Question

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

Short answer

The ratio are,

$$\begin{gathered} \sin \theta =\frac{\sqrt{5}}{5} \\ \cos \theta =\frac{2\sqrt{5}}{5} \\ \tan \theta =\frac{1}{2} \\ \mathrm{co}t\theta =2 \\ csc\theta =\sqrt{5} \\ sec\theta =\frac{\sqrt{5}}{2} \end{gathered}$$

Step-by-step solution

Step 1. Given Information

The given triangle is,

Step 2. Determine the hypotenuse

• In the given triangle, $p=3,b=6$.
• Use Pythagoras theorem, to determine the hypotenuse.

role="math" localid="1646825617277" $\begin{array}{rcl}{h}^2& =& p^2+b^2\\ h^2& =& 3^2+6^2\\ h^2& =& 9+36\\ h& =& \sqrt{45}\\ & =& 3\sqrt{5}\end{array}$

Step 3. Determine the trigonometric ratios.

Determine the trigonometric ratios.

$$\begin{gathered} \sin \theta =\frac{p}{h}=\frac{3}{3\sqrt{5}}=\frac{1}{\sqrt{5}}=\frac{\sqrt{5}}{5} \\ \cos \theta =\frac{b}{h}=\frac{6}{3\sqrt{5}}=\frac{2}{\sqrt{5}}=\frac{2\sqrt{5}}{5} \\ \tan \theta =\frac{p}{b}=\frac{3}{6}=\frac{1}{2} \\ \mathrm{co}t\theta =\frac{b}{p}=\frac{6}{3}=2 \\ csc\theta =\frac{h}{p}=\frac{3\sqrt{5}}{3}=\sqrt{5} \\ sec\theta =\frac{h}{b}=\frac{3\sqrt{5}}{6}=\frac{\sqrt{5}}{2} \end{gathered}$$