f(x)=2sin^2x+cos^2x+sinxcosx,x∈R,f(π/12)f(x)=2sin^2x+cos^2x+sinxcosx,x∈R求:1) f(π/12)的值2) f(x)的最小值及相应x的值3) f(x)的递增区间

f(x)=cos(2x)+2sin(x)化简 f'(sin^2x)=cos^2x,求f(x)f'((sin^x)^2)=(cos^x)^2 f(x)为奇函数,x>0,f(x)=sin 2x+cos x,则x 已知f(x)=cos(x/2)[sin(x/2)-cos(x/2)],其导为? f(x)=cos(2兀-x)cos(2/兀-x)-sin方x化简速求 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）=? 化简：f(x)=sin(x)cos(x)/sin(2x) f(sin x)=cos 2x +1 求 f(cos x) f(sin x)=3-cos 2x求f(cos x) f(cos x)=2-sin 2x,则f(sin x)=? 设f(sin x/2 )-1+cos x ,求f(x)、f(cos x/2 ). 设函数f(x)=p·q,其中向量p=(sin x,cos x+sin x),q=(2cos x,cos sin x) ( f(x)=(6sin^4x-7sin^2x+2)/(sin^2x-cos^2x) f(x)=cos(2x+60)+sin^2x 求f(x)的值域 f(x)=sin(x+π/4)图像怎样变成F(x)=cos(2x) 已知tan=2,求(cos x+sin x)/(cos x-sin x)+sin^2x