Answer: D
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\((90°-\cfrac{A}{2})\)=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\((90°-\cfrac{A}{2})\)=cos\(\cfrac{A}{2}\)