y=sin(x-π/3)cosx=[sin(2x-π/3)+sin(-π/3)]/2
问题描述：
y=sin(x-π/3)cosx=[sin(2x-π/3)+sin(-π/3)]/2
为什么y=sin(x-π/3)cosx,就说明y=[sin(2x-π/3)+sin(-π/3)]/2,望会此题者给小妹说一下,
答
这是积化和差公式：
sinαcosβ=[sin(α+β)+sin(α-β)]/2
还有
sinαsinβ=-[cos(α+β)-cos(α-β)]/2
cosαcosβ=[cos(α+β)+cos(α-β)]/2