<s>[INST] Geometric proof of $\frac{\sin{60^\circ}}{\sin{40^\circ}}=4\sin{20^\circ}\sin{80^\circ}$ It is well-known that $$\sin{20^\circ}\sin{40^\circ}\sin{80^\circ}=\frac{\sqrt{3}}{8}$$
It follows that $$\frac{\sin{60^\circ}}{\sin{40^\circ}}=4\sin{20^\circ}\sin{80^\circ}$$
But how to prove this by geometry?
Thank you.
 [/INST] Another Trigonometric & algebraic proof:
If $\sin3x=\sin3A,3\sin x-4\sin^3x=\sin3A\iff4\sin^3x-3\sin x+\sin3A=0$
$\implies\prod_{r=1}^3\sin x_r=-\dfrac{\sin3A}4$
Again $\sin3x=\sin3A\implies3x=n180^\circ+(-1)^n3A$ where $n$ is any integer
$\implies x=n60^\circ+(-1)^nA$ 
If we choose even $n=2m,x=120^\circ m+A$
$\implies\sin A\sin(120^\circ+A)\sin(240^\circ+A)=-\dfrac{\sin3A}4$
Now use $\sin(180^\circ-y)=\sin y$ to find $\sin(120^\circ+A)=\sin[180^\circ-(60^\circ-A)]=\sin(60^\circ-A)$
and 
use $\sin(180^\circ+y)=-\sin y$ to find $\sin(240^\circ+A)=\sin[180^\circ+(60^\circ+A)]=-\sin(60^\circ+A)$
Here $A=20^\circ$
You can also choose odd $n$ and use the last formula to derive the same identity
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