Q. 21 - Precalculus Enhanced with Graphing Utilities

Question

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

$\tan \theta =\frac{1}{3}, {180}^{°}<\theta <{270}^{°}$

Step-by-Step Solution

Answer

The exact values of the remaining trigonometric functions of $\tan \theta =\frac{1}{3}, {180}^{°}<\theta <{270}^{°}$are,

$$\begin{gathered} \sin \theta =-\frac{\sqrt{10}}{10} \\ \cos \theta =-\frac{3\sqrt{10}}{10} \\ csc \theta =-\sqrt{10} \\ sec \theta =-\frac{\sqrt{10}}{3} \\ cot \theta =3 \end{gathered}$$

Step 1 Given expression is,

$\tan \theta =\frac{1}{3}$ and ${180}^{°}<\theta <{270}^{°}$.

So,$\theta$ lies in the third quadrant and where $\tan \theta$ is positive.

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

So,$\tan \alpha =\frac{1}{3}=\frac{opposite}{adjacent}$.

Use the values $b=1,a=3$ to construct a triangle.

From the figure, we obtain $r=\sqrt{10}$.

Step 2 Now calculate the remaining trigonometric functions.

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

$$\begin{gathered} \sin \alpha =\frac{b}{r} \\ =\frac{1}{\sqrt{10}} \\ \cos \alpha =\frac{a}{r} \\ =\frac{3}{\sqrt{10}} \\ csc \alpha =\frac{r}{b} \\ =\frac{\sqrt{10}}{1} \\ =\sqrt{10} \\ sec \alpha =\frac{\sqrt{10}}{3} \\ cot \alpha =\frac{3}{1} \\ =3 \end{gathered}$$

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

$$\begin{gathered} \sin \theta =-\frac{\sqrt{10}}{10} \\ \cos \theta =-\frac{3\sqrt{10}}{10} \\ csc \theta =-\sqrt{10} \\ sec \theta =-\frac{\sqrt{10}}{3} \\ cot \theta =3 \end{gathered}$$

These are the exact values of the remaining trigonometric functions.