TITLE: How to prove $\frac{1+\sin{6^\circ}+\cos{12^\circ}}{\cos{6^\circ}+\sin{12^\circ}}=\sqrt{3}$?
QUESTION [16 upvotes]: I found this interesting result. Show that
$$\dfrac{1+\sin{6^\circ}+\cos{12^\circ}}{\cos{6^\circ}+\sin{12^\circ}}=\sqrt{3}.$$
See this Wolfram Alpha output.
My attempt is very ugly. We know that
$$\sin{6^\circ}=\dfrac{1}{8}(-1-\sqrt{5})+\dfrac{1}{4}\sqrt{\dfrac{3}{2}(5-\sqrt{5})}$$
$$\cos{6^\circ}=\dfrac{1}{4}\sqrt{\dfrac{1}{2}(5-\sqrt{5})}+\dfrac{1}{8}\sqrt{3}(1+\sqrt{5})$$
$$\sin{12^\circ}=2\sin{6^\circ}\cos{6^\circ}$$$$\cos{12^\circ}=2\cos^2{6^\circ}-1$$ and it is very ugly, maybe there exist simple methods.
Thank you.

REPLY [3 votes]: \begin{align}
x&=\frac{1+\sin6+\cos12}{\cos6+\sin12} \\
 &=\frac{2\sin30+\sin78+\sin6}{\cos78+\cos6} \\
 &=\frac{2(\sin54-\sin18)+2\sin42\cos36}{2\cos42\cos36} \\
 &=\frac{4\cos36\sin18+2\sin42\cos36}{2\cos42\cos36} \\
 &=\frac{2\cos36(2\sin18+\sin42)}{2\cos42\cos36} \\
 &=\frac{2\sin18+\sin42}{\cos42} \\
 &=\frac{\sin42+\sin18+\sin18}{\cos42} \\
 &=\frac{2\sin30\cos12+\cos72}{\cos42} \\
 &=\frac{\cos72+\cos12}{\cos42} \\
 &=\frac{2\cos42\cos30}{\cos42} \\
 &=2\cos30 \\
 &=\sqrt 3
\end{align}