求证sinπ/3+sin2π/3+...+sinnπ/3=2sinnπ/6sin(n+1)π/6

求证sinπ/3+sin2π/3+...+sinnπ/3=2sinnπ/6sin(n+1)π/6

证明：（数学归纳法）
（1）当n=1时，左边=sin(π/3)=√3/2
右边=2sin(π/6)sin(2π/6)=2(1/2)(√3/2)=√3/2
则左边=右边，原命题成立。
（2）假定当n=k时原命题成立，即
sin(π/3)+sin(2π/3)+...+sin(kπ/3)=2sin(kπ/6)sin((k+1)π/6)
那么，当n=k+1时，有
sin(π/3)+sin(2π/3)+...+sin(kπ/3)+sin((k+1)π/3)
=2sin(kπ/6)sin((k+1)π/6)+sin((k+1)π/3)
=2sin(kπ/6)sin((k+1)π/6)+2sin((k+1)π/6)cos((k+1)π/6) (应用倍角公式)
=2sin((k+1)π/6)[sin(kπ/6)+cos((k+1)π/6)]
=2sin((k+1)π/6)[cos(π/2-kπ/6)+cos((k+1)π/6)] (应用诱导公式)
=2sin((k+1)π/6)[cos((3-k)π/6)+cos((k+1)π/6)]
=2sin((k+1)π/6)[2cos(π/3)cos((k-1)π/6)] (应用和差化积公式)
=2sin((k+1)π/6)[2(1/2)cos((k-1)π/6)]
=2sin((k+1)π/6)cos((1-k)π/6)
=2sin((k+1)π/6)sin(π/2-(1-k)π/6) (应用诱导公式)
=2sin((k+1)π/6)sin((k+2)π/6)
则左边=右边，原命题成立。
于是，有数学归纳法知
sin(π/3)+sin(2π/3)+...+sin(nπ/3)=2sin(nπ/6)sin((n+1)π/6)成立，证毕。