Answer: B
\(A+B+C=180°\)
Or, \((B+C)=180°-A\)
Or, \(\cfrac{B+C}{2}=\cfrac{180°-A}{2}\)
Or, \(\cfrac{B+C}{2}=90°-\cfrac{A}{2}\)
∴sin\(\cfrac{B+C}{2}\) = sin\(\left(90°-\cfrac{A}{2}\right)\) = cos\(\cfrac{A}{2}\)
\(A+B+C=180°\)
Or, \((B+C)=180°-A\)
Or, \(\cfrac{B+C}{2}=\cfrac{180°-A}{2}\)
Or, \(\cfrac{B+C}{2}=90°-\cfrac{A}{2}\)
∴sin\(\cfrac{B+C}{2}\) = sin\(\left(90°-\cfrac{A}{2}\right)\) = cos\(\cfrac{A}{2}\)