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sin& x%sin3x

sin3x=sin(x+2x) =sinxcos2x+cosxsin2x =sinx[1-2(sinx)^2]+cosx(2sinxcosx) =sinx[1-2(sinx)^2]+2sinx(cosx)^2 =sinx-2(sinx)^3+2sinx[1-(sinx)^2] =sinx-2(sinx)^3+2sinx-2(sinx)^3 =3sinx-4(sinx)^3 所以 -sin3x=4(sinx)^3-3sinx.

3sin²x cosx-3cos3x 解析: y' =(sin³x-sin3x)' =3sin²x(sinx )'-cos3x*(3x)' =3sin²xcosx-3cos3x

sin3x = sin(x+2x) =sinx*cos2x + cosx*sin2x =sinx*( cosx^2 - sinx^2) + 2 * cosx*sinx*cosx =sinx*cosx^2 - sinx^3 + 2*sinx*cosx^2 =3*sinx*cosx^2 - sinx^3 (像这种题目,应该放在一道题目中,不然展开是没什么用的;)

令a=1/x 则原式=lim(a→0)[sin3a/a+asin(3/a)] 显然lim(a→0)sin3a/a =lim(a→0)=3a/a =3 而lim(a→0)asin93/a)中 3/a趋于无穷 所以sin(3/a)在[-,]震荡,即有界 无穷小乘有界还是无穷小 所以lim(a→0)asin93/a)=0 两部分极限都存在 所以原式=3+0=3

=-1/3*cos3x 换元积分,很简单的题

解法一:等价无穷小 lim sinx/sin3x x→0 =lim x/(3x) x→0 =⅓ 解法二:洛必达法则 lim sinx/sin3x x→0 =lim cosx/(3cos3x) x→0 =cos0/(3·cos0) =1/(3·1) =⅓

无穷小代换: 原式=lim (x-2x)/(x+3x) =lim -x/(4x) =-1/4

sin3x =sin(2x+x) =sin2x×cosx+cos2x×sinx =2×sinx×cosx×cosx+(1-2sin²x)×sinx =2sinx×(1-sin²x)+(1-2sin²x)×sinx =2sinx×(1-sin²x)+(1-2sin²x)×sinx =2sinx-2sin³x+sinx-2sin³x=3sinx-4sin³x

不是的。书上有公式,你想化成什么形式啊

n=1时公式成立; 现在假设对n-1公式成立 那么sinx+sin2x+sin3x+……+sinnx=sinx+sin2x+sin3x+……+sin(n-1)x+sinnx =[sin((n-1)x/2)sin(nx/2)]/sin(x/2)+sinnx =[sin((n-1)x/2)sin(nx/2)+sinnxsin(x/2)]/sin(x/2) =sin(nx/2)[sin((nx/2-x/2)+2cos(nx...

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