f(x)=2cos(x+π/4)cos(x-π/4) =sin(2x-π/2) 其中cos(x-π/4) 是如何变成si

Sqrt[2] * { Sin( x + π/4 ) * cos[ π/4 ] + Cos[ x + π/4] * si 函数f(x)=-√2(sin2x+π/4)+6 sin x cos x-2cos²x+1 已知函数f(x)=cos(2x-π/3)+sin^2 x-cos^2 x 设f(x)=4cos(ωx-π/6)sinωx-cos(2ωx+π),其中ω>0,(1)求函数y=f 已知函数f(x)=cos(2x-π\3)+sin²x-cos²x 已知f(x)=cos^2x/1+sin^2x求f'(π/4) 已知f(x)=cos^2x+sinxcosx g(x)=2sin(x+π/4)sin(x-π/4) 已知f(x)=sinx+2sin(π/4+x/2)cos(π/4+x/2). 化简f(x)=(1+√2*cos(2x-π/4))/sin(π/2-x) f(x)=2SIN^2(x+π\4)-cos(2x+π\6) 化简 谢 已知函数f(x)=2sin(x+π/4)cos(x-π/4) 已知函数f(x)=4cos(π/2-x)-sin^2(π/4+x/2)+cos^4x-sin^4x