Question

If A, B, C are the interior angles of a triangle ABC, $\sin \frac{B+C}{2}$ is :

• A. $\sin \frac{A}{2}$
• B. $\cos \frac{A}{2}$
• C. $1$
• D. $Noneofthese$

Answer 1

The correct option is B $\cos \frac{A}{2}$

$\sin \frac{B+C}{2}=\sin \left(\frac{{180}^o-A}{2}\right)$
$[\because A+B+C={180}^o\Rightarrow B+C={180}^o-A]$
$=\sin ({90}^o-\frac{A}{2})=\cos \frac{A}{2}$
$[\because \sin ({90}^o-\theta )=\cos \theta ]$