# Question #063c7

Jan 9, 2017

$n = 9$

#### Explanation:

We know that

$\frac{\pi}{2} - \frac{4 \pi}{9} = \frac{\pi}{18}$ so

$\left[\begin{matrix}\sin \left(\frac{\pi}{18}\right) - \sin \left(\frac{4 \pi}{9}\right) \\ \sin \left(\frac{4 \pi}{g}\right) \text{ } \sin \left(\frac{\pi}{18}\right)\end{matrix}\right] = \left[\begin{matrix}\sin \left(\frac{\pi}{2} - \frac{4 \pi}{9}\right) & - \sin \left(\frac{4 \pi}{9}\right) \\ \sin \left(\frac{4 \pi}{9}\right) & \sin \left(\frac{\pi}{2} - \frac{4 \pi}{9}\right)\end{matrix}\right] = \left[\begin{matrix}\cos \left(\frac{4 \pi}{9}\right) & - \sin \left(\frac{4 \pi}{9}\right) \\ \sin \left(\frac{4 \pi}{9}\right) & \cos \left(\frac{4 \pi}{9}\right)\end{matrix}\right] = R \left(\frac{4 \pi}{9}\right)$

Here $R \left(\cdot\right)$ represents a rotation. The question now is:

How apart we are from $2 k \pi$? Because $R \left(2 k \pi\right) = {I}_{2} , k = 0 , 1 , 2 , \cdots$ so from

$2 k \pi = n \frac{4 \pi}{9}$ we obtain

$k = 2 \left(\frac{n}{9}\right)$ so $n = 9$