Q. 57 - Precalculus Enhanced with Graphing Utilities

Question

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

$\mathrm{tan\theta }=-\frac{1}{3},\mathrm{sin\theta }>0$

Step-by-Step Solution

Answer

The values of trigonometric functions are,

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

Step 1. Given Information

Given that the trigonometric function is $\mathrm{tan\theta }=-\frac{1}{3},\mathrm{sin\theta }>0.$ To find the exact values of trigonometric function using equation substitution.

Step 2. Solving a equation

It implies that $\mathrm{cos\theta }<0,$ it appears in II quadrant which is $\mathrm{cot\theta }=-3.$ On using the equation to solve,

$$\begin{gathered} {\tan }^2\mathrm{\theta }+1={\sec }^2\mathrm{\theta } \\ \mathrm{sec\theta }=\pm \sqrt{{\tan }^2\mathrm{\theta }+1} \end{gathered}$$

On substituting the equation,

$$\begin{gathered} \mathrm{sec\theta }=-\sqrt{1+{(-\frac{1}{3})}^2} \\ =-\sqrt{1+\frac{1}{9}} \\ \mathrm{sec\theta }=-\frac{\sqrt{10}}{3} \end{gathered}$$

Step 3. Solving the trigonometric function

On solving the rest of trigonometric function,

$$\begin{gathered} \mathrm{cos\theta }=\frac{1}{sec\mathrm{\theta }} \\ =\frac{1}{-{\displaystyle \frac{\sqrt{10}}{3}}} \\ \mathrm{cos\theta }=-\frac{3\sqrt{10}}{10} \end{gathered}$$

On solving the equation of $\mathrm{sin\theta }=\pm \sqrt{1-{\cos }^2\mathrm{\theta }}$

It appears in II quadrant with $\mathrm{sin\theta }>0.$

$$\begin{gathered} \mathrm{sin\theta }=\sqrt{1-{\cos }^2\mathrm{\theta }} \\ =\sqrt{1-{(-\frac{3\sqrt{10}}{10})}^2} \\ \mathrm{sin\theta }=\frac{\sqrt{10}}{10} \\ \mathrm{csc\theta }=\frac{1}{\mathrm{sin\theta }} \\ =\frac{1}{{\displaystyle \frac{\sqrt{10}}{10}}} \\ \mathrm{csc\theta }=\sqrt{10} \end{gathered}$$