Q.50 - Precalculus Enhanced with Graphing Utilities

Question

Find the exact value of each of the remaining trigonometric functions of  $\theta$

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

Step-by-Step Solution

Answer

The values are 

$\cos \theta =\frac{-\sqrt{5}}{3}, csc\theta =\frac{-3}{2},sec\theta =\frac{-3}{\sqrt{5}},\tan \theta =\frac{2}{\sqrt{5}},cot\theta =\frac{\sqrt{5}}{2}$

Step 1 .Given information

$\sin \theta =\frac{-2}{3}, \mathrm{\pi }<\mathrm{\theta }<\frac{3\mathrm{\pi }}{2}$ is given.

 we have to  find the exact value of each of the remaining trigonometric functions  

Step 2.Finding the value of cos &#952;

$$\begin{gathered} Since {\sin }^2\theta +{\cos }^2\theta =1 , here \sin \theta =\frac{-2}{3} \\ \cos \theta =\pm \sqrt{1-{\sin }^2\theta } , here \theta lies on \mathrm{\pi }<\theta <\frac{3\mathrm{\pi }}{2} so \sin \theta is on third quadrant and all other except \tan \theta and cot\theta are will be negative. \\ hence \cos \theta =\sqrt{1-{\sin }^2\theta } \\ =-\sqrt{1-\frac{{(-2)}^2}{3^2}} \\ =-\sqrt{1-\frac{4}{9}} \\ =-\sqrt{\frac{5}{9}} \\ =\frac{-\sqrt{5}}{3} \end{gathered}$$

Step 3. Find the values of c s c &#952; &#160; a n d &#160; &#160; s e c &#952;

$$\begin{gathered} \mathrm{In} \mathrm{this} \mathrm{question} \mathrm{csc\theta }=\frac{1}{\mathrm{sin\theta }} \\ =\frac{1}{{\displaystyle -\frac{2}{3}}} , \mathrm{sin\theta }=-\frac{2}{3} \\ =-\frac{3}{2} \\ \mathrm{cos\theta }=\frac{-\sqrt{5}}{ 3} ,\mathrm{then} \mathrm{sec\theta }=\frac{1}{\mathrm{cos\theta }} \\ =\frac{1}{{\displaystyle \frac{-\sqrt{5}}{3}}} \\ =\frac{-3}{\sqrt{5}} \end{gathered}$$

Step 4. Finding the value of tan &#160; &#952; &#160; a n d &#160; c o t &#952;

$$\begin{gathered} we know that \tan \theta =\frac{\sin \theta }{\cos \theta } ,\sin \theta =\frac{-2}{3} and \cos \theta =\frac{-\sqrt{5}}{3} \\ =\frac{{\displaystyle \frac{{\displaystyle -2}}{{\displaystyle 3}}}}{{\displaystyle \frac{{\displaystyle -\sqrt{5}}}{{\displaystyle 3}}}} \\ =\frac{2}{\sqrt{5}} \\ cot\theta =\frac{1}{\tan \theta } \\ =\frac{1}{{\displaystyle \frac{2}{\sqrt{5}}}} \\ =\frac{\sqrt{}}{2} \end{gathered}$$