Question

If A, B, C are interior angles of triangle ABC, then $\sin \left(\frac{B+C}{2}\right)=$

Answer 1

In $△ABC$,

$A+B+C={180}^o$

$B+C={180}^o-A$

On dividing both sides by 2, we get,

$\frac{B+C}{2}=\frac{{180}^o-A}{2}$

$\frac{B+C}{2}={90}^o-\frac{A}{2}$

$\sin \left(\frac{B+C}{2}\right)=\sin ({90}^o-\frac{A}{2})$

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