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