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Find the value of the following determinant of $n \times n$ matrix:

$$\det A=\begin{vmatrix} 1&\cos \theta_{1}&\cos 2\theta_{1}&\dots&\cos (n-1)\theta_{1}\\ 1&\cos \theta_{2}&\cos 2\theta_{2}&\dots&\cos (n-1)\theta_{2}\\ \vdots&\vdots&\vdots&&\vdots\\ 1&\cos \theta_{n}&\cos 2\theta_{n}&\dots&\cos (n-1)\theta_{n}\\ \end{vmatrix}$$

Here's my idea to evaluate it:
Subtracting the $i^{th}$ column by $(i-1)^{th}$ column and get $-2\sin\frac{\theta_{k}}{2}\sin\frac{2i-1}{2}\theta_{k}$ by sum-to-product identities, and there is common factor of $-2$ on each column (expect the first and second columns) can be pulled out.
And the rest I have stucked in it. Any brilliant idea to continue my work? Or getting a new way to go through it?

weilam06
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1 Answers1

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This is $$\det A=\begin{vmatrix} 1&c_1&T_2(c_1)&\dots&T_{n-1}(c_1)\\ 1&c_2&T_2(c_2)&\dots&T_{n-1}(c_2)\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ 1&c_n&T_2(c_n)&\dots&T_{n-1}(c_n)\\ \end{vmatrix}$$ where $T_r$ is the $r$-th Chebyshev polynomial of the first kind, and $c_j=\cos\theta_j$. By column operations we get $$\det A=\begin{vmatrix} 1&c_1&a_2 c_1^2&\dots&a_{n-1}c_1^{n-1}\\ 1&c_2&a_2 c_2^2&\dots&a_{n-1}c_2^{n-1}\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ 1&c_n&a_2 c_n^2&\dots&a_{n-1}c_n^{n-1}\\ \end{vmatrix} =a_2a_3\cdots a_{n-1}\begin{vmatrix} 1&c_1&c_1^2&\dots&c_1^{n-1}\\ 1&c_2&c_2^2&\dots&c_2^{n-1}\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ 1&c_n& c_n^2&\dots&c_n^{n-1}\\ \end{vmatrix}$$ where $a_r$ is the leading coefficient of $T_r$ (so $a_r=2^{r-1}$). That's a Vandermonde determinant in the last line.

Angina Seng
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  • What if I avoid using Chebyshev's method to solve it? Since I have no fundamental for this kind of polynomials. – weilam06 Oct 07 '18 at 16:18
  • @weilam06 You might use the De Moivre formula to deduce such thing: consider compute $(\cos (t)+\mathrm i \sin(t))^k= \cos(kt) + \mathrm i \sin (kt)$ by binomial theorem, then $\cos(kt) = \mathrm {Re}(\cos(kt)+\mathrm i\sin(kt))$ is actually a polynomial of degree $k$ of $\cos(t)$ whose leading coefficient is $2^k$ [here you might need to use some combinatorical identities]. Then proceed as the answer shown. – xbh Oct 07 '18 at 16:31
  • Can I know what kind of 'column operation' had been done in this determinant? – weilam06 Oct 07 '18 at 17:27