Question

函数f(x）=sin^4x+cos^4x的最小正周期？

Answer 1

因为 ( sin^2 x + cos^2 x ) ^ 2 = sin^4 x + cos^4 x + 2 sin^2 x cos^2 x
所以 f(x) = ( sin^2 x + cos^2 x ) ^ 2 - 2 sin^2 x cos^2 x
f(x) = 1 - 2 sin^2 x cos^2 x
= 1 - 1/2 sin^2 (2x)
= 1 - 1/4 * ( 1 - cos 4x )
cos x 最小正周期为 2pi,cos 4x 的最小正周期为 2pi/4 = pi/2,1,y=sin^4x+cos^4x=[(sinx)^2+(cosx)^2]^2-2(sinxcosx)^2
=1-2(sinxcosx)^2
=1-[(sin2x)^2]/2
=1-[1-(cos4x)]/4
=(3+cos4x)/4
pai
所以周期t=-------
2,1,y=(sin²x+cos²x)²-2sin²xcos²x
=1-(1/2)(2sinxcosx)²
=1-(1/2)sin²2x
所以最小正周期为π,0,