證明：
令S=cos( π/2n+1)+cos( 3π/2n+1)+cos( 5π/2n+1)+cos(7 π/2n+1)+...+cos[(2n-1) π/2n+1)],
則2sin(π/2n+1)×S = sin(2π/2n+1)+[sin(4π/2n+1)-sin(2nπ/2n+1)]+[sin(6π/2n+1)-sin(4π/2n+1)]+...
+{sin(2nπ/2n+1)-sin[(2n-2)π/2n+1]}
=sin2nπ/(2n+1)
得S=1/2