化简f(x)=2cos(x/2)·(sin(x/2)+cos(x/2))-1(1)化简f(x);(2)求f(π/12)的值

f(x)=cos(2x)+2sin(x)化简 化简f(x)=2cos(x/2)·(sin(x/2)+cos(x/2))-1(1)化简f(x);(2)求f(π/12)的值 f(cos)=-cos(2x),则f(sinx) 化简f(x)=cos^2(x+π/12)+sinxcosx 化简f(x)=cos(2x-π/3)-cos2x 化简f(x)=2cos^3 x+sin^2 (360-x)-cos(180-x)-3 / 2+2cos^2(180+x)+cos(-x)求化简图片 化简f(x)=cos(-x/2)+cos(4k+1/2-x/2) 已知f(x)=sin(π/6-x)^2-cos(π/4+x)^2+cos(π/6)cos(π/6-2x)化简f(x) f ' (sinx)=cos^2x,求f（x） f(sinx)=1+cos(2x),求f(x), f(x)为奇函数,x>0,f(x)=sin 2x+cos x,则x 证明：f(x)=x*cos(x)不是周期函数证明：假设y=xcosx是周期函数,因为周期函数有f(x+T)=f(x)xcosx=（x+T)cos(x+T)=xcosx*cosT-xsinx*sinT+Tcosx*cosT-Tsinx*sinT所以cosT=1 T=kπ/2-xsinx*sinT+Tcosx*cosT-Tsinx*sinT=0-xsinx*sinT-Tsinx*si 设f(x)=4cos(ωx-π/6)sinωx-cos(2ωx+π),其中ω>0,求函数y=f(x)的值域,请看问题补充f（x）=4cos（ωx-π/6）sinωx-cos（2ωx+π） =4(coswxcosπ/6+sinwxsinπ6)sinwx+cos2wx =2√3sinwxcoswx+2sin²wx+cos2wx =√3si f'(sin^2x)=cos^2x,求f(x)f'((sin^x)^2)=(cos^x)^2 f(sin x)=3-cos 2x,求f(cos x)=? 设f(sin x)=3-cos 2x,则f(cos x）=? 若F(sin x)=3-cos 2x 则F（cos x）=? f(sin x)=cos 2x +1 求 f(cos x)