Simple trigonometry When solving a complex geometric problem, I came to a trigonometric equality that needs to be proven:
$${{\sin{100^\circ} \over \sin {60^\circ}}+ {1 \over {1+2\sin{50^\circ}}}} = {2 \sin{50^\circ}}$$
I am pretty sure that it's easy to make it, but I can't see the solution right now. Thanks!
 A: Divide by $2\sin50^{\circ}$
$$\frac{\cos50^{\circ}}{\sin60^{\circ}}+\frac1{2\sin50^{\circ}+2-2\cos100^{\circ}}=1\\
\cos50^{\circ}(2\sin50^{\circ}+2+2\sin10^{\circ})+\sin60^{\circ}=2\sin60^{\circ}(\sin50^{\circ}+1+\sin10^{\circ})\\
\sin100^{\circ}+2\cos50^{\circ}+\sin60^{\circ}-\sin40^{\circ}+\sin60^{\circ}=\\
\cos10^{\circ}-\cos110^{\circ}+2\sin60^{\circ}+\cos50^{\circ}-\cos70^{\circ}\\
\sin80^{\circ}+2\cos50^{\circ}-\sin40^{\circ}=\cos10^{\circ}+\cos70^{\circ}+\cos50^{\circ}-\cos70^{\circ}\\
\sin80^{\circ}+\sin40^{\circ}=\cos10^{\circ}+\sin40^{\circ}
$$
A: $$F=2\sin50^\circ-\frac{\sin100^\circ}{\sin60^\circ}$$
$$=2\cos40^\circ-\frac{\sin100^\circ}{\sin60^\circ}=\frac{2\cos40^\circ\sin60^\circ-\sin100^\circ}{\sin60^\circ}$$
Using Werner Formula $2\sin A\cos B,$
$$F=\frac{\sin100^\circ+\sin20^\circ-\sin100^\circ}{\sin60^\circ}=\frac{\sin20^\circ}{\sin60^\circ}$$
Now $\displaystyle\sin3A=3\sin A-4\sin^3A$
and  for $\displaystyle\sin A\ne0,\dfrac{\sin3A}{\sin A}=3-4\sin^2A=3-2(1-\cos2A)=1+2\cos2A$
Here $A=20^\circ$
Can you take it home from here?
