Value of $$\cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ$$​ is:

Correct option is D

Given:
​$$\cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ$$​​
Formula used:
$$\cos 3\theta = 4\cos^{3}\theta - 3\cos\theta$$​​
and
​$$\cos 60^\circ = \frac{1}{2}$$​​
Also,
​$$\cos 3\theta = 4\cos\theta \cos(60^\circ - \theta)\cos(60^\circ + \theta)$$​​
Solution:
Using
​$$\cos 3\theta = 4\cos\theta \cos(60^\circ - \theta)\cos(60^\circ + \theta)$$​​
Put $$\theta = 20^\circ$$​:
​$$\cos 60^\circ = 4\cos 20^\circ \cos 40^\circ \cos 80^\circ\\\frac{1}{2} = 4\cos 20^\circ \cos 40^\circ \cos 80^\circ\\\cos 20^\circ \cos 40^\circ \cos 80^\circ = \frac{1}{8}$$​​
Now multiply by $$\cos 60^\circ:$$​​
​$$\cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ= \frac{1}{8} \times \frac{1}{2}= \frac{1}{16}$$​​
The correct answer is (d) $$\frac{1}{16}.$$​​