Chapter 4.4, Problem 30E

To determine

To find: The values of the six trigonometric functions of $\theta$ if $\cos ec\theta =4$ and $\cot \theta <0$ .

Answer to Problem 30E

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

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

Explanation of Solution

Given information:

  $\cos ec\theta =4$ and $\cot \theta <0$ .

Calculation:

Since, $\cos ec\theta >0$ in the first and the second quadrant. It is given that $\cot \theta <0$ . $\cot \theta <0$ in the second and the fourth quadrant. That’s why $\theta$ lies in the second 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, $\cos ec\theta =4$ lies in the second quadrant.

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

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

  

From the co-ordinate plane, the base is $-x$ and perpendicular is $y=1$ and the hypotenuse of the triangle is $r=4$ .

Apply Pythagoras Theorem to calculate the base $x$ .

  $\begin{array}{c}r=\sqrt{{\left(x\right)}^2+{\left(y\right)}^2}\\ 4=\sqrt{x^2+{\left(1\right)}^2}\\ 16=x^2+1\end{array}$

Further simplify it.

  $\begin{array}{c}{x}^2=16-1\\ x^2=15\\ x=\pm \sqrt{15}\end{array}$

Since, $\theta$ lies in second quadrant. The base is $-\sqrt{15}$ .

Calculate the trigonometric functions.

  $\begin{array}{c}\sin \theta =\frac{y}{r}\\ =\frac{1}{4}\end{array}$

and

  $\begin{array}{c}\cos \theta =\frac{x}{r}\\ =\frac{-\sqrt{15}}{4}\end{array}$

and

  $\begin{array}{c}\tan \theta =\frac{y}{x}\\ =\frac{1}{-\sqrt{15}}\\ =\frac{-1}{\sqrt{15}}\times \frac{\sqrt{15}}{\sqrt{15}}\\ =-\frac{\sqrt{15}}{15}\end{array}$

and

  $\begin{array}{c}\text{cosec}\theta =\frac{1}{\sin \theta }\\ =4\end{array}$

and

  $\begin{array}{c}\sec \theta =\frac{1}{\cos \theta }\\ =\frac{-4}{\sqrt{15}}\\ =\frac{-4}{\sqrt{15}}\times \frac{\sqrt{15}}{\sqrt{15}}\\ =\frac{-4\sqrt{15}}{15}\end{array}$

and

  $\begin{array}{c}\text{cot}\theta =\frac{1}{\tan \theta }\\ =-\sqrt{15}\end{array}$

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

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