What is the value of $\sin 47^{\circ}+\sin 61^{\circ}- \sin25^{\circ} -\sin11^{\circ}$? 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?
 A: 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
$$X = \cos 7.$$
A: Hint: Use the identity: $$\cos\alpha\cos\beta=\dfrac{\cos(\alpha-\beta)+\cos(\alpha+\beta)}{2}.$$ By the way, you should revise your calculations since: $$\sin 47^{\circ}+\sin 61^{\circ}- \sin25^{\circ} -\sin11^{\circ}\neq4\cos36^\circ\cdot\cos7^\circ\cdot\cos18^\circ.$$
A: Using Prosthaphaeresis Formulas
$$\sin47^\circ+\sin61^\circ=2\sin54^\circ\cos7^\circ$$
$$\sin25^\circ+\sin11^\circ=2\sin18^\circ\cos7^\circ$$
$$\implies\sin47^\circ+\sin61^\circ-(\sin25^\circ+\sin11^\circ)=2\cos7^\circ(\sin54^\circ-\sin18^\circ)$$
$$S=\sin54^\circ-\sin18^\circ$$
Method $\#1:$
$$S=\sin54^\circ-\sin18^\circ=2\sin18^\circ\cos36^\circ=2\cos72^\circ\cos36^\circ$$
$$=2\cos72^\circ\cdot\frac{2\sin36^\circ\cdot\cos36^\circ}{2\sin36^\circ}=2\cos72^\circ\cdot\frac{\sin72^\circ}{2\sin36^\circ}$$
$$=\frac{\sin144^\circ}{2\sin(180^\circ-36^\circ)}=\frac12$$
Method $\#2:$
$$S=\sin54^\circ-\sin18^\circ=\sin54^\circ+\sin(-18^\circ)$$
Multiplying either sides by $\displaystyle2\sin\left(\frac{54^\circ-(-18^\circ)}2\right)=2\sin36^\circ$
$$2\sin36^\circ\cdot S=2\sin36^\circ\sin54^\circ-2\sin36^\circ\sin18^\circ$$
Using Werner Formulas
$$2\sin36^\circ\cdot S=\cos18^\circ-\cos90^\circ-(\cos18^\circ-\cos54^\circ)=\cos54^\circ$$
$$\implies2S=\frac{\cos54^\circ}{\sin36^\circ}=1$$
Method $\#3:$
$$S=\sin54^\circ-\sin18^\circ=\cos36^\circ+\cos108^\circ\text{ as }\cos(90^\circ+y)=-\sin y$$
Multiplying either sides by $\displaystyle2\sin\left(\frac{108^\circ-36^\circ}2\right)=2\sin36^\circ$
The rest is like Method$\#2$
