A limit-determinant question 
Let $d_n$ be the determinant of the $n\times n$ matrix whose entries, from left to right and then from top to bottom, are $\cos1,\cos2,\ldots,\cos n^2$. (For example,
$$d_3=
\begin{vmatrix}
\cos1&\cos2&\cos3\\
\cos4&\cos5&\cos6\\
\cos7&\cos8&\cos9\\
\end{vmatrix}$$
The argument of $\cos$ is always in radians, not degrees.) Evaluate $\lim_{n\to\infty}d_n$.

Interesting question, I don't know where to start.
 A: $$\begin{array}{l} \left|\begin{array}{ccc}\cos 1 & \cos 2 & \cos 3 \\ \cos 4 & \cos 5 & \cos 6 \\ \cos 7 & \cos 8 & \cos 9 \end{array}\right| \\ =\left|\begin{array}{ccc}\cos 1 & \cos 2 & \cos 3 \\ \cos 3\cos 1 - \sin 3\sin 1 & \cos 3\cos 2 - \sin 3\sin 2 & \cos 3\cos 3 - \sin 3\sin 3 \\ \cos 7 & \cos 8 & \cos 9  \end{array}\right| \\= \left|\begin{array}{ccc}\cos 1 & \cos 2 & \cos 3 \\  - \sin 3\sin 1 & - \sin 3\sin 2 & - \sin 3\sin 3 \\ \cos 7 & \cos 8 & \cos 9 \end{array}\right|\\= -\sin 3 \left|\begin{array}{ccc}\cos 1 & \cos 2 & \cos 3 \\  \sin 1 & \sin 2 & \sin 3 \\ \cos 7 & \cos 8 & \cos 9 \end{array}\right| \\=-\sin 3 \left|\begin{array}{ccc}\cos 1 & \cos 2 & \cos 3 \\  \sin 1 & \sin 2 & \sin 3 \\ \cos 6\cos 1-\sin 6\sin 1 & \cos 6\cos 2-\sin 6\sin 2 & \cos 6\cos 3-\sin 6\sin 3 \end{array}\right| \\=-\sin 3 \left|\begin{array}{ccc}\cos 1 & \cos 2 & \cos 3 \\  \sin 1 & \sin 2 & \sin 3 \\ -\sin 6\sin 1 & -\sin 6\sin 2 & -\sin 6\sin 3 \end{array}\right| \\=\sin 3 \sin 6 \left|\begin{array}{ccc}\cos 1 & \cos 2 & \cos 3 \\  \sin 1 & \sin 2 & \sin 3 \\ \sin 1 & \sin 2 & \sin 3 \end{array}\right| =0. \end{array}$$
Can you get the value of $d_n$ following the same idea?
A: Hint: find a relation between $\cos(k)$, $\cos(k+1)$ and $\cos(k+2)$.
