答:怀疑n应该是(cos^2x/2,cosx/2):
(1)
f(x)=mXn-1/2
=[√3sin(x/2)]*cos(x/2)-1*[cos(x/2)]^2-1/2
=(√3/2)sinx-(1+cosx)/2-1/2
=sinxcosπ/6-cosxsinπ/6-1
=sin(x-π/6)-1
因为:-π/3
答:怀疑n应该是(cos^2x/2,cosx/2):
(1)
f(x)=mXn-1/2
=[√3sin(x/2)]*cos(x/2)-1*[cos(x/2)]^2-1/2
=(√3/2)sinx-(1+cosx)/2-1/2
=sinxcosπ/6-cosxsinπ/6-1
=sin(x-π/6)-1
因为:-π/3