<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] Hint: Notice that $72 = 2\cdot 36$ and $180 = 36\cdot 5$; use formulaes for halfed or doubled argument.
$$
\sin^2\frac{x}{2} = \frac{1-\cos x}{2}
$$
or
$$
\sin2x = 2\sin x\cos x
$$
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