N2024 P1 Q7

It is given that \(\mathrm{f}(r) = \cos(r\theta)\).

1. Show that \(\mathrm{f}(2r - 1) - \mathrm{f}(2r + 1) = 2 \sin\theta \sin(2r\theta)\).[2]

2. Hence, given that \(\sin\theta \neq 0\), show that

   \(\sum_{r=1}^{n} \sin(2r\theta) = \frac{\cos\theta - \cos\big((2n + 1)\theta\big)}{2 \sin\theta}.\)

   [3]

3. Hence find the three possible values of

   \(\sum_{r=1}^{n} \sin\left(\frac{\pi r}{6}\right) \cos\left(\frac{\pi r}{6}\right).\)

   [3]