已知函数f(x)=sinx+cosx,F(x)=f'(x)[f(x)+f'(x)]-1,f'(x)是f(x)的导函数.(1)若tanx=1/3,求F(X)的值（2）求F（x)的单调减区间

F(x)=(cosx-sinx)[sinx+cosx+cosx-sinx]-1
=2(cosx)^2-sin2x-1
=2(cosx)^2 (1-tanx)-1,
(1)tanx=1/3,
(cosx)^2=1/(1+(tanx)^2)=9/10,
F(X)=2* (9/10 ) *(2/3)-1
=1/5.
(2)
对F(x)求导得-2sin2x-2cos2x,令其

(1)f'(x)=cosx-sinx
F(x)=(cosx-sinx)(sinx+cosx+cosx-sinx)-1
=2(cosx)^2-1-2sinxcosx
=cos2x-sin2x
则F(x)=(2(cosx)^2-2sinxcosx)/((sinx)^2+(cosx)^2)-1
=2(((tanx)^2-2tanx)/((tanx)^2)+1)-1
带入得F(x)=-2
(2)
F(x)=cos2x-sin2x=根号2cos(2x+PI/4)易得
[PI/8+kPI,5PI/8+kPI]

f'(x)=cosx-sinx
F(x)=(cosx-sinx)(sinx+cosx+cosx-sinx)-1
=2cosx(cosx-sinx)-1
=2cos²x-1-2sinxcosx
=cos2x-sin2x
=√2sin(π/4-2x)
(1)tanx=1/3 ∴sinx/cosx=1/3 ∴cosx=3sinx
又∵sin²x+cos²x=1 ∴sin²x+9sin²x=1 sin²x=1/10
则f(x)=4sinx f'(x)=2sinx
F(x)=2sinx(4sinx+2sinx)-1=12sin²x-1=1/5
(2)y=sinx的单调递减区间为[π/2+2kπ,3π/2+2kπ]
则F(x)=√2sin(π/4-2x)的单调递减区间为π/2+2kπ≤π/4-2x≤3π/2+2kπ
[-5π/8+kπ,-π/8+kπ]

楼上正解

∵f（x）=sinx+cosx,∴f'（x）=cosx-sinx,∴F（x）=f'（x）[f（x）+f'（x）]-1=（cosx-sinx）（sinx+cosx+cosx-sinx）-1=2cos^2x-sin2x-1=1+cos2x-sin2x-1= 根号2 cos（2x+π/4 ）,∵tanx=1/3 ,∴cos2x=cos^2x-sin^2...