Question

How do you prove that ${\sin ^2}\left( {\dfrac{x}{2}} \right) = \dfrac{{{{\sin }^2}x}}{{2(1 + \cos x)}}$ ?

Answer 1

Hint: In the given question they have asked us to prove left hand side function is equal to right hand side, this we can achieve by making use of double angle identities which is given by: $\cos 2x = {\cos ^2}x - {\sin ^2}x = 1 - 2{\sin ^2}x$ by making some necessary changes we can arrive at the required answer.

Complete step-by-step answer:
In the given question they have asked us to prove left hand side function is equal to right hand side, this we can achieve by making use of double angle identities which is given by: $\cos 2x = {\cos ^2}x - {\sin ^2}x = 1 - 2{\sin ^2}x$
In the given question we have ${\sin ^2}x$ in the left hand side, so the above formula can be rewrite as below,
$ \Rightarrow 2{\sin ^2}x = 1 - \cos 2x$
By simplifying the above equation, we get
$ \Rightarrow {\sin ^2}x = \dfrac{1}{2}.(1 - \cos 2x)$
Now, if we compare the above equation with the given question that is ${\sin ^2}\left( {\dfrac{x}{2}} \right) = \dfrac{{{{\sin }^2}x}}{{2(1 + \cos x)}}$ , we need to prove for $\dfrac{x}{2}$ .
So put $x = \dfrac{x}{2}$ in the above equation, we get
$ \Rightarrow {\sin ^2}\left( {\dfrac{x}{2}} \right) = \dfrac{1}{2}.\left( {1 - \cos 2\dfrac{x}{2}} \right)$
$ \Rightarrow {\sin ^2}\left( {\dfrac{x}{2}} \right) = \dfrac{1}{2}.(1 - \cos x)$
Now, multiply and divide the right hand side function with $1 + \cos x$, we get
$ \Rightarrow {\sin ^2}\left( {\dfrac{x}{2}} \right) = \dfrac{1}{2}.\left( {\dfrac{{(1 - \cos x).(1 + \cos x)}}{{(1 + \cos x)}}} \right)$
On simplifying the above expression, we get
$ \Rightarrow {\sin ^2}\left( {\dfrac{x}{2}} \right) = \dfrac{1}{2}.\left( {\dfrac{{1 - {{\cos }^2}x}}{{1 + \cos x}}} \right)$
From the trigonometric identities we know that ${\sin ^2}x + {\cos ^2}x = 1$ . So from this we can get that ${\sin ^2}x = 1 - {\cos ^2}x$ . So we can substitute this in the above expression to simplify further, we get
$ \Rightarrow {\sin ^2}\left( {\dfrac{x}{2}} \right) = \dfrac{1}{2}.\left( {\dfrac{{{{\sin }^2}x}}{{1 + \cos x}}} \right)$ or we can rewrite is as below
$ \Rightarrow {\sin ^2}\left( {\dfrac{x}{2}} \right) = \dfrac{{{{\sin }^2}x}}{{2(1 + \cos x)}}$
Hence proved.

Note: This type of question can be solved either by using the left hand side or by using the right hand side. By doing the practice we can easily know through which side the problem can be solved easily. And when simplifying the expression, try to make it as easy as possible so that we can prove the required answer.