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sin(x + π/6)cos(π/3 - x)的最小正周期怎么算?
人气:429 ℃ 时间:2020-01-29 11:31:17
解答
cos[π/2-(π/3+x)]=cos(π/6-x)=sin(π/3+x)
y=sin(x+π/3)cos(π/6-x)
=sin(x+π/3)sin(π/3+x)
=sin²(π/3+x)
=[1-cos(2π/3+2x)]/2
=1/2-1/2cos(2π/3+2x)
T=2π/2=π
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