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.
 A: Hint:
$$
\sin(x)\sin(y)=\frac{\cos(x-y)-\cos(x+y)}2
$$
A: You can use the general formula
$$ \sin\alpha \sin\beta = \frac{\cos(\alpha-\beta) - \cos(\alpha+\beta)}{2}.$$
There are also the similar formulas
$$ \cos\alpha \cos\beta = \frac{\cos(\alpha+\beta) + \cos(\alpha-\beta)}{2}$$
and
$$ \sin\alpha \cos\beta = \frac{\sin(\alpha+\beta) + \sin(\alpha-\beta)}{2}.$$
A: As $\sin50^\circ=\cos(90-50)^\circ$
and $2\sin40^\circ\cdot\cos40^\circ=\sin(2\cdot40)^\circ$
Finally $\sin80^\circ=\cos(90-80)^\circ=?$
A: Taking L.H.S.
$sin40°sin50°$ or $sin50°sin40°$
We know the identity, $sinAsinB=\dfrac{1}{2}\Big(cos(A-B)-cos(A+B)\Big)$
Here we consider A=50° and B=40°
Then, $sin50°sin40°=\dfrac{1}{2}\Big(cos(50°-40°)-cos(50°+40°)\Big)$
$=\dfrac{1}{2}\Big(cos\,10°-cos\,90°\Big)$
$=\dfrac{1}{2}\Big(cos\,10°-0\Big)\;\;\;\;\;(\because cos\,90°=0)$
$=\dfrac{1}{2}cos\,10°$
L.H.S.=R.H.S.
A: Here,
$$L.H.S=sin40°.sin50°$$
$$=\frac {2}{2}.sin40°.sin50°$$
$$=\frac {1}{2} (cos10°-cos90°)$$
$$=\frac {1}{2} cos10°=RH.S$$
                      Proved.
