Prove that $\sin 10^\circ \sin 20^\circ \sin 30^\circ=\sin 10^\circ \sin 10^\circ \sin 100^\circ$? $\sin 10^\circ \sin 20^\circ \sin 30^\circ=\sin 10^\circ \sin 10^\circ \sin 100^\circ$ 
This is a competition problem which I got from the book "Art of Problem Solving Volume 2". I'm not sure how to solve it because there's just so many possibilities of using most of the trig identities that I don't know which path to take. One way I tried is canceling out the $\sin 10$ on both sides:
$\sin 20^\circ \sin 30^\circ= \sin 10^\circ \sin 100^\circ$
$2 \sin 10^\circ \cos 10^\circ \sin 30^\circ=\sin 10^\circ \sin 100^\circ$
$2 \sin 80^\circ \sin 30^\circ= \sin 100^\circ$
$2 \sin 80^\circ \sin 30^\circ= \sin 30^\circ \cos 70^\circ+\sin 70^\circ \cos 30^\circ$
$2 \sin 80^\circ \sin 30^\circ=\sin 20^\circ \sin 30^\circ+\sin 70^\circ \sin 60^\circ$
$2 \sin 80^\circ \sin 30^\circ=\sin 20^\circ \sin 30^\circ + 2\sin 70^\circ \sin 30^\circ \cos 30^\circ$
$2 \sin 80^\circ \sin 30^\circ=\sin 20^\circ \sin 30^\circ + 2\sin 70^\circ \sin 30^\circ \sin 60^\circ$
$1=\large \frac {\sin 20^\circ}{2 \sin 80^\circ}+ \frac {\sin 70^\circ \sin 60^\circ}{\sin 80^\circ}$
Now I set $\sin^2 \theta=\large \frac {\sin 20^\circ}{2 \sin 80^\circ}$ and $\cos^2 \theta =\large \frac {\sin 70^\circ \sin 60^\circ}{\sin 80^\circ}$
I solves for $\sin \theta$ and $\cos \theta$ and set up a right triangle, which then led me to the equation 
$\sin 70^\circ \sin 60^\circ=\sin 80^\circ-\frac 12 \sin 20^\circ$
Another approach I tries is from the beginning having $\sin 10^\circ \sin 20^\circ \sin 30^\circ=\sin^2 10^\circ  \sin 100^\circ$ and the using the power-reducing formula, but that also got me nowhere. Any help is appreciated. Thanks.

Write $\displaystyle\sin10\sin100=\sin10\sin(90+10)=\sin10\cos10=\frac{1}{2}\sin20=\sin20\sin30$.
