<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 key observation is $36 + 72 + 72 = 180$. This means $$\sin 72 = \sin (36 + 72) = \sin 36 \cos 72 + \cos 36 \sin 72$$
Using the identity $\sin 72 = 2\sin 36 \cos 36$, we get
$$2\sin 36 \cos 36 = \sin 36 \cos 72 + 2 \sin 36 \cos^2 36$$
Now notice $\sin 36$ appears everywhere, so we can divide by it...
$$2\cos 36 = \cos 72 + 2\cos^2 36$$
Use the identity $\cos 72 = 2 \cos^2 36 - 1$, we get
$$2\cos 36 = 4\cos^2 36 - 1$$
So $\cos36$ is one of the solutions to $2x = 4x^2 - 1$. Using the quadratic formula, you see that the solutions are $$\frac {2 \pm \sqrt{20}}4 = \frac {1 \pm \sqrt 5}2$$
But you know that $\cos 36$ must be positive, so you know the correct value. Now you can get $\sin 72$ from any identity you wish, and multiply.
How to get such values "in general": This is hard. The reason $36$ and $72$ are easy values is that a small integer multiple of them gives you a multiple of $180$. If you wanted to compute such a value for let's say $\sin 1$, it would be a very nasty expression, as you would have to iterate many times until you reach a "basic" angle like $180$ (or even $30$).
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