So I recently learned about how to check whether functions are independent. As far as I understood it one of the methods is to plug in freely chosen values for x and you can calculate the determinate for the matrix that results from that. $$ f_1(x), f_2(x), f_3(x)$$ $$\begin{vmatrix} f_1(x_1) & f_2(x_1) & f_3(x_1)\\ f_1(x_2) & f_2(x_2) & f_3(x_2)\\ f_1(x_3) & f_2(x_3) & f_3(x_3)\\ \end{vmatrix}$$
Where $x_1, x_2, x_3$ are the three values I chose for $x$. I don't doubt my Professor but for me its still feels kinda "unmathy" if I might call it that way. So I was going to ask whether this a legit strategy or only works in certain scenarios.
Given that I have functions $f_1=\cos^2x$, $f_2=\sin^2ax$, $f_3=1$. And I choose $x_1=0$, $x_2=\pi/2$ $x_3=\pi$
$$\begin{vmatrix} 1 & 0 & 1\\ 0 & \sin^2\frac{a\pi}{2} & 1\\ 1 & \sin^2a\pi &1\\ \end{vmatrix}$$
and the determinant would be $\sin^2a\pi$, so for each $a \in \mathbb{Z}$ the three functions would be dependent. However I could also chose different value for the $x_1$... and I would get a different determinant
Possibly the professor lied; more likely he/she didn't say what you thought he/she said.
In fact if that determinant is non-zero then the functions are independent. But that only goes one way - if the determinant is zero that does not imply the functions are dependent.
(As commented, if the determinant is zero for every choice of $x_1,x_2,x_3$ then the functions are dependent. That's possibly not that useful as a "test", since it might take a while to check every one of the infinitely many possibilities.)