How to prove $\sin 10^\circ = \frac{-1+\sqrt{9-8\sin 50^\circ}}{4}$? How to prove this identity?
$$\sin 10^\circ = \frac{-1+\sqrt{9-8\sin 50^\circ}}{4}$$
Is this a particular case of a more general identity? Also, is it possible to give a geometric proof of this equality?

The way  you would derive this identity is as follows:
Note that $\sin A + \sin B = 2\sin\big(\frac{A+B}{2}\big)\cos\big(\frac{A-B}{2}\big)$
Putting $A=50$ and $B=10$,
$$
\sin 50 + \sin 10 = 2\sin\big(30\big)\cos\big(20\big) = \cos 20
$$
Now, we know that $\cos 20 = 1- 2\sin^2 10$. So we substitute:
$$
\sin 50 + \sin 10 = 1 - 2\sin^2 10
$$
Taking all the $\sin 10$ terms to one side to make a quadratic equation:
$$
2\sin^2 10 + \sin 10 + (\sin 50 - 1) = 0
$$
Solving for $\sin 10$ like a quadratic equation gives you your answer.
This answer contained two important points:
1) $\sin 30 = \frac{1}{2}$, helped us remove one sine factor from the sum.
2) $20 = 2*10$ allowed us to further reduce the equation to just the sines of $10$ and $50$.
I do not know of a general formula, but with respect to these constraints, $10$ and $50$ feel like unique choices, summing to $60$, twice of $30$ (whose sine is nice), and a difference of $40$, half of which is $20$, nicely expressible in terms of trigonometric functions of $10$. I cannot think of generalizations of this problem.
