Q. 55 - Precalculus Enhanced with Graphing Utilities

Question

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

$\mathrm{tan\theta }=\frac{3}{4},\mathrm{sin\theta }<0$

Step-by-Step Solution

Answer

The exact values of trigonometric functions are,

$$\begin{gathered} \mathrm{sec\theta }=-\frac{5}{4} \\ \mathrm{cos\theta }=-\frac{4}{5} \\ \mathrm{sin\theta }=-\frac{3}{5} \\ \mathrm{cot\theta }=\frac{4}{3} \\ \mathrm{csc\theta }=-\frac{5}{3} \end{gathered}$$

Step 1. Given Information

Given trigonometric functions is $\mathrm{tan\theta }=\frac{3}{4},\mathrm{sin\theta }<0.$ To find the exact values of trigonometric functions using substitution.

Step 2. Solve a equation

It implies that $\mathrm{cos\theta }<0$ and angle is III quadrant. On using the equation $$\begin{gathered} {\tan }^2\mathrm{\theta }+1={\sec }^2\mathrm{\theta } \\ {\sec }^2\mathrm{\theta }=\pm \sqrt{{\tan }^2\mathrm{\theta }+1} \end{gathered}$$

Step 3. Substitution to find trigonometric function

On substituting the values,

$$\begin{gathered} \mathrm{sec\theta }=-\sqrt{1+{\left(\frac{3}{4}\right)}^2} \\ =-\sqrt{1+\frac{9}{16}} \\ =-\sqrt{\frac{25}{16}} \\ \mathrm{sec\theta }=-\frac{5}{4} \\ \mathrm{cos\theta }=\frac{1}{\mathrm{sec\theta }} \\ =\frac{1}{-{\displaystyle \frac{5}{4}}} \\ \mathrm{cos\theta }=-\frac{4}{5} \end{gathered}$$

Step 4. Substitution for sin, cot and csc functions

On using the equation of $$\begin{gathered} \mathrm{sin\theta }=\pm \sqrt{1-{\cos }^2\mathrm{\theta }} \\ \mathrm{sin\theta }=-\sqrt{1-{\left(-\frac{4}{5}\right)}^2} \\ =-\sqrt{1-\frac{16}{25}} \\ \mathrm{sin\theta }=-\frac{3}{5} \\ \mathrm{cot\theta }=\frac{1}{\mathrm{tan\theta }} \\ =\frac{1}{{\displaystyle \frac{3}{4}}} \\ \mathrm{cot\theta }=\frac{4}{3} \\ \mathrm{csc\theta }=\frac{1}{\mathrm{sin\theta }} \\ =\frac{1}{-{\displaystyle \frac{3}{5}}} \\ \mathrm{csc\theta }=-\frac{5}{3} \end{gathered}$$