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ǰλãҳ >> 2Cos2xCosx=Cosx+Cos3xΪʲô? >>

2Cos2xCosx=Cosx+Cos3xΪʲô?

⣺ 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/...

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ΪcosX- cos3X =cos(2x-x)-cos(2x+x) =cos2xcosx+sin2xsinx -(cos2xcosx-sin2xsinx) =2sin2xsinx

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

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

1··8ףÿʻ200ף·ʻ6ӣܸĵ棿 26*10ϼһΣʣµǶ٣ 10 3һ30ҳÿҳԷ6Ƭ3һܷŶ...

Sn=cosx+cos2x+cos3x++cosnxУ 2sin(x/2)[cosx+cos2x+cos3x++cosnx ] =2sin(x/2)cosx+2sin(x/2)cos2x+2sin(x/2)cos3x++2sin(x/2)cosnx =sin(3x/2)-sin(x/2)+sin(5x/2)-sin(3x/2)+sin(7x/2)-sin(5x/2)++sin(x/2+nx)-sin(nx-x/2) ...

cosx*cos2x*cos4x = 2 sinx*cosx*cos2x*cos4x / (2sinx) = sin2x * cos2x *cos4x /(2sinx) =......= sin8x / (8sinx) cos3x*cos5x =(1/2) ( cos8x +cos2x) ԭʽ= (1/16) (1/sinx) [ sin8x cos8x + sin8xcos2x ] = (1/32) (1/sinx) [ sin16x + si...

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