Q.52 - Precalculus Enhanced with Graphing Utilities

Question

Find the exact value of each of the remaining trigonometric functions of $\mathrm{\theta }.$

$\mathrm{cos\theta }=-\frac{1}{4},\mathrm{tan\theta }>0.$

Step-by-Step Solution

Answer

The exact values are,

$$\begin{gathered} \mathrm{sin\theta }=-\frac{\sqrt{15}}{4} \\ \mathrm{csc\theta }=-\frac{4\sqrt{15}}{15} \\ \mathrm{sec\theta }=-4 \\ \mathrm{cot\theta }=\frac{\sqrt{15}}{15} \end{gathered}$$

Step 1. Given Information

Given that $\mathrm{cos\theta }=-\frac{1}{4},\mathrm{tan\theta }>0$. To find the exact values for trigonometric functions of angle.

Step 2. Radius of the circle

The angle corresponds to the point $\mathrm{P}=(\mathrm{x},\mathrm{y})$ where terminal point is circle. The circle's radius is ${\mathrm{r}}^2={\mathrm{x}}^2+{\mathrm{y}}^2.$The values are $\mathrm{cos\theta }=\frac{\mathrm{x}}{\mathrm{r}},\mathrm{x}=-1,\mathrm{r}=4.$ The tan functions have sine and cosine which occurs in the III and I quadrant. Mainly, $\mathrm{x}=-1<0$ must in III quadrant. On solving the value of y,

$$\begin{gathered} {\mathrm{y}}^2={\mathrm{r}}^2-{\mathrm{x}}^2 \\ {\mathrm{y}}^2={\left(4\right)}^2-{(-1)}^2=16-1 \\ {\mathrm{y}}^2=15 \\ \mathrm{y}=\sqrt{15} \\ \mathrm{y}=-\sqrt{15} \end{gathered}$$

Step 3. Finding the exact values

To find the exact values are,

$$\begin{gathered} \mathrm{sin\theta }=\frac{\mathrm{y}}{\mathrm{r}} \\ \mathrm{sin\theta }=-\frac{\sqrt{15}}{4} \\ {\mathrm{y}}^2={\mathrm{r}}^2-{\mathrm{x}}^2 \\ ={\left(4\right)}^2-{(-1)}^2{\mathrm{y}}^2=15 \\ \mathrm{y}=\sqrt{15} \\ \mathrm{in} \mathrm{third} \mathrm{quadrant}, \\ \mathrm{y}=-\sqrt{15} \end{gathered}$$

Step 4. Substitution

On substituting the values to get the exact values,

$$\begin{gathered} \mathrm{csc\theta }=\frac{\mathrm{r}}{\mathrm{y}} \\ =-\frac{4}{\sqrt{15}}.\frac{\sqrt{15}}{\sqrt{15}} \\ \mathrm{csc\theta }=-\frac{4\sqrt{15}}{15} \\ \mathrm{sec\theta }=\frac{\mathrm{r}}{\mathrm{x}} \\ =-\frac{4}{1} \\ \mathrm{sec\theta }=-4 \\ \mathrm{cot\theta }=\frac{\mathrm{x}}{\mathrm{y}} \\ =\frac{1}{\sqrt{15}}.\frac{\sqrt{15}}{\sqrt{15}} \\ \mathrm{cot\theta }=\frac{\sqrt{15}}{15} \end{gathered}$$