,

$$\begin{eqnarray*} 8\left(\cos\frac{\pi}{7}+\cos\frac{3\pi}{7}\right)\cos\frac{2\pi}{3}& = & 8\left(2\cos\frac{2\pi}{7}\cos\left(-\frac{\pi}{7}\right)\right)\left(-\frac{1}{2}\right)\\ & = & -8\cos\frac{\pi}{7}\cos\frac{2\pi}{7} \end{eqnarray*}$$

,

เหตุที่ใช้วิธีนี้เพราะว่าอยู่ในรูปผลคูณของ $\cos$ ที่มุมเพิ่มขึ้นเป็นสองเท่า

$$\begin{eqnarray*}  -8\cos\frac{\pi}{7}\cos\frac{2\pi}{7}& = & -\frac{4\left(2\sin\frac{\pi}{7}\cos\frac{\pi}{7}\right)\cos\frac{2\pi}{7}}{\sin\frac{\pi}{7}}\\ & = & -\frac{4\sin\frac{2\pi}{7}\cos\frac{2\pi}{7}}{\sin\frac{\pi}{7}}\\ & = & -\frac{2\sin\frac{4\pi}{7}}{\sin\frac{\pi}{7}} \end{eqnarray*}$$

,

$$\begin{eqnarray*} -\frac{2\sin\frac{4\pi}{7}}{\sin\frac{\pi}{7}}& = & -\frac{2\cos\left(\frac{\pi}{2}-\frac{4\pi}{7}\right)}{\sin\frac{\pi}{7}}\\ & = & -\frac{2\cos\frac{\pi}{14}}{2\sin\frac{\pi}{14}\cos\frac{\pi}{14}}\\ & = & -\frac{1}{\sin\frac{\pi}{14}}\\ & = & -\operatorname{cosec}\frac{\pi}{14} \end{eqnarray*}$$

,

$$\begin{eqnarray*} \tan\frac{9\pi}{16}-\tan\frac{5\pi}{8}& = & \frac{\sin\frac{9\pi}{16}}{\cos\frac{9\pi}{16}}-\frac{\sin\frac{5\pi}{8}}{\cos\frac{5\pi}{8}}\\ & = & \frac{\sin\frac{9\pi}{16}\cos\frac{5\pi}{8}-\cos\frac{9\pi}{16}\sin\frac{5\pi}{8}}{\cos\frac{9\pi}{16}\cos\frac{5\pi}{8}}\\ & = & \frac{\sin\left(\frac{9\pi}{16}-\frac{5\pi}{8}\right)}{\sin\left(\frac{\pi}{2}-\frac{9\pi}{16}\right)\sin\left(\frac{\pi}{2}-\frac{5\pi}{8}\right)}\\ & = & \frac{\sin\left(-\frac{\pi}{16}\right)}{\sin\left(-\frac{\pi}{16}\right)\sin\left(-\frac{\pi}{8}\right)}\\ & = & -\frac{1}{\sin\frac{\pi}{8}} \end{eqnarray*}$$

ซึ่งมีค่าเท่ากับ $-\operatorname{cosec}\frac{\pi}{8}$ ไม่ใช่ $\operatorname{cosec}\frac{\pi}{8}$