Question

求过程，公式，谢谢

Answer 1

1)f(x)=cosx*sin(x+π/3)-√3(cosx)^2+√3/4
=cosx(1/2sinx+√3/2cosx)-√3(cosx)^2+√3/4
=1/4(2sinxcosx+√3/2(cosx)^2-√3(cosx)^2+√3/4
=1/4sin2x-√3/2(cosx)^2+√3/4
=1/4sin2x-√3/4(1+cos2x)+√3/4
=1/4sin2x-√3/4cos2x
=1/2(1/2sin2x-√3/2cos2x)
=1/2sin(2x-π/3)
T=2π/2=π
2)f(x)=cos(π/3+x)*cos(π/3-x)-sinxcosx+1/4
=(1/2cosx-√3/2sinx)(1/2cosx+√3/2sinx)-sinxcosx+1/4
=1/4(cosx)^2-3/4(sinx)^2-1/2*2sinxcosx+1/4
=1/8(1+cos2x)-3/8(1-cos2x)-1/2sin2x+1/4
=1/8+1/8cos2x-3/8+3/8cos2x-1/2sin2x+1/4
=1/2cos2x-1/2sin2x
=√2/2(√2/2cos2x-√2/2sin2x)
=√2/2cos(2x+π/4)
T=2π/2=π