqrt{x^2+}{y}^2\\ =\sqrt{{(-4)}^2+{\left(3\right)}^2}\\ =\sqrt{16+9}\\ =\sqrt{25}\\ =5\end{array}$

In the fourth quadrant, $\cos \theta >0$ and $sec\theta >0$ only. So,

  $\begin{array}{c}\sin \theta =\frac{-y}{r}\\ =\frac{-4}{5}\end{array}$

and,

  $\begin{array}{c}\cos ec\theta =\frac{1}{\sin \theta }\\ =\frac{-5}{4}\end{array}$

and,

  $\begin{array}{c}\cos =\frac{x}{r}\\ =\frac{3}{5}\end{array}$

and,

  $\begin{array}{c}\sec \theta =\frac{1}{\cos \theta }\\ =\frac{5}{3}\end{array}$

and,

  $\begin{array}{c}\cot \theta =\frac{1}{\tan \theta }\\ =\frac{-3}{4}\end{array}$

Therefore, the exact values of six trigonometric functions are as follows.

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