Chapter 9.1, Problem 1GP

Solution

To evaluate: the six trigonometric functions of the angle $\theta$ .

  $\sin \theta =\frac{3}{5}$ , $\cos \theta =\frac{4}{5}$ , $\tan \theta =\frac{3}{4}$ , $\cot \theta =\frac{4}{3}$ , $\csc \theta =\frac{5}{3}$ , $\sec \theta =\frac{5}{4}$

Given information:

The triangle:

  

Definition Used:

Let $\theta$ be an acute angle of a right triangle. The six trigonometric functions of $\theta$ are defined as follows:

  $\sin \theta =\frac{opposite}{hypotenuse}$ , $\cos \theta =\frac{adjacent}{hypotenuse}$ , $\tan \theta =\frac{opposite}{adjacent}$

  $\cot \theta =\frac{adjacent}{opposite}$ , $\csc \theta =\frac{hypotenuse}{opposite}$ , $\sec \theta =\frac{hypotenuse}{adjacent}$

Theorem Used:

Pythagorean Theorem:

In a right triangle, the sum of the squares of the lengths of the legs equals the square of the length of the hypotenuse.

  $Hypotenus{e}^2=opposit{e}^2+adjacen{t}^2$

For the given triangle, this is:

  

  $a^2+b^2=c^2$

Explanation:

Consider the given triangle:

  

Here the length of leg opposite to angle $\theta$ is 3 units and the length of the leg adjacent to angle $\theta$ is 4 units.

Use the Pythagoras theorem to find the length of the hypotenuse.

  $\begin{array}{c}Hypotenuse=\sqrt{opposit{e}^2+adjacen{t}^2}\\ =\sqrt{3^2+4^2}\\ =\sqrt{9+16}\end{array}$

  $\begin{array}{c}=\sqrt{25}\\ =5\text{ units}\end{array}$

Now, find the six trigonometric functions,

  $\begin{array}{c}\sin \theta =\frac{opposite}{hypotenuse}\\ =\frac{3}{5}\end{array}$

  $\begin{array}{c}\cos \theta =\frac{adjacent}{hypotenuse}\\ =\frac{4}{5}\end{array}$

  $\begin{array}{c}\tan \theta =\frac{opposite}{adjacent}\\ =\frac{3}{4}\end{array}$

  $\begin{array}{c}\cot \theta =\frac{adjacent}{opposite}\\ =\frac{4}{3}\end{array}$

  $\begin{array}{c}\csc \theta =\frac{hypotenuse}{opposite}\\ =\frac{5}{3}\end{array}$

  $\begin{array}{c}\sec \theta =\frac{hypotenuse}{adjacent}\\ =\frac{5}{4}\end{array}$