Question

sinbsinc等于cos^2a/2，判断三角形形

Answer 1

sinBsinC=cos²(A/2)
左边通过积化和差公式得：
sinBsinC=-(1/2)[cos(B+C)-cos(B-C)]
= -(1/2)[cos(π-A)-cos(B-C)]
= -(1/2)[-cosA-cos(B-C)]
= (1/2)cosA + (1/2)cos(B-C)

右边通过倍角公式得：
cos²(A/2) = (1/2)[cosA+1]
= (1/2)cosA + 1/2

左边 = 右边：
(1/2)cosA + (1/2)cos(B-C) = (1/2)cosA + 1/2

(1/2)cos(B-C) = 1/2

cos(B-C) = 1

B-C=0
B=C
等腰三角形