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

Given $△ABC$

We know that sum of three angles of a triangle is $180$

Hence $\angle A+\angle B+\angle C={180}^o$

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

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

Multiply both sides by $\frac{1}{2}$

$\frac{1}{2}(B+C)=\frac{1}{2}({180}^o-A)$

$\frac{1}{2}(B+C)={90}^o-\frac{A}{2}...\left(1\right)$

Now

$\frac{1}{2}(B+C)$

Taking sine of this angle

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

$\sin ({90}^o-\frac{A}{2})$

$\cos \frac{A}{2}[\sin ({90}^o-\theta )=\cos \theta ]$

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