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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?

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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.

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  • $\begingroup$ Nice approach. +1. $\endgroup$ – N.S.JOHN Sep 24 '16 at 3:26
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    $\begingroup$ And are you really just 12? $\endgroup$ – N.S.JOHN Sep 24 '16 at 3:26
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    $\begingroup$ @N.S.JOHN Yes, I am twelve. Thank you for your appreciation. $\endgroup$ – астон вілла олоф мэллбэрг Sep 24 '16 at 3:28
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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 $1 - \cos A = 2\sin^{2}(A/2)$.

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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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  • $\begingroup$ Here $$A=40^\circ,\sin10^\circ=\cos?,\sin50^\circ=\cos?$$ $\endgroup$ – lab bhattacharjee Sep 28 '16 at 16:09

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