# Quod Erat Demonstrandum

## 2013/12/21

### 某題行列式

Filed under: NSS — johnmayhk @ 9:56 下午
Tags:

$\left|\begin{array}{rcl}\sin 20^o+\sin 40^o&\sin 80^o&\sin 80^o\\\sin 20^o&\sin 40^o+\sin 80^o&\sin 20^o\\\sin 40^o&\sin 40^o&\sin 80^o+\sin 20^o\\\end{array}\right|$

$\left | \begin{matrix} a+b & c & c\\ a & b+c & a\\ b & b & c+a \end{matrix} \right |$

$f(a)=\left | \begin{matrix} a+b & c & c\\ a & b+c & a\\ b & b & c+a \end{matrix} \right |$

$f(0)=\left | \begin{matrix} b & c & c\\ 0 & b+c & 0\\ b & b & c \end{matrix} \right |$

$\left | \begin{matrix} a+b & c & c\\ a & b+c & a\\ b & b & c+a \end{matrix} \right |=kabc$

$\left | \begin{matrix} 2 & 1 & 1\\ 1 & 2 & 1\\ 1 & 1 & 2 \end{matrix} \right |=k(1)(1)(1)$

$k=4$，亦即

$\left | \begin{matrix} a+b & c & c\\ a & b+c & a\\ b & b & c+a \end{matrix} \right |=4abc$

$\left|\begin{array}{rcl}\sin 20^o+\sin 40^o&\sin 80^o&\sin 80^o\\\sin 20^o&\sin 40^o+\sin 80^o&\sin 20^o\\\sin 40^o&\sin 40^o&\sin 80^o+\sin 20^o\\\end{array}\right|$

$=4\sin 20^o\sin 40^o\sin 80^o$

$=2\sin 20^o(\cos 40^o - \cos 120^o)$

$=2\sin 20^o\cos 40^o+\sin 20^o$

$=(\sin 60^o - \sin 20^o)+\sin 20^o$

$=\frac{\sqrt{3}}{2}$