দেওয়া আছে, $9\theta = \pi$, অর্থাৎ $\theta = 20^\circ$। এখন, $$ \begin{aligned} &\cos \theta \cos 2\theta \cos 4\theta \\ &= \cos 20^\circ \cos 40^\circ \cos 80^\circ \\ &= \frac{1}{2\sin 20^\circ} (2\sin 20^\circ \cos 20^\circ) \cos 40^\circ \cos 80^\circ \\ &= \frac{1}{2\sin 20^\circ} \sin 40^\circ \cos 40^\circ \cos 80^\circ \\ &= \frac{1}{4\sin 20^\circ} (2\sin 40^\circ \cos 40^\circ) \cos 80^\circ \\ &= \frac{1}{4\sin 20^\circ} \sin 80^\circ \cos 80^\circ \\ &= \frac{1}{8\sin 20^\circ} (2\sin 80^\circ \cos 80^\circ) \\ &= \frac{\sin 160^\circ}{8\sin 20^\circ} \\ &= \frac{\sin(180^\circ - 20^\circ)}{8\sin 20^\circ} \\ &= \frac{\sin 20^\circ}{8\sin 20^\circ} \\ &= \frac{1}{8} \end{aligned} $$