trigonometry
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.
Just for fun, a complicated proof by dissection:
The sine theorem gives that the length of the red segment is $\frac{\sin 40}{\sin 80}$, hence:
$$ 3+\frac{\sin 20}{\sin 40} = \frac{\sin 80}{\sin 20}+\frac{\sin 40}{\sin 80}$$
as wanted.
That last equation is trivial because $\sin 3\theta=3\sin \theta-4\sin^3\theta$.
Best Answer
Hint: $$ \sin(x)\sin(y)=\frac{\cos(x-y)-\cos(x+y)}2 $$