<ol class="abc-list"><li data-list-item-id="imported-item-block-0-item-0" data-list-item-path="imported-item-block-0-item-0"><div class="list-item-paragraph">Show that&nbsp;\(\sin 2 nx = \sin((2n + 1)x) \cos x - \cos((2n + 1)x) \sin x\).<span class="inline-marks">[2]</span></div></li><li data-list-item-id="imported-item-block-0-item-1" data-list-item-path="imported-item-block-0-item-1"><div class="list-item-paragraph"><p><strong>Hence</strong> prove, by induction, that \[\cos x + \cos 3x + \cos 5x + ... + \cos((2n - 1)x) = \frac{\sin 2nx}{2 \sin x},\] for all \(n \in \mathbb{Z}^+\), \(\sin x \neq 0\).</p><span class="inline-marks">[12]</span></div></li></ol>

Show that \(\sin 2 nx = \sin((2n + 1)x) \cos x - \cos((2n + 1)x) \sin x\).

<p><strong>Hence</strong> prove, by induction, that \[\cos x + \cos 3x + \cos 5x + ... + \cos((2n - 1)x) = \frac{\sin 2nx}{2 \sin x},\] for all \(n \in \mathbb{Z}^+\), \(\sin x \neq 0\).</p>

Mathematical Induction

International Baccalaureate Mathematics: Analysis and Approaches Higher Level