If $\displaystyle A=36^o$, show that $\displaystyle \sin 3A= \sin 2A$, and deduce that $\displaystyle \cos 36^o=\frac{\sqrt{5}+1}{4}$

I have already shown $\displaystyle \sin 3A= \sin 2A$ when $\displaystyle A=36^o$, since $\displaystyle 108^o=180^o-72^o$.

But I don't know how to deduce that $\displaystyle \cos 36^o=\frac{\sqrt{5}+1}{4}$.

Thanks!