求证三角函数.2cos（wt1+θ）*s(wt2+θ)=cosw(t2-t1)+cos[w(t1+t2)+2θ]

这不是积化和差吗：cosacosb=(cos(a+b)+cos(a-b))/2p表示theta：2cos(wt1+p)*cos(wt2+p)=cos(wt1+wt2+2p)+cos(wt1+p-wt2-p)=cos(w(t1-t2))+cos(w(t1+t2)+2p)=cos(w(t2-t1))+cos(w(t1+t2)+2p)