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ǰλãҳ >> ΪʲôCos2x=1%2(sinx)^2 >>

ΪʲôCos2x=1%2(sinx)^2

cos2x=(cosx)^2-(sinx)^2 =1-(sinx)^2-(sinx)^2 =1-2(sinx)^2

cos(x+y)=cosxcosy-sinxsiny Ȼ Ƶ cos2x=cos(x+x) cos(x+x)=cosxcosx-sinxsinx=(cosx)^2-(sinx)^2 ϣﵽ㡣 ֻǸһѧд©ָ̡

⣺ cos2x=cos(x+x) =cosx*cosx-sinx*sinx =(cosx)^2-(sinx)^2 ʽӦõcos(x+y)=cosx*cosy-sinx*sinyx=y

sinx̩չxx3/6+ox3Լʣ2/61/3

ҵĺͽǹʽɵãcos2x=cos(x+x)=(cosx)^2-(sinx)^2 (cosx)^2=1-(sinx)^2 ԣcos2x=1-(sinx)^2-(sinx)^2=1-2(sinx)^2

## ̩չʽ ˼·ȫȷӦsinx̩չˣοͼ

sinxcos^2x/1+cos^2xdx =cos^2x/1+cos^2xdcosx cosx=t ԭʽ=t^2/(1+t^2)dt =[1-1/(1+t^2)]dt =t-arctant+c =cosx-arctancosx+c

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this is tks

(sinx)^2*(cos2x)dx =1/2(1-cos2x)*cos2xdx =1/2[cos2x-(cos2x)^2]dx =1/4sin2x-1/4(1+cos4x)dx =1/4sin2x-1/4x-1/16sin4x+C

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