In any triangle ABC,{sin ^2}frac{A}{2} + {sin | vedclass

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(C) We know that $\sin^2 	heta = \frac{1 - \cos 2	heta}{2}$.Applying this to each term: $\sin^2 \frac{A}{2} + \sin^2 \frac{B}{2} + \sin^2 \frac{C}{2

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) We know that $\sin^2 	heta = \frac{1 - \cos 2	heta}{2}$.Applying this to each term: $\sin^2 \frac{A}{2} + \sin^2 \frac{B}{2} + \sin^2 \frac{C}{2} = \frac{1 - \cos A}{2} + \frac{1 - \cos B}{2} + \sin^2 \frac{C}{2}$.$= 1 - \frac{1}{2}(\cos A + \cos B) + \sin^2 \frac{C}{2}$.$= 1 - \cos \frac{A+B}{2} \cos \frac{A-B}{2} + \sin^2 \frac{C}{2}$.Since $A+B = 180^o - C$,then $\frac{A+B}{2} = 90^o - \frac{C}{2}$,so $\cos \frac{A+B}{2} = \sin \frac{C}{2}$.$= 1 - \sin \frac{C}{2} \cos \frac{A-B}{2} + \sin^2 \frac{C}{2}$.$= 1 - \sin \frac{C}{2} (\cos \frac{A-B}{2} - \sin \frac{C}{2})$.$= 1 - \sin \frac{C}{2} (\cos \frac{A-B}{2} - \cos \frac{A+B}{2})$.Using $\cos X - \cos Y = 2 \sin \frac{X+Y}{2} \sin \frac{Y-X}{2}$,we get $2 \sin \frac{A}{2} \sin \frac{B}{2}$.Thus,the expression equals $1 - 2 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$.

In any triangle $ABC$,${\sin ^2}\frac{A}{2} + {\sin ^2}\frac{B}{2} + {\sin ^2}\frac{C}{2}$ is equal to:

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