Chapter 12.2, Problem 4WE

To determine

Find the values of the six trigonometric functions of $\theta$ .

Answer to Problem 4WE

  $\sin \theta =\frac{\sqrt{5}}{5}$

  $\cos \theta =\frac{2\sqrt{5}}{5}$

  $\tan \theta =\frac{1}{2}$

  $\csc \theta =\sqrt{5}$

  $\sec \theta =\frac{\sqrt{5}}{2}$

  $\cot \theta =2$

Explanation of Solution

Given:

  

Formula Used:

  $\sin \theta =\frac{\text{length of side opposite to }\theta }{\text{length of the hypotenuse}}$

  $\cos \theta =\frac{\text{length of side adjacent to }\theta }{\text{length of the hypotenuse}}$

  $\tan \theta =\frac{\text{length of side opposite to }\theta }{\text{length of side adjacent to }\theta }$

  $\csc \theta =\frac{1}{\sin \theta }$

  $\sec \theta =\frac{1}{\cos \theta }$

  $\cot \theta =\frac{1}{\tan \theta }$

Calculation:

In the triangle ,

x = length of side adjacent to $\theta$ = 2 units

y = length of side opposite to $\theta$= 1 units

r = length of the hypotenuse = $\sqrt{x^2+y^2}=\sqrt{4+1}=\sqrt{5}$

So,

  $\begin{array}{l}\sin \theta =\frac{\text{length of side opposite to }\theta }{\text{length of the hypotenuse}}=\frac{y}{r}=\frac{1}{\sqrt{5}}\\ \Rightarrow \sin \theta =\frac{1}{\sqrt{5}}\cdot \frac{\sqrt{5}}{\sqrt{5}}\mathrm{....}[rationalize]\\ \Rightarrow \sin \theta =\frac{\sqrt{5}}{5}\end{array}$

  $\begin{array}{l}\cos \theta =\frac{\text{length of side adjacent to }\theta }{\text{length of the hypotenuse}}=\frac{x}{r}=\frac{2}{\sqrt{5}}\\ \Rightarrow \cos \theta =\frac{2\sqrt{5}}{5}\end{array}$

  $\begin{array}{l}\tan \theta =\frac{\text{length of side opposite to }\theta }{\text{length of side adjacent to }\theta }=\frac{1}{2}\\ \Rightarrow \tan \theta =\frac{1}{2}\end{array}$

  $\csc \theta =\frac{1}{\sin \theta }=\sqrt{5}$

  $\sec \theta =\frac{1}{\cos \theta }=\frac{\sqrt{5}}{2}$

  $\cot \theta =\frac{1}{\tan \theta }=2$