求解一道数学题~~化简:sin(α+β)cos(r-β)-cos(β+α)sin(β-r)。。。过程详细点 谢谢！！

化简:sin(α+β)cos(r-β)-cos(β+α)sin(β-r)

我用A代表α B代表β C代表r（因为那几个希腊字母难打）
sin(A+B)cos(C-B)-cos(B+A)sin(B-C)
=sin(A+B)con(C-B)+cos(A+B)sin(C-B)
=sin((A+B)+(C-B))=sin(A+C)

f(x)=sin(ωx+π/6)+sin(ωx-π/6)-2cos^2(ωx/2), =sinωxcosπ/6+cosωx sinπ/6+sinωxcosπ/6-cosωx sinπ/6-1+cosωx =√3sinωx+cosωx-1 =2sin(ωx+π/6)-12sin(ωx+π/6)-1=-1sin(ωx+π/6)=0在一个π内有且仅有两个不同的交点，即周期为πω＝2单调增区间：2kπ-π/2<=2x+π/6<=2kπ+π/2kπ-π/3<=x<=kπ+π/6单调增区间 [kπ-π/3,kπ+π/6]