内部的函数sin（π－x）+sin（π/2+x)=sin(x) +cos(x)2sinx［sin（π－x）+sin（π/2+x）］= 2sinx(sinx +cosx) = 2(sinx)^2 +2sinxcosx = 1-cos2x +sin2x= 1-根号2sin(2x-pi/4)所以振幅根号2,周期pi,初相=f(0) = 0...