<s>[INST] Proving $\frac{\sin 30^\circ}{\sin 50^\circ}= \frac{\sin 40^\circ}{\sin 80^\circ}$ 
Prove that
$$\frac{\sin 30^\circ}{\sin 50^\circ}= \frac{\sin 40^\circ}{\sin 80^\circ}$$

I tried making $\sin 80^\circ=\sin(50^\circ+30^\circ)$, but it didn't go well. I also tried using, maybe, trig periodicities, but I still can't get it.
 [/INST] Let us first prove that $\sin(30^\circ) \cdot \sin(80^\circ) = \sin(40^\circ) \cdot \sin(50^\circ)$.
\begin{align}
LHS & = \sin(30^\circ) \cdot \sin(80^\circ) \\
&= \cos(60^\circ) \cdot \cos(10^\circ) \\
& = \frac{1}{2} \cos(10^\circ) \\
\end{align}
\begin{align}
RHS & = \sin(40^\circ) \cdot\sin(50^\circ) \\
& = \frac{1}{2}[\cos(40^\circ-50^\circ)-\cos(40^\circ+50^\circ)] \\
& = \frac{1}{2}[\cos(10^\circ) - \cos(90^\circ)] \\
& = \frac{1}{2} \cos(10^\circ) \\
\therefore LHS & = RHS
\end{align}
So we have $$\sin(30^\circ) \cdot \sin(80^\circ) = \sin(40^\circ) \cdot \sin(50^\circ)$$
And hence it follows that $$\frac{\sin(30^\circ)}{\sin(50^\circ)} = \frac{\sin(40^\circ)}{\sin(80^\circ)} $$
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