Something quick cause I have nothing else. Yeah I had to scrape the barrel deeply for this one. But, I was a friend’s hero, twice, for my troubles solving this.

sin(pi/100) + sin(2pi/100) + sin(3pi/100) + … + sin(199pi/100) = ?

For each angle “a” such that 0 <= a <= pi, Sine(a) = -Sine(a + pi). (this also holds for all “a”)

It follows that Sine(a) + Sine(a + pi) = 0

The series can be re-written as

[sin(pi/100) + sin(101pi/100)] + [sin(2pi/100) + sin(102pi/100)] + … + [sin(99pi/100) + sin(199pi/100)]

Notice that the sum of each square brackets is 0, and 0 + 0 + … + 0 = 0