Q 54. - Precalculus Enhanced with Graphing Utilities

Question

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

$\mathrm{csc\theta }=3,\mathrm{cot\theta }<0.$

Step-by-Step Solution

Answer

The exact values are,

$$\begin{gathered} \mathrm{cos\theta }=-\frac{2\sqrt{2}}{3} \\ \mathrm{tan\theta }=-\frac{\sqrt{2}}{4} \\ \mathrm{cot\theta }=-2\sqrt{2} \\ \mathrm{sec\theta }=-\frac{3\sqrt{2}}{4} \end{gathered}$$

Step 1. Given information

Given that the trigonometric function is $\mathrm{csc\theta }=3,\mathrm{cot\theta }<0.$ To find the exact values using substitution.

Step 2. Pythagorean theorem

It implies that $\mathrm{sin\theta }=\frac{1}{\mathrm{csc\theta }}=\frac{1}{3}.$ Since $\mathrm{cot\theta }<0,\mathrm{sin\theta }=\frac{1}{3}>0$ which implies that $\mathrm{cos\theta }<0.$ On using the Pythagorean theorem,

$$\begin{gathered} {\sin }^2\mathrm{\theta }+{\cos }^2\mathrm{\theta }=1 \\ {\cos }^2\mathrm{\theta }=1-{\sin }^2\mathrm{\theta } \\ \mathrm{cos\theta }=\pm \sqrt{1-{\sin }^2\mathrm{\theta }} \end{gathered}$$

Step 3. Substitution to find the values

On substituting the values in the above equation,

$$\begin{gathered} \mathrm{cos\theta }=-\sqrt{1-{\left(\frac{1}{3}\right)}^2} \\ =-\sqrt{1-\frac{1}{9}} \\ \mathrm{cos\theta }=-\frac{2\sqrt{2}}{3} \\ \mathrm{tan\theta }=\frac{\mathrm{y}}{\mathrm{x}} \\ =\frac{\mathrm{sin\theta }}{\mathrm{cos\theta }} \\ =\frac{{\displaystyle \frac{1}{3}}}{-{\displaystyle \frac{2\sqrt{2}}{3}}} \\ \mathrm{tan\theta }=-\frac{\sqrt{2}}{4} \end{gathered}$$

Step 4. Substitution to find the values of cot and sec

The reciprocal of cot is tan. On substituting the values,

$$\begin{gathered} \mathrm{cot\theta }=\frac{1}{\mathrm{tan\theta }} \\ =\frac{1}{-{\displaystyle \frac{\sqrt{2}}{4}}} \\ \mathrm{cot\theta }=-2\sqrt{2} \\ \mathrm{sec\theta }=\frac{1}{\mathrm{cos\theta }} \\ =\frac{1}{-{\displaystyle \frac{2\sqrt{2}}{3}}} \\ \mathrm{sec\theta }=-\frac{3\sqrt{2}}{4} \end{gathered}$$