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?
 A: 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.
A: 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 $1 - \cos A = 2\sin^{2}(A/2)$.
A: If $(4\cos2A+1)^2=9-8\cos A,$
$9-8\cos A=16\cos^22A+8\cos2A+1=8(1+\cos4A)+8\cos2A+1$
$\iff0=\cos A+\cos2A+\cos4A$
$=\cos A+2\cos A\cos3A=\cos A(1+2\cos3A)$
If $\cos A=0,A=(2n+1)90^\circ$ where $n$ is any integer
Otherwise, $$\cos3A=-\dfrac12=\cos120^\circ$$
$\implies3A=360^\circ m\pm120^\circ$ where $m$ is any integer
The set of values of $A$ can be chosen as $\{40^\circ,80^\circ,160^\circ\}$
For $A=40^\circ,160^\circ;\cos2A>0\implies4\cos2A+1>0$
Consequently, $4\cos2A+1=+\sqrt{9-8\cos A}$
For $A=80^\circ,\cos2A=-\cos20^\circ;$
$4\cos2A+1=1-4\cos20^\circ$ which is $<0$ as $\cos20^\circ>\cos60^\circ=\dfrac12>\dfrac14$
Consequently, $4\cos2A+1=-\sqrt{9-8\cos A}$
