化简sin^8x+cos^8x,并求其周期

Question

化简sin^8x+cos^8x,并求其周期

Answer 1

1)
已知：f、g的周期都是：T
f(x+T)=f(t)
g(x+T)=g(t)
h(t)=f(t)+g(t)
h(t+T)=f(t+T)+g(t+T)=f(t)+g(t)=h(t)
那么f+g的周期还是：T.
2) sinx 和 cosx 的周期均为：2π；
sin^8x 和 cos^8x 的周期都是：π；
那么由1）：sin^8x+cos^8x 的周期还是：π
3）化简：
sin^8x+cos^8x=
=(sin^4x-cos^4x)^2+2sin^4xcos^4x
=[(sin^2x-cos^2x)(sin^2x+cos^2x)]^2+2sin^4Xcos^4x
=(sin^2x-cos^2x)^2+2sin^4Xcos^4x
=sin^4x-2sin^2Xcos^2x+cos^4x+2sin^4Xcos^4x
=sin^4x+cos^4x+2sin^4Xcos^4x-2sin^2Xcos^2x=(sin^4x+cos^4x)+2sin^2Xcos^2x(sin^2xcos^2x-1)
=[(sinx-cosx)(sinx+cosx)]^2+2sin^4xcos^4x
=(1-2sinxcosx)(1+2sinxcosx)+2sin^4xcos^4x
=1-4sin^2xcos^2x+2sin^4xcos^4x
= 2(sin^2xcos^2x-1)^2-1
即：sin^8x+cos^8x = 2(sin^2xcos^2x-1)^2-1