Claim: \(\sum_{k=1}^n \sin(k) = \frac{\sin(n/2) \sin((n+1)/2)}{\sin(1/2)}\).

Proof: Observe that \(\sum_{k=1}^n k = \frac{(n/2)((n+1)/2)}{1/2}\). Now apply $\sin$ to both sides.

I saw this somewhere but have forgotten where.