Cos 3pi/5

Consider the right-angled triangle with acute angles $\frac{3\pi}{10}$ and $\frac{2\pi}{10}$ and note that $\sin{\left(\frac{3\pi}{10}\right)}=\cos{ \left( \frac{2\pi}{10}\right) }.$

To ease typing, let $\frac{\pi}{10}=A,$ then $\sin 3A=\cos 2A.$

Use the standard identities, $\sin 3A =3\sin A-4\sin^{3} A .............(1) \text{ and } \cos 2A= 1-2\sin^{2} A ............(2)$

and we have the cubic $4\sin^{3} A -2\sin^{2}A-3\sin A +1 = 0.$

$(\sin A -1)$ is a factor, remove this and we are left with the quadratic

$4\sin^{2} A +2\sin A -1=0,$ and solving this, (for the angle in the $0,\pi/2$ range),

$\sin A = \frac{-1+\sqrt{5}}{4}.$

Substituting this into (1) and simplifying leads to $\sin 3A = \sin{ \ \left( \frac{3\pi}{10} \right)} = \frac{1+\sqrt{5}}{4}.$