怎么化解cosx+cos2x+.+cosnx
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怎么化解cosx+cos2x+.+cosnx
同上
同上
![怎么化解cosx+cos2x+.+cosnx](/uploads/image/z/6927028-52-8.jpg?t=%E6%80%8E%E4%B9%88%E5%8C%96%E8%A7%A3cosx%2Bcos2x%2B.%2Bcosnx)
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
=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
怎么化解cosx+cos2x+.+cosnx
cosx+cos2x+cos3x+.+cosnx=
高中三角函数题:化简cosx+cos2x+...+cosnx
化简:cosx+cos2x+cos3x+……cosnx=?
求和Sn=cosx+cos2x+cos3x+……+cosnx
三角恒等变换证明cosx+cos2x+…+cosnx=[cos(n+1/2)·sinnx/2]/sinx/2怎么证明?
COSX+COS2X+COS3X+COS4X+COS5X+COS6X+...+COSNX=1/2|{SIN(N+1/2)
求证:cosx+cos2x+…+cosnx=cosn+12x•sinn2xsinx2
lim[(1-cosx*cos2x****cosnx)/x^2]在x趋于0时
C(N,1)COSX+C(N,2)COS2X+-----+C(N,N)COSNX
0.5+cosx+cos2x+cos3x…………cosnx,把这个式子化简成分子和分母的形式
求证:cosx+cos2x+...+cosnx={[cos(n+1)x/2]*[sin(n/2)x]}/[sin(x/2