已知△ABC中,a,b,c为角A.B.C的对边,且a+c=2b,A-C=π/3求sinB的值2sin[(A+C)/2]*cos[(A-C)/2]=2sinBsin[(A+C)/2]*cos(pi/6)=sinB这一步我看不懂，为什么1/2cos[(A-C)/2]=cos(pi/6)

Question

已知 △ABC中,a,b,c为角A.B.C的对边,且a+c=2b,A-C=π/3 求sinB的值
2sin[(A+C)/2] * cos[(A-C)/2] = 2sinB
sin[(A+C)/2] * cos(pi/6) = sinB
这一步我看不懂，为什么1/2 cos[(A-C)/2] =cos(pi/6)

Answer 1

因为 a + c = 2b
由正弦定理,知：
sinA +sinC = 2sinB
2sin[(A+C)/2] * cos[(A-C)/2] = 2sinB
sin[(A+C)/2] * cos(pi/6) = sinB
因为A + B + C = 180
所以：(A+C)/2 = pi/2 - B/2
所以：
cos(B/2) * √3/2 = 2sin(B/2)cos(B/2)
显然B/2不等于pi/2,cos(B/2)不等于0
所以：
sin(B/2) = √3/4
cos(B/2) = √13/4
sinB = 2sin(B/2)cos(B/2) = √39/8