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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.

Rodrigo de Azevedo
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arnav007
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    Hint: Try to prove your conjecture by induction, using e.g. expansion by the first row: such an $n\times n$ determinant is $\frac{\sin((n+1)\theta)}{\sin\theta}$. – Berci Dec 27 '19 at 21:18
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    https://math.stackexchange.com/questions/266998/how-to-compute-the-determinant-of-a-tridiagonal-matrix-with-constant-diagonals – StubbornAtom Dec 28 '19 at 07:16

2 Answers2

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By an expansion on the first column one gets

\begin{align}\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} &= 2 \cos(\theta) \begin{vmatrix} 2 \cos(\theta) & 1 & 0 \\ 1 & 2 \cos(\theta) & 1 \\ 0 & 1 & 2 \cos(\theta) \end{vmatrix}\\ &- \begin{vmatrix} 1 & 0 & 0 \\ 1 & 2 \cos(\theta) & 1 \\ 0 & 1 & 2 \cos(\theta) \end{vmatrix} \\ &=2\cos(\theta) \bigg ( 2 \cos(\theta) \begin{vmatrix} 2\cos(\theta) & 1 \\ 1 & 2 \cos(\theta) \end{vmatrix} - \begin{vmatrix} 1 & 0 \\ 1 & 2 \cos(\theta) \end{vmatrix} \bigg)\\ &= 4 \cos^2(\theta) (4\cos^2(\theta) - 1) - 4 \cos^2(\theta) - (4 \cos^2\theta) - 1) \\ &= 16 \cos^4(\theta) - 12 \cos^2(\theta) + 1.\end{align}

Now we use the trigonometric Pythagoras $\sin^2(\theta) + \cos^2(\theta) = 1$. With this formula, the above expression can be rewritten as

\begin{align} 16 (1 - \sin^2(\theta))^2 - 12 (1 - \sin^2(\theta)) + 1 &= 16 - 32 \sin^2(\theta) + \sin^4(\theta)-12 + 12 \sin^2(\theta) + 1 \\ &= \sin^4(\theta) - 20 \sin^2(\theta) + 5.\end{align}

The last formula is valid for every value of $\theta$. The result which you have to show is true if and only if $\theta \neq k\pi$, where $k \in \mathbb{Z}$. For those $\theta$, the last term is equivalent to

$$\frac{\sin^5(\theta) - 20 \sin^3(\theta) + 5\sin(\theta)}{\sin(\theta)}.$$

Now it is a basic trigonometric addition formula that $\sin^5(\theta) - 20 \sin^3(\theta) + 5\sin(\theta) = \sin(5\theta)$. This shows that the last term is equal to

$$\frac{\sin(5\theta)}{\sin(\theta)},$$

which was to be proven.

Jan
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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)} $.

marty cohen
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