Q.46 - Precalculus Enhanced with Graphing Utilities

Question

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

$\sin \theta =-\frac{5}{13} ,where \theta lies on 3rd quadrant.$

Step-by-Step Solution

Answer

 The exact values are $\cos \theta =\frac{-12}{13},csc\theta =-\frac{13}{5},sec\theta =-\frac{13}{12} ,\tan \theta =\frac{5}{12} , cot \theta =\frac{12}{5}$

Step 1. Given information

          Here $\sin \theta =-\frac{5}{13 } where \theta lies on 3rd quadrant$

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

Step 2 . description of finding the value of cos θ

Here $$\begin{gathered} \sin \theta =-\frac{5}{13} \\ We know that {\sin }^2\theta +{\cos }^2\theta =1 \\ \cos \theta =\pm \sqrt{1-{\sin }^2\theta } and \sin ce \sin \theta is on third quadrant all other except \tan \theta and cot\theta are will be negative. \\ hence \cos \theta =-\sqrt{1-{\sin }^2\theta } \\ =-\sqrt{1-\frac{{(-5)}^2}{{13}^2}} \\ =-\sqrt{1-\frac{25}{169}} \\ =-\frac{12}{13} \end{gathered}$$

Step 3. Finding the values of c s c θ   a n d   s e c θ

in this question 

$$\begin{gathered} \cos \theta =-\frac{12}{13} ,then sec\theta =\frac{1}{\cos \theta } \\ =\frac{1}{-{\displaystyle \frac{12}{13}}} \\ =-\frac{13}{12} \\ csc\theta =\frac{1}{\sin \theta } \\ =\frac{1}{-{\displaystyle \frac{5}{13}}} \\ =-\frac{13}{5} \end{gathered}$$

Step 4. Finding the values of tan   θ ,   c o t θ

Since

$$\begin{gathered} \sin \theta =-\frac{5}{13} and \cos \theta =-\frac{12}{13} \\ \tan \theta =\frac{\sin \theta }{\cos \theta } \\ =\frac{-\frac{5}{13}}{-\frac{12}{13}} \\ =\frac{5}{12} \\ cot\theta =\frac{1}{\tan \theta } \\ =\frac{1}{{\displaystyle \frac{5}{12}}} \\ =\frac{12}{5} \end{gathered}$$