2025 SJI P1 Q9

Using the compound angle identity, prove that \[\sin(2r+1)\theta-\sin(2r-1)\theta\equiv2\sin\theta\cos2r\theta.\][2]

Using mathematical induction and the result from part (a), prove that \[\sum_{r=1}^n\cos2r\theta=\frac{\sin(2n+1)\theta-\sin\theta}{2\sin\theta},\quad\text{for}\hspace{0.5em}n\in\mathbb Z^+,\ n\geq1.\][7]

Hence express \(2\sin\theta\cos(4\theta)+2\sin\theta\cos(6\theta)+\cdots+2\sin\theta\cos(2024\theta)\) in the form \(\sin(m\theta)-\sin(n\theta)\), where the integer values of \(m\) and \(n\) are to be found.[4]