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求高手解三角函数,sin3X+cos2X的周期
人气:141 ℃ 时间:2020-04-15 04:50:38
解答
Cos2a=Cos^2(a)-Sin^2(a)=2Cos^2(a)-1=1-2Sin^2(a)
sin(3a)  =sin(a+2a)  =sin2acosa+cos2asina  =2sina(1-sin²a)+(1-2sin²a)sina  =3sina-4sin^3a
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