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Cos3x

cos3x-1 =cos(2x+x)-1 =cos2xcosx-sin2xsinx-1 =(2cos²x-1)cosx-2(1-cos²x)cosx-1 =4cos³x-3cosx-1

lim(x->0) ( cosx - cos3x) /x^2 (0/0) =lim(x->0) ( -sinx +3sin3x) /(2x) (0/0) =lim(x->0) ( -cosx +9cos3x) /2 =( -1+9)/2 =4

为sinx-1/3*sin^3x+C 具体过程看图,有不懂可以问我~~~

令y=cosu,u=3x,根据复合函数的求导规则dy/dx=(dy/du)*(du/dx) y'=(cosu)'(3x)'=-3sinu,u替换为3x,最终结果y'=-3sin3x

cosx周期是2π cos3x周期是 2/3 π 所以周期应该是两者的最小公倍数 也就是 2π

cos3x=cos(2x+x)=cos2xcosx-sin2xsinx =[(cosx)^2-(sinx)^2]cosx-2sinxcosxsinx =(cosx)^3-(sinx)^2cosx-2(sinx)^2cosx =(cosx)^3-3(sinx)^2cosx sin3x=sin(2x+x)=sin2xcosx+cos2xsinx =2sinxcosxcosx+[(cosx)^2-(sinx)^2]sinx =2sinx(cosx)^2+(...

你好!可以用分部积分法如图求出原函数。经济数学团队帮你解答,请及时采纳。谢谢!

cosx-cos3x=cosx-[cos2xcosx-sin2xsinx] =cosx-cos2xcosx+sin2xsinx =cosx(1-cos2x)+sin2xsinx =cosx*2sin^2x+sin2xsinx =sin2xsinx+sin2xsinx =2sin2xsinx

你好! 数学之美团为你解答 积化和差公式:cosαcosβ = 1/2 [ cos(α+β) + cos(α - β) ] cosx cos2x cos3x = 1/2 ( cos3x + cosx ) cos3x = 1/2 cos²(3x) + 1/2 cosx cos3x = 1/4 ( 1 + cos6x ) + 1/4 ( cos4x + cos2x ) ∴ ∫ cosx cos2x cos3...

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