Chapter 8, Problem 1CT

To determine

To calculate: Theexact value of the six trigonometric functions of the angle $\theta$ for the figurebelow.

  

Answer to Problem 1CT

Solution:

The exact values of six trigonometric functions are

  $\sin \theta =\frac{\sqrt{5}}{5},\cos \theta =\frac{2\sqrt{5}}{5},\tan \theta =\frac{1}{2},\csc \theta =\sqrt{5},\sec \theta =\frac{\sqrt{5}}{2},\cot \theta =2$ .

Explanation of Solution

Given Information:

The figure

  

Formula used:

Ratios of the sides of a right triangle:

  $\begin{array}{cc}\sin \theta =\frac{\text{Opposite}}{\text{Hypotenuse}}& \csc \theta =\frac{\text{Hypotenuse}}{\text{Opposite}}\\ \cos \theta =\frac{\text{Adjacent}}{\text{Hypotenuse}}& \sec \theta =\frac{\text{Hypotenuse}}{\text{Adjacent}}\\ \tan \theta =\frac{\text{Opposite}}{\text{Adjacent}}& \cot \theta =\frac{\text{Adjacent}}{\text{Opposite}}\end{array}$

Pythagorean Theorem: In a right angle triangle, the square of the length of the hypotenuse is the sum of the squares of the other two sides.

  ${\left(\text{Hypotenuse}\right)}^2={\left(\text{Opposite}\right)}^2+{\left(\text{Adjacent}\right)}^2$

Calculation:

In the given triangle, the length of the opposite side is $3$, and the length of the adjacent side is $6$ .

By using Pythagorean Theorem ${\left(\text{Hypotenuse}\right)}^2={\left(\text{Opposite}\right)}^2+{\left(\text{Adjacent}\right)}^2$ ,

  $\Rightarrow {\left(\text{Hypotenuse}\right)}^2=3^2+6^2$

  $\Rightarrow {\left(\text{Hypotenuse}\right)}^2=9+36$

  $\Rightarrow {\left(\text{Hypotenuse}\right)}^2=45$

  $\Rightarrow \text{Hypotenuse}=\sqrt{45}=3\sqrt{5}$

Thus, the length of the hypotenuse is $3\sqrt{5}$ .

Now, by using ratios of the sides of the right triangle,

  $\sin \theta =\frac{\text{Opposite}}{\text{Hypotenuse}}$

  $\Rightarrow \sin \theta =\frac{3}{3\sqrt{5}}$

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

  $\cos \theta =\frac{\text{Adjacent}}{\text{Hypotenuse}}$

  $\Rightarrow \cos \theta =\frac{6}{3\sqrt{5}}$

  $\Rightarrow \cos \theta =\frac{2}{\sqrt{5}}=\frac{2\sqrt{5}}{5}$

  $\tan \theta =\frac{\text{Opposite}}{\text{Adjacent}}$

  $\Rightarrow \tan \theta =\frac{3}{6}$

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

  $\csc \theta =\frac{\text{Hypotenuse}}{\text{Opposite}}$

  $\Rightarrow \csc \theta =\frac{3\sqrt{5}}{3}$

  $\Rightarrow \csc \theta =\sqrt{5}$

  $\sec \theta =\frac{\text{Hypotenuse}}{\text{Adjacent}}$

  $\Rightarrow \sec \theta =\frac{3\sqrt{5}}{6}$

  $\Rightarrow \sec \theta =\frac{\sqrt{5}}{2}$

  $\cot \theta =\frac{\text{Adjacent}}{\text{Opposite}}$

  $\Rightarrow \cot \theta =\frac{6}{3}$

  $\Rightarrow \cot \theta =2$

Thus, the exact values of six trigonometric functions are

  $\sin \theta =\frac{\sqrt{5}}{5}$, $\cos \theta =\frac{2\sqrt{5}}{5}$, $\tan \theta =\frac{1}{2}$, $\csc \theta =\sqrt{5}$, $\sec \theta =\frac{\sqrt{5}}{2}$, and $\cot \theta =2$ .