2022-M2-11

1. $\begin{array}{cl}
   & I-A \\
   = & \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}-\begin{pmatrix} \cos \theta &-\sin\theta \\ \sin\theta &\cos\theta \end{pmatrix}\\
   = & \begin{pmatrix} 1-\cos\theta & \sin\theta \\ -\sin\theta & 1-\cos\theta \end{pmatrix}
   \end{array}$

   由此，可得

   $\begin{array}{cl}
   & (I-A)^{-1} \\
   = & \dfrac{1}{\begin{vmatrix} 1-\cos\theta & \sin\theta \\ -\sin\theta & 1-\cos\theta \end{vmatrix}} \begin{pmatrix} 1-\cos\theta & -(-\sin\theta) \\ -\sin\theta & 1-\cos\theta \end{pmatrix}^T \\
   = & \dfrac{1}{(1-\cos\theta)^2-\sin^2\theta}\begin{pmatrix} 1-\cos\theta & -\sin\theta \\ \sin\theta & 1-\cos\theta \end{pmatrix} \\
   = & \dfrac{1}{1-2\cos\theta+\cos^2\theta+\sin^2\theta}\begin{pmatrix} 1-\cos\theta & -\sin\theta \\ \sin\theta & 1-\cos\theta \end{pmatrix} \\
   = & \dfrac{1}{2-2\cos\theta}\begin{pmatrix} 1-\cos\theta & -\sin\theta \\ \sin\theta & 1-\cos\theta \end{pmatrix}\\
   = & \dfrac{1}{2} \begin{pmatrix} 1 & \dfrac{-\sin\theta}{1-\cos\theta} \\ \dfrac{\sin\theta}{1-\cos\theta} & 1 \end{pmatrix} \\
   = & \dfrac{1}{2} \begin{pmatrix} 1 & \dfrac{-\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\sin^2\frac{\theta}{2}} \\ \dfrac{\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\sin^2\frac{\theta}{2}} & 1 \end{pmatrix} \\
   = & \dfrac{1}{2\sin\frac{\theta}{2}} \begin{pmatrix} \sin\dfrac{\theta}{2} & -\cos\dfrac{\theta}{2} \\ \cos\dfrac{\theta}{2} & \sin\dfrac{\theta}{2} \end{pmatrix}
   \end{array}$

2. $\begin{array}{cl}
   & I+A+A^2+\cdots+A^n \\
   = & (I-A)^{-1}(I-A^{n+1})\\
   = & \dfrac{1}{2\sin\frac{\theta}{2}} \begin{pmatrix} \sin\dfrac{\theta}{2} & -\cos\dfrac{\theta}{2} \\ \cos\dfrac{\theta}{2} & \sin\dfrac{\theta}{2} \end{pmatrix} \left( \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}-\begin{pmatrix} \cos(n+1)\theta & \sin(n+1)\theta \\ -\sin(n+1)\theta & \cos(n+1)\theta \end{pmatrix}\right) \\
   = & \dfrac{1}{2\sin\frac{\theta}{2}} \begin{pmatrix} \sin\dfrac{\theta}{2} & -\cos\dfrac{\theta}{2} \\ \cos\dfrac{\theta}{2} & \sin\dfrac{\theta}{2} \end{pmatrix}\begin{pmatrix} 1-\cos(n+1)\theta & \sin(n+1)\theta \\ -\sin(n+1)\theta & 1-\cos(n+1)\theta \end{pmatrix} \\
   = & \dfrac{1}{2\sin\frac{\theta}{2}} \begin{pmatrix} \sin\dfrac{\theta}{2} & -\cos\dfrac{\theta}{2} \\ \cos\dfrac{\theta}{2} & \sin\dfrac{\theta}{2} \end{pmatrix}\left(2\sin\dfrac{(n+1)\theta}{2}\right)\begin{pmatrix} \sin\dfrac{(n+1)\theta}{2} & \cos\dfrac{(n+1)\theta}{2} \\ -\cos\dfrac{(n+1)\theta}{2} & \sin\dfrac{(n+1)\theta}{2} \end{pmatrix} \text{ , 利用 (a)(ii)(1) 中的方法。} \\
   = & \dfrac{\sin\frac{(n+1)\theta}{2}}{\sin\frac{\theta}{2}}\begin{pmatrix} \sin\dfrac{\theta}{2}\sin\dfrac{(n+1)\theta}{2}+\cos\dfrac{\theta}{2}\cos\dfrac{(n+1)\theta}{2} & \sin \dfrac{\theta}{2}\cos\dfrac{(n+1)\theta}{2}-\cos\dfrac{\theta}{2}\sin\dfrac{(n+1)\theta}{2} \\ \cos\dfrac{\theta}{2}\sin\dfrac{(n+1)\theta}{2}-\sin\dfrac{\theta}{2}\cos\dfrac{(n+1)\theta}{2} & \cos\dfrac{\theta}{2}\cos\dfrac{(n+1)\theta}{2}+\sin\dfrac{\theta}{2}\sin\dfrac{(n+1)\theta}{2} \end{pmatrix} \\
   = & \dfrac{\sin\frac{(n+1)\theta}{2}}{\sin\frac{\theta}{2}}\begin{pmatrix} \cos\dfrac{n\theta}{2} &-\sin\dfrac{n\theta}{2} \\ \sin\dfrac{n\theta}{2} & \cos \dfrac{n\theta}{2} \end{pmatrix}
   \end{array}$

