This following is a problem from B. S. Grewal's Higher Engineering Mathematics.
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$$\begin{vmatrix} 2\cos(\theta) & 1 & 0 & 0 \\ 1 & 2 \cos(\theta) & 1
& 0 \\ 0 & 1 & 2 \cos(\theta) & 1 \\ 0 & 0 & 1 & 2 \cos(\theta)
\end{vmatrix} = \frac{\sin(5\theta)}{\sin(\theta)}.$$
If I take the $3 \times 3$ matrix after deleting the first row and first column, the value is $\frac{\sin(4\theta)}{\sin(\theta)}$, but I am unable to solve this $4\times 4$ matrix. I tried solving the RHS but I am still unable to solve.
Best Answer
https://en.wikipedia.org/wiki/Tridiagonal_matrix
$f_n = \left| \begin{array}{llll} a_1 & b_1 &0 &0 &0 \\ c_1 & a_2 & b_2 &0 & 0\\ 0 & c_2 & \ddots & \ddots & 0\\ 0 &0 & \ddots & \ddots & b_{n-1}\\ 0 &0 &0 & c_{n-1} & a_n \end{array} \right| $
with $f_0 = 1, f_{-1} = 0$.
$f_n = a_n f_{n-1}-c_{n-1}b_{n-1}f_{n-2} $.
If all $a_i = a, b_i = b, c_i = c$ then $f_n = a f_{n-1}-cbf_{n-2} $.
In this case, $b = c = 1, a = 2\cos(t) $ so $f_n = 2\cos(t)f_{n-1}-f_{n-2} $ so
$f_n = \left| \begin{array}{ccccc} 2\cos(t) & 1 &0& 0&0\\ 1 & 2\cos(t) & 1 & 0& 0\\ 0 & 1 & \ddots & \ddots &0 \\ 0 &0 & \ddots & \ddots & 1\\ 0&0 &0 & 1 & 2\cos(t) \end{array} \right| $
Therefore $f_n\sin(t) = 2f_{n-1}\sin(t)\cos(t)-\sin(t)f_{n-2} $.
Let $g_n =f_n\sin(t) $, so $g_n = 2\cos(t)g_{n-1}-g_{n-2} $.
Since $f_0 = 1$ and $f_1 = 2\cos(t) $,
$g_0 = \sin(t)$ and $g_1 =2\sin(t)\cos(t) =\sin(2t) $.
We have
$\begin{array}\\ \sin((n+1)t)+\sin((n-1)t) &=\sin(nt)\cos(t)+\cos(nt)\sin(t)+\sin(nt)\cos(t)-\cos(nt)\sin(t)\\ &=2\sin(nt)\cos(t)\\ \end{array} $
or $\sin((n+1)t)=2\sin(nt)\cos(t)-\sin((n-1)t) $.
Therefore $g_n(t) =\sin((n+1)t)$, so $f_n(t) =\dfrac{\sin((n+1)t)}{\sin(t)} $.