If A + B + C = π, prove that

Question:

If A + B + C = π, prove that

$\sin ^{2} \frac{A}{2}+\sin ^{2} \frac{B}{2}+\sin ^{2} \frac{C}{2}=1-2 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$

Solution:

$=\sin ^{2} \frac{A}{2}+\sin ^{2} \frac{B}{2}+\sin ^{2} \frac{C}{2}$

Using formula ,

$\frac{1-\cos 2 \mathrm{~A}}{2}=\sin ^{2} \mathrm{~A}$

$=\frac{1-\cos A}{2}+\frac{1-\cos B}{2}+\frac{1-\cos C}{2}$

$=\frac{1-\cos A+1-\cos B+1-\cos C}{2}$

$=\frac{3-\cos A-\cos B-\cos C}{2}$

Using,

$\cos A+\cos B=2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)$

$=\frac{3-\cos A-\left\{2 \cos \left(\frac{B+C}{2}\right) \cos \left(\frac{B-C}{2}\right)\right\}}{2}$

$=\frac{3-\cos A-2 \cos \left(\frac{B+C}{2}\right) \cos \left(\frac{B-C}{2}\right)}{2}$

Using, since $A+B+C=\pi$

$\rightarrow \mathrm{B}+\mathrm{C}=180-\mathrm{A}$

And, $\cos (\pi-A)=-\cos A$

$=\frac{3-\cos A-2 \cos \left(\frac{\pi}{2}-\frac{A}{2}\right) \cos \left(\frac{B-C}{2}\right)}{2}$

$=\frac{3-\cos A-2 \sin \left(\frac{A}{2}\right) \cos \left(\frac{B-C}{2}\right)}{2}$

Using, $\cos 2 A=1-2 \sin ^{2} A$

$=\frac{\left.3-1+2 \sin ^{2} \frac{A}{2}-2 \sin \frac{A}{2} \cos \frac{B-C}{2}\right)}{2}$

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

Since $A+B+C=\pi$

And Using ,

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

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

$=\frac{2-2 \sin \frac{A}{2}\left\{2 \sin \left(\frac{\frac{2 B}{2}}{2}\right) \sin \left(\frac{\frac{2 C}{2}}{2}\right)\right\}}{2}$

Using, since $A+B+C=\pi$

$=\frac{2-2 \sin \frac{A}{2}\left\{2 \sin \left(\frac{B}{2}\right) \sin \left(\frac{C}{2}\right)\right\}}{2}$

$=1-2 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$

= R.H.S