If $A , B$ and $C$ are interior angles of a triangle $ABC ,$ then show that
$\sin \left(\frac{B+C}{2}\right)=\cos \frac{A}{2}$
If $A , B$ and $C$ are interior angles of a triangle $ABC ,$ then show that
$\sin \left(\frac{B+C}{2}\right)=\cos \frac{A}{2}$
We know that for a triangle $ABC$
$\angle A+\angle B+\angle C=180^{\circ}$
$\angle B+\angle C=180^{\circ}-\angle A$
$\frac{\angle B+\angle C}{2}=90^{\circ}-\frac{\angle A}{2}$
$\sin \left(\frac{B+C}{2}\right)=\sin \left(90^{\circ}-\frac{A}{2}\right)$
$=\cos \left(\frac{ A }{2}\right)$
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