(15) \( \cos 20^{\circ} \cdot \cos 40^{\circ} \cdot \cos 60^{\circ} \cdot \cos 80^{\circ}=\frac{1}{16} \)

Yes, the identity \[ \cos 20^{\circ} \cdot \cos 40^{\circ} \cdot \cos 60^{\circ} \cdot \cos 80^{\circ} = \frac{1}{16} \] is **correct**. Let's verify and prove this step by step. ### Step 1: Simplify Using Known Values We know that: \[ \cos 60^{\circ} = \frac{1}{2} \] So, the original expression becomes: \[ \cos 20^{\circ} \cdot \cos 40^{\circ} \cdot \cos 60^{\circ} \cdot \cos 80^{\circ} = \left( \cos 20^{\circ} \cdot \cos 40^{\circ} \cdot \cos 80^{\circ} \right) \cdot \frac{1}{2} \] ### Step 2: Use a Trigonometric Identity There's a useful trigonometric identity for the product of three cosines in an arithmetic progression: \[ \cos A \cdot \cos (60^{\circ} - A) \cdot \cos (60^{\circ} + A) = \frac{\cos 3A}{4} \] Let's set \( A = 20^{\circ} \): \[ \cos 20^{\circ} \cdot \cos 40^{\circ} \cdot \cos 80^{\circ} = \frac{\cos 60^{\circ}}{4} = \frac{\frac{1}{2}}{4} = \frac{1}{8} \] ### Step 3: Combine the Results Substituting back: \[ \left( \cos 20^{\circ} \cdot \cos 40^{\circ} \cdot \cos 80^{\circ} \right) \cdot \frac{1}{2} = \frac{1}{8} \cdot \frac{1}{2} = \frac{1}{16} \] ### Conclusion Therefore: \[ \cos 20^{\circ} \cdot \cos 40^{\circ} \cdot \cos 60^{\circ} \cdot \cos 80^{\circ} = \frac{1}{16} \] is indeed a valid identity.