Evaluating $\frac{\cos 10^\circ \cos 50^\circ\cos 70^\circ}{\cos 30^\circ}$

Question

$$\frac{\cos 10^\circ \cos 50^\circ\cos 70^\circ}{\cos 30^\circ} = ?$$ I use double sine formula $$\sin2A=2\sin A \cos A$$ But isn't help to reduce any fraction but more degree of $\sin 40^\circ$

How should I do after that?

Answer 1

$$\frac{\cos10^{\circ}\cos50^{\circ}\cos70^{\circ}}{\cos30^{\circ}}=\frac{(\cos60^{\circ}+\cos40^{\circ})\cos70^{\circ}}{2\cos30^{\circ}}=$$ $$=\frac{\cos10^{\circ}+\cos130^{\circ}+\cos110^{\circ}+\cos30^{\circ}}{4\cos30^{\circ}}=$$ $$=\frac{\cos10^{\circ}+2\cos120^{\circ}\cos10^{\circ}}{4\cos30^{\circ}}+\frac{1}{4}=\frac{1}{4}.$$

Answer 2

Hints:

The result is 1/4.

$$ \frac{\cos(10°)\cos(50°)\cos(70°)}{\cos(30°)} = \frac{\sin(80°)\sin(40°)\sin(20°)}{\sin(60°)} $$

You might have to use trigonometric arc tri-section.

Good luck.