Question

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

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

Short 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-by-step solution

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,

role="math" localid="1646414505856" $$\begin{gathered} \cos \alpha =\frac{a}{r} \\ =\frac{12}{13} \\ \tan \alpha =\frac{b}{a} \\ =\frac{5}{12} \\ csc\alpha =\frac{r}{b} \\ =\frac{13}{5} \\ sec\alpha =\frac{r}{a} \\ =\frac{13}{12} \\ \mathrm{co}t\alpha =\frac{a}{b} \\ =\frac{12}{5} \end{gathered}$$

Step 3 Now find the exact values for the remaining trigonometric function of the reference angle θ.

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