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sin2x=2sinxCosx怎么推导出来的

可以从和角公式么推导出来。 sin2x=sin(x+x) =sinxcosx+cosxsinx =2sinxcosx

sin2x =sin(x+x) =sinxcosx+cosxsinx =2sinxcosx

参考

以此类推,还可以得到三倍角公式.....

∵y=sin2x+2sinxcosx=1?cos2x2+sin2x=sin2x-12cos2x+12=52sin(2x+φ)+12,(tanφ=-12)∴其周期T=2π2=π.故答案为:π.

解: ∵f(cosx)=2-sin2x=2-2sinxcosx 设cosx=t ∴sint=√(1-t^2) ∴f(t)=2-2t·√(1-t^2) 再设t=sinx ∴f(sinx)=2-2sinxcosx f(sinx)=2-sin2x

lim[(sin2x)/x]=lim[(2sinxcosx)/x]=lim{[(2sinx)/x]•cosx}

sin2x=sin2x/[(sinx)^2+(cosx)^2] =2sinxcosx/[(sinx)^2+(cosx)^2] ,(分子和分母同除以(cosx)^2得 =2(sinx/cosx)/[1+(sinx/cosx)^2] =2tanx/[1+(tanx)^2] 所以,sin2x=2tanx/[1+(tanx)^2]

因为cosX- cos3X =cos(2x-x)-cos(2x+x) =cos2xcosx+sin2xsinx -(cos2xcosx-sin2xsinx) =2sin2xsinx

=∫(1-2sin²x-2sinxcosx)/(cosx+sinx)dx =∫1/√2sin(x+π/4)-2sinxdx =-1/√2∫1/(1-cos²(x+π/4))dcos(x+π/4)+2cosx =-(1/2√2)ln(1+cos(x+π/4))/(1-cos(x+π/4))+2cosx+C

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