<s>[INST] Prove this trigonometry equation: $\sin 40^\circ \cdot \sin 50^\circ$ is equal to $\frac{1}{2} \cos 10^\circ$. 
Prove that $\sin 40^\circ \cdot \sin 50^\circ$ is equal to $\frac{1}{2} \cos 10^\circ$.

I've tried writing $\sin 40^\circ$ as $\sin(40^\circ+10^\circ)$, then wrote $\sin(50^\circ+10^\circ)$ as $\sin 40^\circ \cos 10^\circ + \cos 40^\circ \sin 10^\circ$, but I don't know what to do next.
 [/INST] As $\sin50^\circ=\cos(90-50)^\circ$
and $2\sin40^\circ\cdot\cos40^\circ=\sin(2\cdot40)^\circ$
Finally $\sin80^\circ=\cos(90-80)^\circ=?$
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