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sinX的6次方+cosX的6次方的周期怎么算?
人气:120 ℃ 时间:2019-09-09 17:51:51
解答
y=sin^6 x+cos^6 x
=(sin^2 x+cos^2 x)*(sin^4 x+cos^4 x-
sin^2 x*cos^2 x)
=sin^4 x+cos^4 x+2sin^2 x*cos^2 x-3sin^2 x*cos^2 x
=(sin^2 x+cos^2 x)^2-3*sin^2 x*cos^2 x
=1-3*sin^2 x*cos^2 x
=-3*[(1-cos2x)/2]*[(1+cos2x)/2]+1
=-3*[(1-cos^2 2x)/4]+1
=-3/4*sin^2 2x+1
=-3/4*[(1-cos4x)/2]+1
=-3/8*(1-cos4x)+1
=3/8*cos4x+5/8
则T=2兀/4=兀/2
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