已知函数f(x)=(1+cos2x)sin平方x,x属于R,则f(x)的周期?奇偶?=(1/2)[sin(2x)]^2这一步看不懂，
问题描述：
已知函数f(x)=(1+cos2x)sin平方x,x属于R,则f(x)的周期?奇偶?
=(1/2)[sin(2x)]^2
这一步看不懂，
答
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
答
=（2cosx乘sinx)的
=(sin2x)的平方乘二分之一
答
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....