Show that $$\sin47^\circ+\sin61^\circ-\sin11^\circ-\sin25^\circ=\cos7^\circ$$
NOTE: I have seen the other questions and solutions for this problem. I have a particular question if my idea has a potential.
I decided to rearrange the LHS (without a particular reason, it just felt right to me) as follows $$\begin{align}(&\sin47^\circ-\sin11^\circ)+(\sin61^\circ-\sin25^\circ)\\&=2\cos\frac{47^\circ+11^\circ}{2}\sin\frac{47^\circ-11^\circ}{2}+2\cos\frac{61^\circ+25^\circ}{2}\sin\frac{61^\circ-25^\circ}{2}\\&=2\cos29^\circ\sin18^\circ+2\cos43^\circ\sin18^\circ\\&=2\sin18^\circ(\cos29^\circ+\cos43^\circ)\\&=4\sin18^\circ\cos36^\circ\cos7^\circ\end{align}$$
If this won't work, what is the intuition that leads to the appropriate rearranging?
Best Answer
$$4\sin18^\circ\cos36^\circ\cos7^\circ=\frac{4\sin18^\circ\cos18^\circ\cos36^\circ\cos7^\circ}{\cos18^\circ}=\frac{2\sin36^\circ\cos36^\circ\cos7^\circ}{\cos18^\circ}=\\ \frac{\sin72^\circ\cos7^\circ}{\cos18^\circ}=\cos7^\circ$$