若f（x）=（sinx+sin3x）⼀（cosx+cos3x），则f（x）的最小正周期为-----

1.f(x)=(sinx+sin3x)/(cosx+cos3x)
=(sinx+3sinx-4sin^3 x)/(cosx+4cos^3 x-3cosx)
=(4sinx-4sin^3 x)/(4cos^3 x-2cosx)
=[4sinx(1-sin^2 x)]/[2cosx(cos^2 x-1)]
=(4sinxcos^ x)/(2cosxcos2x)
=(2sin2xcosx)/(2cosxcos2x)
=tan2x
T=π/2

2.f(x)=f(x-1)+f(x+1)……(1)
f(x+1)=f(x)+f(x+2)……(2)
由(1)(2)得:f(x-1)=-f(x+2)
设t=x-1 x=t+1
f(t)=-f(t+1+2)
f(t+3)=-f(t)
f(t+6)=f(t)
T=6

3.

4.-sin^2 x+2sinx+a=0
-(sinx-1)^2+1+a=0
(sinx-1)^2=a+1
-1≤sinx≤1
-2≤sinx-1≤0
0≤(sinx-1)^2≤4
0≤a+1≤4
-1≤a≤3

5.x∈(-π/4,π/4)
sinx∈(-√2/2,√2/2)
f(x)=cos^2 x+sinx
=(1-sin^2 x)+sinx
=-sin^2 x+sinx+1
设sinx=t
f(x)=-t^2+t+1 t∈(-√2/2,√2/2)
=-(t-1/2)^2+5/4
f(x)min=f(-√2/2)=(√2-1)/2

6.f(5)=f(5-6)=f(-1)=-f(1)=-2

f(x)=(sinx+sin3x)/(cosx+cos3x)
=2sin(2x)cos(-x)/[2cos(2x)cos(-x)]
=sin(2x)/cos(2x)
=tan(2x)
最小正周期T=π/2

运用到的公式：
sinα+sinβ=2sin[(α+β)/2]cos[(α-β)/2]
cosα+cosβ=2cos[(α+β)/2]cos[(α-β)/2]