Chapter 6, Problem 46RE

To determine

Describe all the trigonometric ratios of $angle\theta$ .

Answer to Problem 46RE

  $\sin \theta =\frac{1}{\sqrt{5}},\cos \theta =\frac{2}{\sqrt{5}},\tan \theta =\frac{1}{2},\csc \theta =\sqrt{5},\sec \theta =\frac{\sqrt{5}}{2},\cot \theta =2$

Explanation of Solution

Given information:

If lies in the third quadrant and parallel to the line $4y-2x-1=0$ .

Calculation:

Here, to find all the trigonometric ratios we will first sketch the figure according to the problem;

  

Now, as the terminal side of the given angle is parallel to the line $4y-2x-1=0$ .

Thus, the tangent of the $angle\theta$ will be equal to the $slope\left(m\right)$ of the line.

  $\therefore m=\tan \theta$

Now, we will find the slope of the line:

  $\begin{array}{l}y=mx+c\\ 4y-2x-1=0\\ 4y=2x+1\\ y=\left(\frac{2}{4}\right)x+\frac{1}{4}\\ y=\frac{1}{2}x+\frac{1}{4}\end{array}$

Now, on comparing $y=\frac{1}{2}x+\frac{1}{4}withy=mx+c$ we get:

  $\begin{array}{l}m=\frac{1}{2}\\ Onputtingm=\tan \theta ,weget\\ tan\theta =\frac{1}{2}\\ Thus,\\ \theta ={\tan }^{-1}\left(\frac{1}{2}\right)\end{array}$

Now, we will find $\sin \theta$ :

  $\begin{array}{l}\theta ={\tan }^{-1}\frac{1}{2}\\ \sin \theta =\sin \left({\tan }^{-1}\frac{1}{2}\right)\\ Weknowthat,\\ {\tan }^{-1}x={\sin }^{-1}\frac{x}{\sqrt{1+x^2}}\\ Thus,\\ \sin \theta =\sin \left({\sin }^{-1}\frac{\frac{1}{2}}{\sqrt{1+\frac{1}{4}}}\right)\\ \sin \theta =\sin \left({\sin }^{-1}\frac{\frac{1}{2}}{\sqrt{1+\frac{1}{4}}}\right)\\ \sin \theta =\sin \left({\sin }^{-1}\frac{1}{\sqrt{5}}\right)\\ Now,wewilluse:\\ \sin \left({\sin }^{-1}x\right)=x\\ \sin \theta =\frac{1}{\sqrt{5}}\\ \end{array}$

Now, to find:

  $\begin{array}{l}\cos \theta ,weput\theta ={\tan }^{-1}\frac{1}{2}\\ \cos \theta =\cos \left({\tan }^{-1}\frac{1}{\sqrt{2}}\right)\\ Weknowthat,\\ {\tan }^{-1}x={\cos }^{-1}x\frac{1}{\sqrt{1+x^2}}\\ Thus,\\ \cos \theta =\cos \left({\cos }^{-1}\frac{1}{\sqrt{1+x^2}}\right)\\ \cos \theta =\cos \left({\cos }^{-1}\frac{1}{\sqrt{1+\frac{1}{4}}}\right)\\ \cos \theta =\cos \left({\cos }^{-1}\frac{1}{\frac{\sqrt{5}}{2}}\right)\\ \cos \theta =\cos \left({\cos }^{-1}\frac{2}{\sqrt{5}}\right)\\ \end{array}$

Now, by using:

  $\begin{array}{l}\cos \left({\cos }^{-1}x\right)=x,weget\\ \cos \theta =\frac{2}{\sqrt{5}}\\ \end{array}$

Now, to find:

  $\begin{array}{l}\tan \theta ,weput\theta ={\tan }^{-1}\frac{1}{2}\\ \tan \theta =\tan \left({\tan }^{-1}\frac{1}{2}\right)\\ Now,use\tan \left({\tan }^{-1}x\right)=x\\ Thus,\\ \tan \theta =\frac{1}{2}\end{array}$

Now, we are having:

  $\sin \theta =\frac{1}{\sqrt{5}}$

Here, we will use reciprocal identity $\csc \theta =\frac{1}{\sin \theta }$ :

  $\csc \theta =\sqrt{5}$

Now, we are having:

  $\cos \theta =\frac{2}{\sqrt{5}}$

Now, again we will use reciprocal identity $\sec \theta =\frac{1}{\cos \theta }$ :

  $\sec \theta =\frac{\sqrt{5}}{2}$

Now, we are having:

  $\tan \theta =\frac{1}{2}$

Now, again we will use reciprocal identity $\cot \theta =\frac{1}{\tan \theta }$ :

  $\cot \theta =2$

Hence, $\sin \theta =\frac{1}{\sqrt{5}},\cos \theta =\frac{2}{\sqrt{5}},\tan \theta =\frac{1}{2},\csc \theta =\sqrt{5},\sec \theta =\frac{\sqrt{5}}{2},\cot \theta =2$ .