Q. 22 - Precalculus Enhanced with Graphing Utilities

Question

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

$sec \theta =3$,$\frac{3\mathrm{\pi }}{2}<\theta <2\mathrm{\pi }$

Step-by-Step Solution

Answer

The exact values of the remaining trigonometric functions of $sec \theta =3$ are,

$$\begin{gathered} \sin \theta =-\frac{2\sqrt{2}}{3} \\ \cos \theta =\frac{1}{3} \\ \tan \theta =-2\sqrt{2} \\ csc \theta =-\frac{3\sqrt{2}}{4} \\ cot \theta =-\frac{1}{2\sqrt{2}} \\ =-\frac{\sqrt{2}}{4} \end{gathered}$$

Step 1 Given expression is,

$sec \theta =3$ and $\frac{3\mathrm{\pi }}{2}<\theta <2\pi$.

So,$\theta$ lies in the fourth quadrant where $sec \theta$ is positive.

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

So,$sec \alpha =\frac{3}{1}=\frac{Opposite}{Adjacent}$

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

From the figure, we obtain $b=2\sqrt{2}$.

Step 2 Now calculate the remaining trigonometric functions.

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

$$\begin{gathered} \sin \alpha =\frac{b}{r} \\ =\frac{2\sqrt{2}}{3} \\ \cos \alpha =\frac{a}{r} \\ =\frac{1}{3} \\ \tan \alpha =\frac{b}{a} \\ =2\sqrt{2} \\ csc \alpha =\frac{r}{b} \\ =\frac{3\sqrt{2}}{4} \\ cot \alpha =\frac{a}{b} \\ =\frac{1}{2\sqrt{2}} \\ =\frac{\sqrt{2}}{4} \end{gathered}$$

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

$$\begin{gathered} \sin \theta =-\frac{2\sqrt{2}}{3} \\ \cos \theta =\frac{1}{3} \\ \tan \theta =-2\sqrt{2} \\ csc \theta =-\frac{3\sqrt{2}}{4} \\ cot \theta =-\frac{1}{2\sqrt{2}} \\ =-\frac{\sqrt{2}}{4} \end{gathered}$$

These are the exact values of the remaining trigonometric functions.