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Cos2x=Cosx

∫cosxcos2xdx =∫cos2xd(sinx) =∫1-2sin^2xdsinx =sinx-1/3(sinx)^3+C

f(cosx)=f[sin(x+π/2)]=cos2(x+π/2)+1=cos(2x+π)+1=-cos2x+1

cos2x=1是cosx=1的什么条件 必要条件。 如果没有事物情况A,则必然没有事物情况B;如果有事物情况A而未必有事物情况B,A就是B的必要而不充分的条件,简称必要条件。 假设A是条件,B是结论 (1)由A可以推出B,由B可以推出A,则A是B的充要条件(A=B)...

求y=cosx/cos2x的二阶导数 ,并求二阶导数的零点 解:dy/dx=(-cos2xsinx+2cosxsin2x)/cos²2x 令d²y/dx²=[cos²2x(2sin2xsinx-cos2xcosx-2sinxsin2x+4cosxcos2x)+4cos2xsin2x(-cos2xsinx+2cosxsin2x)]/cos⁴(2x)=0 约分...

这是二倍角公式: cos2x=cos²x-sin²x =2cos²x-1 =1-2sin²x 祝你开心!希望能帮到你,如果不懂,请追问,祝学习进步!O(∩_∩)O

cosx+cos2x+......+cosnx =1/2sin(x/2)*(cosx*2sin(x/2)+cos2x*2sin(x/2)+......+cosnx*2sin(x/2)) =1/2sin(x/2)*(sin(3x/2)-sin(x/2)+sin(5x/2)-sin(3x/2)+......+sin(n+1/2)x-sin(n-1/2)x) =1/2sin(x/2)*(sin(n+1/2)x-sin(x/2)) =1/2sin(x/2)*...

lim(x->0) (cosx-cos2x)/(1-cosx) (0/0) =lim(x->0) (-sinx+2sin2x)/sinx (0/0) =lim(x->0) (-cosx+4cos2x)/cosx =(-1+4)/1 =3

解答如图

首先对等式左边进行化简: 2cosxcosx2x =2cosxcos(x+x) =2cosx(cosxcosx-sinxsinx) =2cosxcosxcosx-2cosxsinxsinx 除以cosx后为: 2cosxcosx-2sinxsinx 然后对等式右边进行化简: cosx+cos3x =cosx+cos(x+2x) =cosx+cosxcos2x-sinxsin2x =cosx+c...

cos(3π/11) =cos(π-8π/11) =-cos(8π/11) cos(5π/11) =cos(16π/11-π) =-cos(16π/11) 所以, cos(π/11)·cos(2π/11)·cos(3π/11) ·cos(4π/11)·cos(5π/11) =cos(π/11)·cos(2π/11)·cos(4π/11) ·cos(8π/11)·cos(16π/11) =32sin(π/11)·cos(π/11)·cos(2π/...

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