Q. 20 - Precalculus Enhanced with Graphing Utilities

Question

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

$\sin \theta =-\frac{5}{13},\frac{3\mathrm{\pi }}{2}<\theta <2\mathrm{\pi }$

Step-by-Step Solution

Answer

The exact values of the remaining trigonometric functions of $\sin \theta =-\frac{5}{13}$are,

$$\begin{gathered} \sin \theta =-\frac{5}{13} \\ \cos \theta =\frac{12}{13} \\ \tan \theta =-\frac{5}{12} \\ csc \theta =-\frac{13}{5} \\ sec \theta =\frac{13}{12} \\ \mathrm{co}t \theta =-\frac{12}{5} \end{gathered}$$

Step 1 Given expression is,

$\sin \theta =-\frac{5}{13}$ and $\frac{3\mathrm{\pi }}{2}<\theta <2\mathrm{\pi }$.

So,$\theta$ lies in the fourth quadrant, where $\cos \theta , sec \theta$ are positive and the remaining trigonometric functions are negative.

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

So,$\sin \alpha =\frac{5}{13}=\frac{Opposite}{Hypotenuse}$.

Use the values $b=5,r=13$ to construct a triangle.

From the figure, we obtain $a=12$.

Step 2 Now calculate the remaining trigonometric functions.

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

$$\begin{gathered} \cos \alpha =\frac{a}{r} \\ =\frac{12}{13} \\ \tan \alpha =\frac{b}{a} \\ =\frac{5}{12} \\ csc \alpha =\frac{r}{b} \\ =\frac{13}{5} \\ sec \alpha =\frac{r}{a} \\ =\frac{13}{12} \\ \mathrm{co}t \alpha =\frac{a}{b} \\ =\frac{12}{5} \end{gathered}$$

Step 3 Now find the exact values for the remaining trigonometric function of the reference angle &#952; .

$$\begin{gathered} \sin \theta =-\frac{5}{13} \\ \cos \theta =\frac{12}{13} \\ \tan \theta =-\frac{5}{12} \\ csc \theta =-\frac{13}{5} \\ sec \theta =\frac{13}{12} \\ \mathrm{co}t \theta =-\frac{12}{5} \end{gathered}$$