Show that \(2\cos \left( r\theta \right)\sin \theta \equiv \sin \left[ \left( r+1 \right)\theta \right]-\sin \left[ \left( r-1 \right)\theta \right]\).[1]
By considering the method of differences, find \(\sum\limits_{r=1}^{n}{\cos \left( r\theta \right)}\) where \(0<\theta <\frac{\pi }{2}\).
(You need not simplify your answer.)[3]
Hence evaluate the sum \(\cos \left( \frac{19}{6}\pi \right)+\cos \left( \frac{20}{6}\pi \right)+\cos \left( \frac{21}{6}\pi \right)+...+\cos \left( \frac{56}{6}\pi \right)+\cos \left( \frac{57}{6}\pi \right)\),
leaving your answer in exact form.[4]
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