[Math] the value of $\sin 47^{\circ}+\sin 61^{\circ}- \sin25^{\circ} -\sin11^{\circ}$

Question

After simplification using sum to product transformation equations I keep ending up with

$$4\cos36^\circ\cdot\cos7^\circ\cdot\cos18^\circ$$

How do I simplify this to a single term?

Answer 1

First we rearrange \begin{align} X = \sin 47 + \sin 61 - \sin 25 - \sin 11 & = \sin 47 - \sin 25 + \sin 61- \sin 11 \end{align} In general, we have $\sin a - \sin b = 2\cos(\frac{a+b}{2})\sin(\frac{a-b}{2})$, and therefore, $\sin 47 - \sin 25 = 2\cos 36 \sin 11$, and $\sin 61 - \sin 11 = 2\cos 36\sin 25$. Substituting to above, we obtain \begin{align} X = 2\cos 36(\sin 11 + \sin 25). \end{align} We now have in general $\sin a + \sin b = 2\cos(\frac{a-b}{2})\sin(\frac{a+b}{2})$, and thus \begin{align} X = 4\cos 36\sin 18 \cos 7. \end{align} On the other hand, $\cos 36 = \frac{1+\sqrt{5}}{4}$ and $\sin 18 = \frac{\sqrt{5}-1}{4}$, and therefore, in fact $$X = \cos 7.$$