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 $M \cos 10^\circ$.

 A: From the triple-angle formula $\sin (3\theta) = - 4\sin^3\theta + 3\sin\theta$ when $\sin (3\theta) = 1/2$, we get that
$\sin(10^\circ)$, $\sin(50^\circ)$, $\sin(-70^\circ)$ are the roots of $8x^3-6x+1$. Therefore
$$
\sin(10^\circ) \sin(50^\circ) \sin(70^\circ)
=-\sin(10^\circ) \sin(50^\circ) \sin(-70^\circ)
=-(-\frac18)
=\frac18.
$$
A: $$\begin{eqnarray*}\sin(A)\sin(B)\sin(C) &=& \frac{1}{2}\left(\cos(A-B)-\cos(A+B)\right)\sin C\\&=&\frac{1}{4}\left(\sin(C+A-B)+\sin(C-A+B)-\sin(C+A+B)-\sin(C-A-B)\right)\end{eqnarray*}$$
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:

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

A: Here is a method that uses Gelfand's hint.
Let us first transform all the sines into cosines of the complementary angle:
$\tag{1}M:=\sin(10^\circ)\sin(50^\circ)\sin(70^\circ)=\cos(80^\circ)\cos(40^\circ)\cos(20^\circ)$
Let us multiply LHS and RHS of $(1)$ by $\sin(20^\circ):$
$M \sin(20^\circ)=\cos(80^\circ)\cos(40^\circ)(\cos(20^\circ)\sin(20^\circ).$
$=\cos(80^\circ)\cos(40^\circ)\frac12\sin(40^\circ).$
(applying the so useful formula $\sin(a)\cos(a)=\frac12\sin(2a).$)
Using the same trick again, we get:
$M \sin(20^\circ)=\frac14\cos(80^\circ)\sin(80^\circ).$
And once more:
$M \sin(20^\circ)=\frac18\sin(160^\circ).$
But $\sin(160^\circ)=\sin(20^\circ).$
We can thus conclude that

$M=\dfrac18.$

Edit: in fact, I just discover that it is a direct consequence of Morrie's law.
A: I would use a variation of Jean Marie's approach to this. Let 
$$
 =\sin(10)\sin(50)\sin(70)
$$
(multiply both sides by $\cos(10)$ as per the hint)
$$
\cos(10)M = \cos(10)\sin(10)\sin(50)\sin(70)
$$
(apply double angle rule $\sin(a)\cos(a)=1/2\sin(2)$)
$$
\cos(10)M = 1/2\sin(20)\sin(50)\sin(70)
$$
(apply $\sin(x) = \cos(90-x)$)
$$
\cos(10)M = 1/2\sin(20)\sin(50)\cos(20)
$$
(rearrange)
$$
\cos(10)M = 1/2 \sin(20)\cos(20)\sin(50)
$$
(double angle rule)
$$
\cos(10)M = 1/4 \sin(40) \sin(50)
$$
(complimentary cosine and double angle again)
$$
\cos(10)M = 1/8 \sin(80)
$$
(cancel out because $\cos(10) = \sin(80)$)
$$
M = 1/8
$$
A: Using $\sin(90^\circ-x)=\cos x$
$$\sin10^\circ\sin50^\circ\sin70^\circ=\cos20^\circ\cos40^\circ\cos80^\circ$$
Now $S=\cos x\cos2x\cos4x=\dfrac{\sin2x\cos2x\cos4x}{2\sin x}$ if $\sin x\ne0\iff x\ne180^\circ m$ where $m$ is any integer
$\implies8\sin x\cdot S=\sin8x$
If $\sin8x=\sin x\iff8S=1$
We need $8x=n180^\circ+(-1)^nx$ where $n$ is any integer
If $n$ is even, $=2m$(say), $7x=360^\circ m$ where $7\nmid m$
If $n$ is odd, $=2m+1$(say), $x=(2r+1)20^\circ$  where $r$ is any integer , but $2r+1\not\equiv9\pmod{18}\iff r\not\equiv4\pmod9$
In this problem $r=0$
A: We can prove $$4\sin(60^\circ-x)\sin x\sin(60^\circ+x)=\sin3x$$
Proof $\#1:$
Using Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $,
$$\sin(60^\circ-x)\sin(60^\circ+x)=\sin^260^\circ-\sin^2x?$$
and $\sin3x=3\sin x-4\sin^3x$
Proof$\#2:$
If $\sin3x=\sin3A, 3x=180^\circ n+(-1)^n3A$ where $n$ is any integer
$\implies x=60^\circ n+(-1)^nA$ where $n\equiv-1,0,1\pmod3$
$\implies x=-(60^\circ+A), A,60^\circ-A$
As $\sin3x=3\sin x-4\sin^3x,3\sin x-4\sin^3x=\sin3A\iff4\sin^3x-3\sin x-\sin3A=0$
Using Vieta's formula, $$\prod_{n=-1}^1\sin\left(60^\circ n+(-1)^nA\right)=\dfrac{\sin3A}4$$
Use $\sin\left\{-(60^\circ+A)\right\}=-\sin(60^\circ+A)$
A: Here's the strategy that I would use.
Note that
$$\sin 10^\circ\sin 50^\circ\sin 70^\circ = \frac{\sin 10^\circ\sin 30^\circ\sin 50^\circ\sin 70^\circ}{\sin 30^\circ}$$
And that 
$$\sin 10^\circ\sin 30^\circ\sin 50^\circ\sin 70^\circ = \frac{\sin 10^\circ\sin 20^\circ\sin 30^\circ\sin 40^\circ \sin 50^\circ\sin 60^\circ\sin 70^\circ\sin 80^\circ}{\sin 20^\circ\sin 40^\circ\sin 60^\circ\sin 80^\circ}$$
See where this is going?
