Q.47 - Precalculus Enhanced with Graphing Utilities

Question

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

$\sin \theta =\frac{5}{13} , {90}^o<\theta <{180}^o$

Step-by-Step Solution

Answer

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

Step 1. Given information

Here       $\sin \theta =\frac{5}{13} , {90}^o<\theta <{180}^o$are given.

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

Step 2.Description of finding the value cos &#952; .

In this question $$\begin{gathered} \sin \theta =\frac{5}{13} \\ We know that {\sin }^2\theta +{\cos }^2\theta =1 \\ \cos \theta =\pm \sqrt{1-{\sin }^2\theta } , here \theta lies on {90}^o<\theta <{180}^o so \sin \theta is on second quadrant all other except \sin \theta and csc\theta are will be negative. \\ hence \cos \theta =-\sqrt{1-{\sin }^2\theta } \\ =-\sqrt{1-\frac{5^2}{{13}^2}} \\ =-\sqrt{1-\frac{25}{169}} \\ =-\frac{12}{13} \end{gathered}$$

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

Here

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

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

We know that 

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