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

Given, A, B and C are the interior angles of a \( ∆ \ ABC \)

We know that, sum of interior angles of a triangle is 180°

\( \because A + B + C \ = \ 180° \)

=> \( \frac{A+B+C}{2} \ = \ 90° \)

=> \( \ \frac{B+C}{2} \ = \ 90° - \frac{A}{2} \)

=> \( sin( \frac{B+C}{2} ) \ = \ sin( 90° - \frac{A}{2} ) \)

=> \( sin( \frac{B+C}{2} ) \ = \ cos \frac{A}{2} \)    [ ∵ \( sin(90° - \theta) \ = \ cos\theta \) ]