<s>[INST] 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?
 [/INST] 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, $4\cos2A+1=-\sqrt{9-8\cos A}$
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