Question

In triangle ABC ,prove that ${\sin }^2\frac{A}{2}+{\sin }^2\frac{B}{2}-{\sin }^2\frac{C}{2}=1-2\cos \frac{A}{2}\cos \frac{B}{2}sin\frac{C}{2}$

Answer 1

Let us rearrange the given expression as follows:

$si{n}^2\frac{A}{2}$ $-$ $si{n}^2\frac{C}{2}$ $+$$si{n}^2\frac{B}{2}$

We know that $si{n}^{2}x$ $-$$si{n}^{2}y$ = $sin(x+y)sin(x-y)$

Hence,

$si{n}^2\frac{A}{2}$ $-$$si{n}^2\frac{C}{2}$ $+$$si{n}^2\frac{B}{2}$ = $sin\left(\frac{A+C}{2}\right)$ $sin\left(\frac{A-C}{2}\right)$+ $1$ $-$$co{s}^2\frac{B}{2}$

=$cos\left(\frac{B}{2}\right)$ $sin\left(\frac{A-C}{2}\right)$ $-$ $co{s}^2\frac{B}{2}$ $+$ $1$

=$cos\left(\frac{B}{2}\right)$ [$sin\left(\frac{A-C}{2}\right)$ $-$ $cos\left(\frac{B}{2}\right)$] +$1$

=$cos\left(\frac{B}{2}\right)$ [$sin\left(\frac{A-C}{2}\right)$ $-$ $sin\left(\frac{A+C}{2}\right)$]+$1$

We know that $sin(x-y)-sin(x+y)=-2cosxsiny$

Using the above equality, the expression can be reduced to,

$si{n}^2\frac{A}{2}$ $-$$si{n}^2\frac{C}{2}$ $+$$si{n}^2\frac{B}{2}$ = $1$ $-$ 2$cos\left(\frac{A}{2}\right)$ $cos\left(\frac{B}{2}\right)$ $sin\left(\frac{C}{2}\right)$