Question

Find the exact value of the six trigonometric functions of the angle $\theta$in each figure.

Short answer

The required six trigonometric functions are:

$$\begin{gathered} \sin \vartheta =\frac{3}{\sqrt{13}}\cos \theta =\frac{2}{\sqrt{13}} \\ \tan \theta =\frac{3}{2}csc\theta =\frac{\sqrt{13}}{3} \\ sec\theta =\frac{\sqrt{13}}{2}cot\theta =\frac{2}{3} \end{gathered}$$

Step-by-step solution

Step 1. Given Information

From the figure, we have

$adjacent,a=2$ and$opposite,b=3$

Step 2. Formula Used

Pythagorean Theorem: $a^2+b^2=c^2$

where, a, b and c are adjacent, opposite and hypotenuse respectively.

Trigonometric Functions:

$$\begin{gathered} \sin \theta =\frac{b}{c}csc\theta =\frac{c}{b} \\ \cos \theta =\frac{a}{c}sec\theta \frac{c}{a} \\ \tan \theta =\frac{b}{a}cot\theta =\frac{a}{b} \end{gathered}$$

Step 3. Calculation

We have, $adjacent,a=2$and $opposite,b=3$

We have to find hypotenuse, c

Using Pythagorean Theorem,

$c^2=a^2+b^2$

$c^2=2^2+3^2$

$c^2=4+9$

$c^2=13$

$c=\sqrt{13}$

Now, the value of the trigonometric functions are:

$\sin \theta =\frac{b}{c}=\frac{3}{\sqrt{13}}$ $\cos \theta =\frac{a}{c}=\frac{2}{\sqrt{13}}$

$\tan \theta =\frac{b}{a}=\frac{3}{2}$ $csc\theta =\frac{c}{b}=\frac{\sqrt{13}}{3}$

$sec\theta \frac{c}{a}=\frac{\sqrt{13}}{2}$ $cot\theta \frac{a}{b}=\frac{2}{3}$