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sin3x sinx

解:∫sin3XsinXdx =∫-(1/2)[cos(3x+x)-cos(3x-x)]dx =-(1/2)∫(cos4x-cos2x)dx =(-1/2)[(1/4)sin4x-(1/2)sin2x]+C =-(1/8)sin12x+(1/4)sin2x+C 不懂再问懂请采纳

sin3x =sin(2x+x) =sin2xcosx+cos2xsinx =2sinxcosxcosx+cos2xsinx =2sinxcos²x+cos2xsinx sin3x-sin2x+sinx =2sinxcos²x+cos2xsinx-sin2x+sinx=0 若sinx=0 此时x=kπ 若sinx≠0时,得: 2cos^2x+cosx+1=0 相当于一元二次方程2x^2+x+1=0...

sin3x=sinx3x=x+2kπ或(2k+1)π-xx=kπ或[(2k+1)/4]π

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...

记住在x趋于0的时候, sinx就等价于x, 那么这里的sin3x/sinx就 等价于3x/x, 显然极限值就是3

不是的。书上有公式,你想化成什么形式啊

lim(x->0) (sin5x - sin3x)/sinx =lim(x->0) (5x - 3x)/x =2

f ′(0)=0

令 f(x)=sin4xsin2x-sinxsin3x=?12(cos6x?cos2x)+12(cos4x?cos2x)=12(cos4x-cos6x),则有f′(x)=3sin6x-2sin4x,令f′(x)=0,可得x=0 或 x=π2,即f′(0)=0,f′(π2)=0,而且还有f′(π)=0.由于f′(x)在x=0的左侧小于0,右侧大于0,故f...

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