Find the value of $\cos 20^\circ \cos 40^\circ \cos 80^\circ$.

We use the product identity $\cos \theta \cos(60^\circ - \theta) \cos(60^\circ + \theta) = \frac{1}{4}\cos 3\theta$.
Let $\theta = 20^\circ$. Then:

• $60^\circ - \theta = 40^\circ$
• $60^\circ + \theta = 80^\circ$

The given expression matches the formula exactly:

$$\cos 20^\circ \cos 40^\circ \cos 80^\circ = \frac{1}{4}\cos(3 \times 20^\circ) = \frac{1}{4}\cos 60^\circ$$

Since $\cos 60^\circ = \frac{1}{2}$:

$$Value = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8}$$

Answer: $\frac{1}{8}$