Question

已知函数f(x)=(1+cos2x)sin^2x,x属于R,则f(x)是

Answer 1

f(x)=(1+cos2x)sin^2x
=2cos^2xsin^2x
=1/2(sin2x)^2
=1/4(1-cos4x)
f(x)是偶函数，周期[kπ/2,(k+1)π/2],值域[0,1/2]

Answer 2

1+cos2x=sin²x+cos²x+(cos²x-sin²x)=2cos²x
f(x)=(1+cos2x)sin²x=2cos²xsin²x=(2sinxcosx)*(sinxcosx)=sin2x*1/2sin2x =1/2*(sin2x)²
T=2π/2/2=π/2
最小正周期为二分之派的偶函数

Answer 3

f(x)=(1+cos2x)sin^2x
=2cos²xsin²x
=(1/2)sin²2x

f(x)是偶函数

Answer 4

f(x)=(1+cos2x)sin^2x=2cosx^2sinx^2=1/2(sin2x)^2
最小周期为派

Answer 5

原式=（1+2cos^2x-1)sin^2x=(2sinxcosx)^2/2=sin^2(2x)/2