Chapter 4, Problem 4CT

To determine

To discuss: the exact values of the other five trigonometric functions when $\tan \theta =\frac{3}{2}$ .

Answer to Problem 4CT

  $\begin{array}{l}\sin \theta =\frac{3\sqrt{13}}{13}\\ \cos \theta =\frac{2\sqrt{13}}{13}\end{array}$

  $\begin{array}{l}\csc \theta =\frac{\sqrt{13}}{3}\\ \sec \theta =\frac{\sqrt{13}}{2}\\ \cot \theta =\frac{2}{3}\end{array}$

Explanation of Solution

It is known that,

  $\begin{array}{c}\tan \theta =\frac{3}{2}\\ \frac{opp}{adj}=\frac{3}{2}\end{array}$

So, a right triangle can be imagined for which the side opposite to $\theta$ is 3 units and adjacent side to $\theta$ is 2 units in length. Then the hypotenuse of that triangle is given by,

Then,

  $\begin{array}{c}hyp=\sqrt{op{p}^2+ad{j}^2}\\ =\sqrt{3^2+2^2}\\ =\sqrt{9+4}\\ =\sqrt{13}\end{array}$

Now,

  $\begin{array}{c}\sin \theta =\frac{opp}{hyp}\\ =\frac{3}{\sqrt{13}}\\ =\frac{3\sqrt{13}}{13}\end{array}$

  $\begin{array}{c}\cos \theta =\frac{adj}{hyp}\\ =\frac{2}{\sqrt{13}}\\ =\frac{2\sqrt{13}}{13}\end{array}$

  $\begin{array}{c}\csc \theta =\frac{hyp}{opp}\\ =\frac{\sqrt{13}}{3}\end{array}$

  $\begin{array}{c}\sec \theta =\frac{hyp}{adj}\\ =\frac{\sqrt{13}}{2}\end{array}$

  $\begin{array}{c}\cot \theta =\frac{adj}{opp}\\ =\frac{2}{3}\end{array}$