y=sin^2(x+π/2)-sin^2(x-π/4)的最小正周期和值域
问题描述:
y=sin^2(x+π/2)-sin^2(x-π/4)的最小正周期和值域
答
y=sin²(x+π/2) - sin²(x-π/4)=cos²x + [1-2sin²(x-π/4)-1]/2=(2cos²x-1+1)/2 + [cos2(x-π/4)-1]/2=1/2cos2x+1/2 + 1/2cos(2x-π/2)-1/2=1/2cos2x + 1/2sin2x=√2/2(√2/2cos2x+√2/2si...