How can the trigonometric equation be proven? This question :
https://math.stackexchange.com/questions/1411700/whats-the-size-of-the-x-angle
has the answer $10°$. This follows from the equation
$$2\sin(80°)=\frac{\sin(60°)}{\sin(100°)}\times \frac{\sin(50°)}{\sin(20°)}$$
which is indeed true , which I checked with Wolfram.


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*How can this equation be proven ?

 A: $$\begin{align}\frac{\sin 80^\circ \sin 20^\circ\sin 100^\circ }{\sin 50^\circ}&=\frac{\sin 80^\circ \sin 20^\circ\cdot 2\sin 50^\circ \cos 50^\circ }{\sin 50^\circ}\\&=2\sin 80^\circ\sin 20^\circ \cos 50^\circ\\&=2\left(-\frac 12(\cos 100^\circ-\cos 60^\circ)\right)\cos 50^\circ\\&=(\cos 60^\circ-\cos 100^\circ)\cos 50^\circ\\&=\frac 12\cos 50^\circ-\cos 100^\circ\cos 50^\circ\\&=\frac 12\cos 50^\circ-\frac 12(\cos 150^\circ+\cos 50^\circ)\\&=-\frac 12\times\left(-\frac{\sqrt 3}{2}\right)\\&=\frac{\sqrt 3}{4}\\&=\frac 12\sin 60^\circ\end{align}$$
A: $$
2\cdot\sin80^\circ\cdot \underbrace{\sin100^\circ}_{\sin80^\circ}\cdot\sin20^\circ=2\cdot \sin80^\circ\cdot2\sin40^\circ\underbrace{\cos40^\circ}_{\sin50^\circ}\cdot\sin20^\circ=\\
=\underbrace{4\cdot\sin80^\circ\cdot\sin40^\circ\cdot\sin20^\circ}_{\sqrt{3}/2}\cdot\sin50^\circ=\sin60^\circ\cdot\sin50^\circ.
$$
The only non-obvious identity is Morrie's law for sine.
A: $$\frac{\sin 80^\circ \sin 20^\circ\sin 100^\circ }{\sin 50^\circ}=\frac{\sin 80^\circ \sin 20^\circ\cdot 2\sin 50^\circ \cos 50^\circ }{\sin 50^\circ}=2\sin 80^\circ\sin 20^\circ\cos 50^\circ=2\sin 80^\circ\sin 20^\circ\sin40^\circ$$
as $\cos 50^\circ=\sin40^\circ$
Now by Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $, $$\sin\left(60^\circ-x\right)\sin\left(60^\circ+x\right)=\sin^260^\circ-\sin^2x$$
$$\implies\sin x\sin\left(60^\circ-x\right)\sin\left(60^\circ+x\right)=\dfrac{\sin3x}4$$
Here $x=20^\circ$
