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Answer :
Given, A, B and C are the interior angles of a \( ∆ \ ABC \)
We know that, sum of interior angles of a triangle is 180°
\( \because A + B + C \ = \ 180° \)
=> \( \frac{A+B+C}{2} \ = \ 90° \)
=> \( \ \frac{B+C}{2} \ = \ 90° - \frac{A}{2} \)
=> \( sin( \frac{B+C}{2} ) \ = \ sin( 90° - \frac{A}{2} ) \)
=> \( sin( \frac{B+C}{2} ) \ = \ cos \frac{A}{2} \) [ ∵ \( sin(90° - \theta) \ = \ cos\theta \) ]