Question

If $\sin \alpha =-\dfrac{3}{5}$, where $\pi <\alpha <\dfrac{3\pi }{2}$, then $\cos \dfrac{1}{2\alpha }$is equal to
A. $-\dfrac{1}{\sqrt{10}}$
B. $\dfrac{1}{\sqrt{10}}$
C. $\dfrac{3}{\sqrt{10}}$
D. \[-\dfrac{3}{\sqrt{10}}\]

Answer 1

Hint: This question is based on inverse trigonometric equations. In this question, first we find out in which our quadrant lies and check it for positive and negative signs. Then the value of $\cos \alpha $$\sin \alpha $is given. By putting Pythagoras theorem, we are able to find out the $\cos \alpha $. Then by putting the formulas of trigonometry, we are able to find out our desired answer.

Complete step by step solution: 
Given $\sin \alpha =-\dfrac{3}{5}$ where $\pi <\alpha <\dfrac{3\pi }{2}$
Thus $\alpha $lies in Quadrant III and we know $\cos \alpha $is negative in this quadrant.
Also $\pi <\alpha <\dfrac{3\pi }{2}$( given )
That is $\dfrac{\pi }{2}<\dfrac{\alpha }{2}<\dfrac{3\pi }{4}$
Thus $\alpha $lies in Quadrant II and we know $\cos \alpha $is negative in this quadrant.
Given that $\sin \alpha =-\dfrac{3}{5}$
By putting Pythagoras theorem, we find out $\cos \alpha =-\dfrac{4}{5}$
We know the formula of cos 2A = 2 ${{\cos }^{2}}A$- 1
$\cos \alpha =2{{\cos }^{2}}\dfrac{\alpha }{2}-1$
Now put all the values in the above equation, we get
$-\dfrac{4}{5}=2{{\cos }^{2}}\dfrac{\alpha }{2}-1$
By solving the above equation, we get
$2{{\cos }^{2}}\dfrac{\alpha }{2}=1-\dfrac{4}{5}$
By taking the LCM and solving the above equation, we get
$2{{\cos }^{2}}\dfrac{\alpha }{2}=\dfrac{1}{5}$
Then we get ${{\cos }^{2}}\dfrac{\alpha }{2}=\dfrac{1}{10}$
Now we take square root on both sides, we get
$\cos \dfrac{\alpha }{2}=\dfrac{1}{\sqrt{10}}$
As it lies in Quadrant II, we get
$\cos \dfrac{\alpha }{2}=-\dfrac{1}{\sqrt{10}}$
Hence , value of $\cos \dfrac{\alpha }{2}=-\dfrac{1}{\sqrt{10}}$

Note: In these types of Questions, students made mistakes in finding out the quadrant in which the function lies. It is to be remembered that in quadrant I all are positive and in quadrant II only sine is positive while others are negative. In Quadrant III, only tangent is positive while others are negative and in Quadrant IV, only cosine is positive while others are negative. Take care of these types of things while solving the Questions.