Q. 23 - Precalculus Enhanced with Graphing Utilities

Question

In Problems 16–23, find the exact value of each of the remaining trigonometric functions.

$cot \theta =-2,\frac{\mathrm{\pi }}{2}<\theta <\pi$.

Step-by-Step Solution

Answer

The exact values of the remaining trigonometric functions of $cot \theta =-2$ are,

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

Step 1 Given expression is,

$cot \theta = -2$ and $\frac{\mathrm{\pi }}{2}<\theta <2\pi$.

So,$\theta$ lies in the second quadrant where $cot \theta$ is negative.

Let $\alpha$ be the reference angle for $\theta$ then $cot \theta =\frac{a}{b}=cot \alpha$.

So,$cot \alpha =\frac{2}{1}=\frac{Adjacent}{Opposite}$

Use the values $a=2,b=1$ to construct a triangle.

From the figure, we can obtain $r=\sqrt{5}$. 

Step 2 Now find the exact values of the remaining trigonometric functions.

The exact values of the remaining trigonometric functions of the reference angle $\alpha$ are,

$$\begin{gathered} \sin \alpha =\frac{b}{r} \\ =\frac{1}{\sqrt{5}} \\ =\frac{\sqrt{5}}{5} \\ \cos \alpha =\frac{a}{r} \\ =\frac{2}{\sqrt{5}} \\ =\frac{2\sqrt{5}}{5} \\ \tan \alpha =\frac{b}{a} \\ =\frac{1}{2} \\ csc \alpha =\frac{r}{b} \\ =\frac{\sqrt{5}}{1} \\ =\sqrt{5} \\ sec \alpha =\frac{r}{a} \\ =\frac{\sqrt{5}}{2} \end{gathered}$$

Step 3 The exact values of the remaining trigonometric functions of the angle &#952; are,

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

These are the exact values of the remaining trigonometric functions.