求函数f(x)=cosx的四次方-2cos2x的二次方+sinx的平方的值域和最小正周期2cos2x的二次方是2(cos2x的二次方)

原式=(cosx)^4 - 2*(cos2x)^2 + (sinx)^2
=(cosx)^4 - 1 + (1 - 2 * (cos2x)^2) + (sinx)^2
=[(cosx)^2 - 1][(cosx)^2 + 1] - cos4x + (sinx)^2
=-(sinx)^2 * [(cosx)^2 + 1] - cos4x + (sinx)^2
=-(sinx)^2 * cosx^2 -(sinx)^2 - cos4x + (sinx)^2
=-(sinx)^2 * cosx^2 - cos4x
=-1/4 * (sin2x)^2 - cos4x
=1/8 *(cos4x - 1) - cos4x
=-7/8 * cos4x - 1/8
T = 2π / 4 = π /2
由-1

由于 2cos2x^2=cos4x+1 所以
f(x)=cosx^4-2cos2x^2+sinx^2
=cosx^4-1-2cos2x^2+1+sinx^2
=(cosx^2+1)(cosx^2-1)-2cos2x^2+1+sinx^2
=-(cosx^2+1)sinx^2-2cos2x^2+1+sinx^2
=-cosx^2*sinx^2-2cos2x^2+1
=-1/4sin2x^2-2cos2x^2+1
=1/4cos2x^2-1/4-2cos2x^2+1
=-7/4(cos2x^2)+3/4
=-7/8(cos4x+1)+3/4
所以 可以确定T=Pi/2
当cos4x=-1时 可以求得其最大值 f(x)=3/4
当cos4x=1时 可以求得其最小值 f(x)=-1
所以f(x)的值域为[-1，3/4]

f(x)=cos^4x-2(cos2x)^2+sin^2x=cos^4x-2(2cos^2x-1)^2+1-cos^2x=cos^4x-8cos^4x+8cos^2x-cos^2x-1=-7(cos^2x-1/2)^2+3/4=-7/4* cos^2*2x+3/4所以f(x)∈〔-1,3/4〕T=π/2