Question

sinA*sinC=1/2*[cos(A-C)-cos(A+C)]=cos^2B=1/4这一步怎么算。求详解

sinA*sinC怎么得出的1/2*[cos(A-C)-cos(A+C)]

Answer 1

推导过程：
sinA*sinC=sin[(A+C)/2+(A-C)/2]*sin[(A+C)/2-(A-C)/2]
可令x=(A+C)/2,y=(A-C)/2
则sinA*sinC=sin(x+y)*sin(x-y)
=(sinxcosy+cosxsiny)*(sinxcosy-cosxsiny)
=(sinxcosy)^2-(cosxsiny)^2
=[1-(cosx)^2](cosy)^2-(cosx)^2[1-(cosy)^2]
=(cosy)^2-(cosx)^2
=(cos2y+1)/2-(cos2x+1)/2
=1/2*[cos2y-cos2x]
=1/2*[cos(A-C)-cos(A+C)]

Answer 2

sinasinc
=[(cosacosa+sinasinc)-(cosacosa-sinasinc)]/2
=[cos(a-c)-cos(a+c)]/2

追问

为什么要用sinasinc=[cos(a-c)-cos(a+c)]/2

追答

需要

采纳吧

Answer 3

sinAsinC怎么得出的1/2[cos(A-C)-cos(A+C)]？

这是积化和差公式
逆着推你应该会
正着推很麻烦，不好打，其实就是逆着在正过来