প্রমাণ কর যে, 

  $16 \cos{\left ( \frac{2 \pi}{15} \right )} \cos{\left ( \frac{4 \pi}{15} \right )} \cos{\left ( \frac{8 \pi}{15} \right )} \cos{\left ( \frac{14 \pi}{15} \right )} = 1$

$16 \cos \frac{2 \pi}{15} \cdot \cos \frac{4 \pi}{15} \cos \frac{8 \pi}{15} \cos \frac{14 \pi}{15}$. Let. $\frac{2 \pi}{15}=\theta$

$\begin{array}{l} =16 \cos \theta \cos 2 \theta \cos 4 \theta \cos 7 \theta \\ =\frac{8}{\sin \theta} 2 \sin \theta \cos \theta \cos 2 \theta \cos 4 \theta \cos 7 \theta \\ =\frac{4}{\sin \theta} 2 \sin 2 \theta \cos 2 \theta \cos 4 \theta \cos 7 \theta \\ =\frac{2}{\sin \theta} 2 \sin 4 \theta \cos 4 \theta \cdot \cos 7 \theta \\ =\frac{1}{\sin \theta} 2 \sin 8 \theta \cdot \cos 7 \theta \\ =\frac{1}{\sin \theta}(\sin 15 \theta+\sin \theta) \\ =\frac{1}{\sin \theta}(\sin 2 \pi+\sin \theta)=1 \end{array}$