Since \(\{\sin0,\sin(\pm\pi),\sin(\pm2\pi),\ldots\}\) all amount to 0, the first term disappears. Meanwhile, \(\{\cos0,\cos(\pm\pi),\cos(\pm2\pi),\ldots\}=\{1,-1,1,\ldots\}\) in an alternating pattern, which agrees with the sequence \((-1)^n\). Hence \(\sin(n\pi+\theta)=(-1)^n\sin\theta\) for integers \(n\).