Q.49 - Precalculus Enhanced with Graphing Utilities

Question

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

$\cos \theta =\frac{-1}{3} , \frac{\mathrm{\pi }}{2}<\theta <\mathrm{\pi }$

Step-by-Step Solution

Answer

 The values are

$\sin \theta =\frac{2\sqrt{2}}{3}, csc\theta =\frac{3}{2\sqrt{2}},sec\theta =-3,\tan \theta =-2\sqrt{2},cot\theta =-\frac{1}{2\sqrt{2}}$

Step 1 .Given information

$\cos \theta =\frac{-1}{3} , \frac{\mathrm{\pi }}{2}<\theta <\mathrm{\pi }$is given.

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

Step 2. Finding the value sin &#952;

$$\begin{gathered} We know that {\sin }^2\theta +{\cos }^2\theta =1 , here \cos \theta =\frac{-1}{3} \\ \sin \theta =\pm \sqrt{1-{\cos }^2\theta } , here \theta lies on \frac{\mathrm{\pi }}{2}<\theta <\mathrm{\pi } so \cos \theta is on second quadrant and all other except \sin \theta and csc\theta are will be negative. \\ hence \sin \theta =\sqrt{1-co{s}^2\theta } \\ =\sqrt{1-\frac{{(-1)}^2}{3^2}} \\ =\sqrt{1-\frac{1}{9}} \\ =\sqrt{\frac{8}{9}} \\ =\frac{2\sqrt{2}}{3} \end{gathered}$$

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

$$\begin{gathered} Since \cos \theta =\frac{-1}{3} ,then sec\theta =\frac{1}{\cos \theta } \\ =\frac{1}{{\displaystyle \frac{-1}{3}}} \\ =-3 \\ csc\theta =\frac{1}{\sin \theta } \\ =\frac{1}{{\displaystyle \frac{2\sqrt{2}}{3}}} \\ =\frac{3}{2\sqrt{2}} \end{gathered}$$

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

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