Question

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

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

Short 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-by-step solution

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,

role="math" localid="1646419068024" $$\begin{gathered} \sin \alpha =\frac{b}{r} \\ =\frac{1}{\sqrt{5}} \\ =\frac{\sqrt{5}}{5} \\ \cos \alpha =\frac{a}{r} \\ =\frac{2}{\sqrt{5}} \\ =\frac{2\sqrt{5}}{5} \\ \tan \alpha =\frac{b}{a} \\ =\frac{1}{2} \\ csc\alpha =\frac{r}{b} \\ =\frac{\sqrt{5}}{1} \\ =\sqrt{5} \\ sec\alpha =\frac{r}{a} \\ =\frac{\sqrt{5}}{2} \end{gathered}$$

Step 3 The exact values of the remaining trigonometric functions of the angle θ 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.