sinx+sin2x+sin3x+...+sin(nx)

=1/2*(cot(x/2)-cot(x/2)*cos(nx)+sin(nx))。

解：因為cos(A-B)-cos(A+B)=2sinAsinB，

那麼2sinx*sin(x/2)=cos(x-x/2)-cos(x+x/2)=cos(x/2)-cos(3x/2)

2sin(2x)*sin(x/2)=cos(2x-x/2)-cos(2x+x/2)=cos(3x/2)-cos(5x/2)

2sin(3x)*sin(x/2)=cos(3x-x/2)-cos(3x+x/2)=cos(5x/2)-cos(7x/2)

......

2sin(nx)*sin(x/2)=cos(nx-x/2)-cos(nx+x/2)=cos((2n-1)x/2)-cos((2n+1)x/2)

等式兩邊全部相加得，

2sinx*sin(x/2)+2sin(2x)*sin(x/2)+2sin(3x)*sin(x/2)+...+2sin(nx)*sin(x/2)

=cos(x/2)-cos(3x/2)+cos(3x/2)-cos(5x/2)+cos(5x/2)-cos(7x/2)+...+cos((2n-1)x/2)-cos((2n+1)x/2)

化簡可得，

2sin(x/2)*(sinx+sin2x+sin3x+...+sinnx)=cos(x/2)-cos((2n+1)x/2)

那麼，

sinx+sin2x+sin3x+...+sinnx=(cos(x/2)-cos((2n+1)x/2))/(2sin(x/2))

=1/2*(cot(x/2)-cot(x/2)*cos(nx)+sin(nx))