<s>[INST] $4 \sin 72^\circ \sin 36^\circ = \sqrt 5$ How do I establish this and similar values of trigonometric functions?
$$
4 \sin 72^\circ \sin 36^\circ = \sqrt 5
$$
 [/INST] A different approach
$ a = n\pi/5$
$5a = n\pi$
$3a = n\pi-2a$
$\sin 3a =  \sin{(n\pi-2a)}$
$\sin 3a = \sin2a$
Expand using identities
Remove a sin
Square both sides
You get a quadratic in $\sin^2$
Let $\sin^2 a = x$
So
$16x^2 -20 a +5 = 0$
So 
$\sin^2 72 \cdot\sin^2 36 = 5/16$
Hence
$4\sin72\sin36 = \sqrt5$
</s>