Chapter 4.4, Problem 36E

To determine

Tofind:The values of the six trigonometric functions of $\theta$ if $y=\frac{1}{3}x$ and $\theta$ lies in first quadrant.

Answer to Problem 36E

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

  $\begin{array}{l}\sin \theta =\frac{\sqrt{10}}{10},\cos \theta =\frac{3\sqrt{10}}{10},\tan \theta =\frac{1}{3}\\ \cos ec\theta =\sqrt{10},\sec \theta =\frac{\sqrt{10}}{3},\cot \theta =3\end{array}$

Explanation of Solution

Given information:

  $y=\frac{1}{3}x$ where $\theta$ lies in first quadrant.

Calculation:

Since, the equation of line is $y=\frac{1}{3}x$ . Compare the given equation with standard equation of line $y=mx+c$ .

  $m=\frac{1}{3}$ and $c=0$

The slope of the line can be given by,

  $\begin{array}{c}m=\tan \theta \\ m=\frac{1}{3}\end{array}$

The slope of the equation is,

  $\begin{array}{l}\tan \theta =\frac{1}{3}\\ \tan \theta =\frac{y}{x}\end{array}$

Compare the above equations.

  $y=1$ and $x=3$

Applying Pythagoras theorem to calculate hypotenuse $r$ ,

  $\begin{array}{c}r=\sqrt{x^2+y^2}\\ =\sqrt{3^2+{\left(1\right)}^2}\\ =\sqrt{10}\end{array}$

In the first quadrant, $\sin \theta >0,\csc \theta >0,\cos \theta >0,\sec \theta >0,\tan \theta >0$ and $\cot \theta >0$ .

So,

  $\begin{array}{c}\sin \theta =\frac{y}{r}\\ =\frac{1}{\sqrt{10}}\times \frac{\sqrt{10}}{\sqrt{10}}\\ =\frac{\sqrt{10}}{10}\end{array}$

and,

  $\begin{array}{c}\cos ec\theta =\frac{1}{\sin \theta }\\ =\sqrt{10}\end{array}$

and,

  $\begin{array}{c}\cos \theta =\frac{x}{r}\\ =\frac{3}{\sqrt{10}}\times \frac{\sqrt{10}}{\sqrt{10}}\\ =\frac{3\sqrt{10}}{10}\end{array}$

and,

  $\begin{array}{c}\sec \theta =\frac{1}{\cos \theta }\\ =\frac{\sqrt{10}}{3}\end{array}$

and,

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

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

  $\begin{array}{l}\sin \theta =\frac{\sqrt{10}}{10},\cos \theta =\frac{3\sqrt{10}}{10},\tan \theta =\frac{1}{3}\\ \cos ec\theta =\sqrt{10},\sec \theta =\frac{\sqrt{10}}{3},\cot \theta =3\end{array}$