Chapter 4.4, Problem 31E

To determine

To find: The values of the six trigonometric functions of $\theta$ if $\sec \theta =-2$ where $0\le \theta \le \pi$ .

Answer to Problem 31E

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

  $\begin{array}{l}\sin \theta =\frac{\sqrt{3}}{2},\cos \theta =\frac{-1}{2},\tan \theta =-\sqrt{3}\\ \cos ec\theta =\frac{2\sqrt{3}}{3},\sec \theta =-2,\cot \theta =\frac{-\sqrt{3}}{3}\end{array}$

Explanation of Solution

Given information:$\sec \theta =-2$ where $0\le \theta \le \pi$ .

Calculation:

Since, $\sec \theta <0$ in the second and the third quadrant. It is given that $0\le \theta \le \pi$ .That’s why $\theta$ lies in the second quadrant because $\theta >0$ in the first quadrant.

In second quadrant, $\sin \theta >0$ and $\cos ec\theta >0$ only whereas $\tan \theta <0,\cot \theta <0,\cos \theta <0$ and $\sec \theta <0.$

Since, $\sec \theta =-2$ lies in the second quadrant.

  $\begin{array}{l}\sec \theta =\frac{1}{\cos \theta }\\ \cos \theta =\frac{-1}{2}\end{array}$

Let us suppose point A $\left(-1,y\right)$ is drawn on the line with angle $\theta$ .

  

From the co-ordinate plane, the base is $-1$ and perpendicular is $y$ and the hypotenuse of the triangle is $r=2$ .

Apply Pythagoras Theorem to calculate the perpendicular $y$ .

  $\begin{array}{c}r=\sqrt{{\left(x\right)}^2+{\left(y\right)}^2}\\ 2=\sqrt{{(-1)}^2+y^2}\\ 4=y^2+1\end{array}$

Further simplify the above equation.

  $\begin{array}{c}{y}^2=4-1\\ x^2=3\\ x=\pm \sqrt{3}\end{array}$

Since, $\theta$ lies in the second quadrant. The perpendicular is $\sqrt{3}$ .

Calculate the trigonometric functions.

  $\begin{array}{c}\sin \theta =\frac{y}{r}\\ =\frac{\sqrt{3}}{2}\end{array}$

and

  $\begin{array}{c}\cos \theta =\frac{x}{r}\\ =\frac{-1}{2}\end{array}$

and

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

and

  $\begin{array}{c}\text{cosec}\theta =\frac{1}{\sin \theta }\\ =\frac{2}{\sqrt{3}}\\ =\frac{2}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}\\ =\frac{2\sqrt{3}}{3}\end{array}$

and

  $\begin{array}{c}\sec \theta =\frac{1}{\cos \theta }\\ =-2\end{array}$

and

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

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

  $\begin{array}{l}\sin \theta =\frac{\sqrt{3}}{2},\cos \theta =\frac{-1}{2},\tan \theta =-\sqrt{3}\\ \cos ec\theta =\frac{2\sqrt{3}}{3},\sec \theta =-2,\cot \theta =\frac{-\sqrt{3}}{3}\end{array}$