怎么化解cosx+cos2x+......+cosnx

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)*(2*sinnx*cos(n+1)x)
=(sinnx*cos(n+1)x)/sin(x/2)
或用，[sin(n+1/2)x/sin(x/2)]/2-1/2本回答被提问者采纳