Question

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

Answer 1

Step-1: Obtain a relationship between angles:

By angle sum property of a triangle, the sum of all interior angles of a triangle is ${180}^{\circ }$

$\Rightarrow$$A+B+C={180}^{\circ }$

$\Rightarrow$ $B+C={180}^{\circ }-A$

Dividing throughout by $2$ we get

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

$\Rightarrow$ $\frac{B+C}{2}={90}^{\circ }-\frac{A}{2}$ $...\left(i\right)$

Step-2: Prove given statement:

Taking sine of the angles obtained in equation $\left(i\right)$ we get

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

We know that $\sin \left({90}^{\circ }-\theta \right)=\cos \theta$ …[property of complementary angles]

$\Rightarrow$ $\sin \left({90}^{\circ }-\frac{A}{2}\right)=\cos \left(\frac{A}{2}\right)$

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

Hence, the given statement is proved.