Find the numerical value of $\sin 10^\circ \sin 50^\circ \sin 70^\circ$. Prerequisite
This problem is found in "Trigonometry" by I. M. Gelfand [in English].
It is asked in the section "Double the angle". So, assume that I know the sin/cos angle additions [i.e.: $\sin(A + B) = \sin A \cos B + \cos A \sin B$, etc.] as well as everything learned prior.
I've check other sources and they say to use Morrie's Law, however I have not actually learned it in the book.

Problem
Find the numerical value of $\sin 10^\circ \sin 50^\circ \sin 70^\circ$.
Hint: If the value of the given expression is $M$, find $M \cos 10^\circ$.


$$\begin{eqnarray*}\sin(A)\sin(B)\sin(C) &=& \frac{1}{2}\left(\cos(A-B)-\cos(A+B)\right)\sin C\\&=&\frac{1}{4}\left(\sin(C+A-B)+\sin(C-A+B)-\sin(C+A+B)-\sin(C-A-B)\right)\end{eqnarray*}$$
hence:
$$\begin{eqnarray*} \sin(10^\circ)\sin(50^\circ)\sin(70^\circ)&=&\frac{1}{4}\left(\sin(110^\circ)+\sin(30^\circ)+\sin(-10^\circ)-\sin(130^\circ)\right) \\&=&\frac{1}{4}\left(\cos(20^\circ)+\sin(30^\circ)-\sin(10^\circ)-\cos(40^\circ)\right)\end{eqnarray*}$$
but:
$$ \cos(20^\circ)-\cos(40^\circ) = 2\sin(30^\circ)\sin(10^\circ) = \sin(10^\circ)$$
hence:

$$ \sin(10^\circ)\sin(50^\circ)\sin(70^\circ) = \color{red}{\frac{1}{8}}.$$

