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ǰλãҳ >> 2Cos2xCosx=Cosx+Cos3xΪʲô? >>

2Cos2xCosx=Cosx+Cos3xΪʲô?

Ͳʽ 2cosAcosB = cos(A+B)+cos(A-B) A=2xB=x뼴

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cos2x-x+cos2x+x =cos2xcosx+sin2xsinx+cos2xcosx-sin2xsinx =2cos2xcosx

⣺ ǺͲʽתġ£ cos(+)=coscos-sinsin cos(-)=coscos+sinsin ʽӣ cos(+)+cos(-) = 2coscos coscos=1/2[cos(+)+cos(-)] Ӷ cosx/2cos3x/2=1/2[cos(x/2+3x/2)+cos(x/...

cosxcos3x =1/2*(cosxcos3x+cosxcos3x) =1/2*(cosxcos3x+sinxsin3x+cosxcos3x-sinxsin3x) =1/2*[cos(3x-x)+cos(3x+x)] =1/2*(cos4x+cos2x)

cos2xcos3x =cos3xcos2x =(1/2){cos(3x+2x)+cos(3x-2x)} =1/2(cos5x+cosx)

ϧͲʽɾˣƽ˼ɡ cosx+cos3x=cos(2x-x)+cos(2x+x) =[cos2xcosx+sin2xsinx]+[cos2xcosx-sin2xsinx] =2cos2xcosx ͬ 2ã cosxcos2x=(1/2)(cosx+cos3x)

ͲʽڵĽ̲Ѿɾˣ ˼· ҵ x3x ƽֵ2x ƽֵ 2x ƽֵΪߣԭĽǵʽɵߵĽ x,2x, ˼·ڵȲУ cosx+cos3x=cos(2x-x)+cos(2x+x) =[cos2xc...

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