F(x,n) = cos x + cos2x + cos3x + cos4x + ...+ cos(nx)
2sin(x/2) F = 2sinx/2cosx + 2sinx/2cos2x + ...+ 2sinx/2 cos(nx)
= sin(3x/2) - sin(x/2) + sin(5x/2) - sin(3x/2) + ...+ sin(n+1/2)x - sin(n-1/2)x
= sin(n+1/2)x - sin(x/2)
令 x = 1,n = 89得
2sin(0.5)F(1,89) = cos1 + cos2 + ...+ cos89 = sin(89.5) - sin(0.5) = cos0.5 - sin0.5
所以cos90+cos89+cos88+……cos0 = 1 + 1/2 cot0.5 - 1/2 = 1/2 ( cot0.5 + 1)