透過比較 (b)(i) 及 (a)(ii)(2) 的元素，可得

$\begin{array}{rcl}
1+\cos\theta+\cos2\theta+\cdots+\cos n\theta & = & \dfrac{\sin\frac{(n+1)\theta}{2}\cos\frac{n\theta}{2}}{\sin\frac{\theta}{2}} \ldots \unicode{x2460}
\end{array}$

把 $n=90$ 及 $\theta=\dfrac{5\pi}{18}$ 代入 $\unicode{x2460}$，可得

$\begin{array}{rcl}
1+\cos\dfrac{5\pi}{18}+\cos2\left(\dfrac{5\pi}{18}\right)+\cdots+\cos 90\left(\dfrac{5\pi}{18}\right) & = & \dfrac{\sin\left(\frac{90+1}{2}\times\frac{5\pi}{18}\right)\cos\left(\frac{90}{2}\times \frac{5\pi}{18}\right)}{\sin\left(\frac{1}{2}\times\frac{5\pi}{18}\right)} \\
1+\cos \dfrac{5\pi}{18}+\cos\dfrac{5\pi}{9}+\cos\dfrac{5\pi}{6}+\cdots+\cos 25\pi & = & \dfrac{\sin\frac{445\pi}{36}\cos\frac{25\pi}{2}}{\sin\frac{5\pi}{36}} \\
\cos \dfrac{5\pi}{18}+\cos\dfrac{5\pi}{9}+\cos\dfrac{5\pi}{6}+\cdots+\cos 25\pi & = & \dfrac{\sin\frac{445\pi}{36}\cos\left(12\pi+\frac{\pi}{2}\right)}{\sin\frac{5\pi}{36}} -1 \\
\cos \dfrac{5\pi}{18}+\cos\dfrac{5\pi}{9}+\cos\dfrac{5\pi}{6}+\cdots+\cos 25\pi & = & \dfrac{\sin\frac{445\pi}{36}\cos\frac{\pi}{2}}{\sin\frac{5\pi}{36}} -1 \\
\cos \dfrac{5\pi}{18}+\cos\dfrac{5\pi}{9}+\cos\dfrac{5\pi}{6}+\cdots+\cos 25\pi & = & \dfrac{\sin\frac{445\pi}{36}\times 0}{\sin\frac{5\pi}{36}} -1 \\
\cos \dfrac{5\pi}{18}+\cos\dfrac{5\pi}{9}+\cos\dfrac{5\pi}{6}+\cdots+\cos 25\pi & = & -1
\end{array}$

把 $n=49$ 及 $\theta=\dfrac{2\pi}{7}$ 代入 $\unicode{x2460}$，可得

$\begin{array}{rcl}
1+\cos\dfrac{2\pi}{7}+\cos\dfrac{4\pi}{7}+\cos\dfrac{6\pi}{7}+\cdots+\cos 14\pi & = & \dfrac{\sin\left(\frac{49+1}{2}\times\frac{2\pi}{7}\right)\cos \left(\frac{49}{2}\times \frac{2\pi}{7}\right)}{\sin\left(\frac{1}{2}\times \frac{2\pi}{7}\right)} \\
1+\cos\dfrac{2\pi}{7}+\cos\dfrac{4\pi}{7}+\cos\dfrac{6\pi}{7}+\cdots+\cos 14\pi & = & \dfrac{\sin\frac{50\pi}{7}\cos 7\pi}{\sin\frac{\pi}{7}} \\
\cos\dfrac{2\pi}{7}+\cos\dfrac{4\pi}{7}+\cos\dfrac{6\pi}{7}+\cdots+\cos 14\pi & = & \dfrac{\sin\frac{50\pi}{7}\cos (6\pi+\pi)}{\sin\frac{\pi}{7}} -1 \\
\cos\dfrac{2\pi}{7}+\cos\dfrac{4\pi}{7}+\cos\dfrac{6\pi}{7}+\cdots+\cos 14\pi & = & \dfrac{\sin\left(7\pi+\frac{\pi}{7}\right)\times -1}{\sin\frac{\pi}{7}} -1 \\
\cos\dfrac{2\pi}{7}+\cos\dfrac{4\pi}{7}+\cos\dfrac{6\pi}{7}+\cdots+\cos 14\pi & = & \dfrac{-(-\sin\frac{\pi}{7})}{\sin\frac{\pi}{7}} -1 \\
\cos\dfrac{2\pi}{7}+\cos\dfrac{4\pi}{7}+\cos\dfrac{6\pi}{7}+\cdots+\cos 14\pi & = & 0 \ \ldots \unicode{x2462}
\end{array}$

由此，把 $\unicode{x2462}$ 代入 $\unicode{x2461}$，可得

$\begin{array}{rcl}
\cos^2\dfrac{\pi}{7}+\cos^2\dfrac{2\pi}{7}+\cos^2\dfrac{3\pi}{7}+\cdots+\cos^2 7\pi & = & \dfrac{1}{2}(49+0) \\
\cos^2\dfrac{\pi}{7}+\cos^2\dfrac{2\pi}{7}+\cos^2\dfrac{3\pi}{7}+\cdots+\cos^2 7\pi & = & \dfrac{49}{2}
\end{array}$