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?

From the expression given for $\sin 10^{\circ}$ it is obvious that it is a root of a quadratic equation. Comparing $$\frac{-1 + \sqrt{9 - 8\sin 50^{\circ}}}{4}$$ with $$\frac{-b + \sqrt{b^{2} - 4ac}}{2a}$$ we can see that it almost fits with $a = 2, b = 3, c = \sin 50^{\circ}$ and we have then $$\frac{-1 + \sqrt{9 - 8 \sin 50^{\circ}}}{4} = \frac{1}{2} + \frac{-b + \sqrt{b^{2} - 4ac}}{2a}$$ Our job is now complete if we can show that $\alpha = \sin 10^{\circ} - (1/2) = \beta - (1/2)$ is a root of the equation $$ax^{2} + bx + c = 0$$ We have then $$a(\beta - 1/2)^{2} + b(\beta - 1/2) + c = 0$$ or $$4a\beta^{2} + 4(b - a)\beta + a - 2b + 4c = 0$$ or $$2\sin^{2}10^{\circ} + \sin 10^{\circ} + \sin 50^{\circ} = 1$$ or $$2\sin^{2}10^{\circ} + 2\sin 30^{\circ}\cos 20^{\circ} = 1$$ or $$2\sin^{2}10^{\circ} = 1 - \cos 20^{\circ}$$ which is true via the identity $1 - \cos A = 2\sin^{2}(A/2)$.
