Question

已知函数f(x)=(1+cos2x)sin平方x,x属于R,则f(x)的周期?奇偶?

请给我详细的解答过程，谢谢
=(1/2)[sin(2x)]^2
这一步看不懂，能再详细点吗

Answer 1

f(x)=(1+cos2x)(sinx)^2
=2(cosx)^2(sinx)^2
=(1/2)[sin(2x)]^2
=(1/2)[1-cos(4x)]/2
=(1/4)[1-cos(4x)]
=1/4-cos(4x)/4

最小正周期为π/2 ，周期为kπ/2.

f(-x)=1/4-cos(-4x)/4=1/4-cos(4x)/4=f(x)
函数是偶函数。

Answer 2

f(x)=(1+cos2x)(sinx)^2
=2(cosx)^2(sinx)^2
=（2cosxsinx）（2cosxsinx）/2=(1/2)[sin(2x)]^2
=(1/2)[1-cos(4x)]/2
=(1/4)[1-cos(4x)]
=1/4-cos(4x)/4

Answer 3

=（2cosx乘sinx)的
=(sin2x)的平方乘二分之一