Find $\sin 10^\circ \cdot \sin 50^\circ \cdot \sin 70^\circ$

Question

Please read the screen captured image for the problem. I have actually solved the problem however I was not able to use the hint by the author. Give another hint or solve the problem USING THE HINT.

Here is how I did it. Do note all angles are in degrees. $$\sin10^{\circ}\sin(60^{\circ}-10^{\circ})\sin(60^{\circ}+10^{\circ})=(3\sin10^{\circ}-4(\sin10^{\circ})^3)/4=\sin(3\cdot10^{\circ})/4=1/8.$$ I used triple angle formula which at this point has not been introduced and nowhere in my solution have i used the hint.

Answer 1

Hint: $\;\sin 10^\circ \sin 50^\circ \sin 70^\circ = \sin 10^\circ \cos 40^\circ \cos 20^\circ$, then remember the double angle formulas.

Answer 2

Let a value of our expression be $M$.

Thus,

$$M\cos10^{\circ}=\cos10^{\circ}\sin10^{\circ}\sin50^{\circ}\sin70^{\circ}=$$ $$=\cos10^{\circ}\sin10^{\circ}\cos20^{\circ}\cos40^{\circ}=$$ $$=\frac{1}{2}\sin20^{\circ}\cos20^{\circ}\cos40^{\circ}=$$ $$=\frac{1}{4}\sin40^{\circ}\cos40^{\circ}=$$ $$=\frac{1}{8}\sin80^{\circ}=\frac{1}{8}\cos10^{\circ},$$ which gives $M=\frac{1}{8}.$

Answer 3

I recall needing to figure this out when training for contests (35+ years ago). My solution was to observe that the angles $\alpha_1=10^\circ$, $\alpha_2=50^\circ$ and $\alpha_3=-70^\circ$ all satisfy $\sin3\alpha=1/2$. As the three sines are distinct, they are, by the triplication formula, the solutions of $$ \frac12=(\sin3\alpha=3\sin\alpha-4\sin^3\alpha=)3x-4x^3. $$ But Vieta relations then imply that the product of the zeros of this cubic is equal to $$ -\frac18=\sin10^\circ\sin 50^\circ\sin(-70^\circ). $$