2014/070) Prove that $\displaystyle\sum_{n=1}^{15} \sin(4m-2) = \dfrac{1}{4\sin\ 2^\circ}$

we have

LHS = $\dfrac{\displaystyle\sum_{n=1}^{15}2 \sin(4m-2)^\circ\sin\ 2^\circ}{2\sin\ 2^\circ}$
= $\dfrac{\displaystyle\sum_{n=1}^{15} \cos(4m-2-2)^\circ-\cos(4m-2+2)^\circ}{2\sin\ 2^\circ}$
=  $\dfrac{\displaystyle\sum_{n=1}^{15} \cos(4m-4)^\circ-\cos(4m)^\circ}{2\sin\ 2^\circ}$
=  $\dfrac{\cos\ 0^\circ-\cos\ 60^\circ}{2\sin\ 2^\circ}$
=  $\dfrac{1 -\dfrac{1}{2}}{2\sin\ 2^\circ}$
=  $\dfrac{1}{4\sin\ 2^\circ}$