Chapter 9, Problem 3T

Solution

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

  $\sin \theta =\frac{\sqrt{13}}{7},\cos \theta =\frac{6}{7},\tan \theta =\frac{\sqrt{13}}{6},\cot \theta =\frac{6\sqrt{13}}{13},\csc \theta =\frac{7\sqrt{13}}{13},\sec \theta =\frac{7}{6}$

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:

  $\begin{array}{l}\sin \theta =\frac{opposite}{hypotenuse},\cos \theta =\frac{adjacent}{hypotenuse},\tan \theta =\frac{opposite}{adjacent}\\ \cot \theta =\frac{adjacent}{opposite},\csc \theta =\frac{hypotenuse}{opposite},\sec \theta =\frac{hypotenuse}{adjacent}\end{array}$

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 adjacent.

  $\begin{array}{c}Hypotenus{e}^2=opposit{e}^2+adjacen{t}^2\\ adjacen{t}^2=Hypotenus{e}^2-opposit{e}^2\\ adjacent=\sqrt{Hypotenus{e}^2-opposit{e}^2}\end{array}$

So,

  $\begin{array}{c}adjacent=\sqrt{7^2-{\left(\sqrt{13}\right)}^2}\\ =\sqrt{49-13}\\ =\sqrt{36}\\ =6\end{array}$

Now, find the six trigonometric functions,

  $\begin{array}{c}\sin \theta =\frac{opposite}{hypotenuse}\\ =\frac{\sqrt{13}}{7}\\ \cos \theta =\frac{adjacent}{hypotenuse}\\ =\frac{6}{7}\\ \tan \theta =\frac{opposite}{adjacent}\\ =\frac{\sqrt{13}}{6}\\ \cot \theta =\frac{adjacent}{opposite}\\ =\frac{6}{\sqrt{13}}\\ =\frac{6}{\sqrt{13}}\times \frac{\sqrt{13}}{\sqrt{13}}\\ =\frac{6\sqrt{13}}{13}\\ \csc \theta =\frac{hypotenuse}{opposite}\\ =\frac{7}{\sqrt{13}}\\ =\frac{7}{\sqrt{13}}\times \frac{\sqrt{13}}{\sqrt{13}}\\ =\frac{7\sqrt{13}}{13}\\ \sec \theta =\frac{hypotenuse}{adjacent}\\ =\frac{7}{6}\end{array}$