三角形ABC中，sinAcos^(C/2)+sinCcos^(A/2)=3/2sinB,求cos[(A-C)/2]-2sin(B/2)的值
參考答案:答：cos[(A-C)/2]-2sin(B/2)=0
解：
cosC=2(cosC/2)^2-1,
(cosC/2)^2=(1+cosC)/2
(cosA/2)^2=(1+cosA)/2
sin(A+C)=sinB
sin(90°-B/2)=cosB/2
sinAcos^(C/2)+sinCcos^(A/2)=3/2sinB
sinA*[(1+cosC)/2]+sinC*[(1+cosA)/2]=3/2sinB
sinA+sinC+sin(A+C)=3sinB
sinA+sinC+sin(A+C)=3sinB
sinA+sinC=2sinB
2sin[(A+C)/2]*cos[(A-C)/2]=4sin(B/2)*cos(B/2)
sin[(A+C)/2]*cos[(A-C)/2]=2sin(B/2)*cos(B/2)
sin[(180°-B)]/2]*cos[(A-C)/2]=2sin(B/2)*cos(B/2)
sin(90°-B/2)*cos[(A-C)/2]=2sin(B/2)*cos(B/2)
cosB/2*cos[(A-C)/2]=2sin(B/2)*cos(B/2)
cos[(A-C)/2]=2sin(B/2)
cos[(A-C)/2]-2sin(B/2)=0
