Something quick because I have nothing else and social media math is fun.

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

Solution

For each angle θ such that 0 <= θ <= π, sin(θ) = -sin(θ + π). (this also holds for all θ)

It follows that sin(θ) + sin(θ + π) = 0

The series can be re-written as

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

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

QED