Are the functions $\sin^2(x)$ and $\cos^2(x)$ linearly independent? $a\sin^2(x)+b\cos^2(x)=0$, such that the domain is the set $\{0\}$. (Only has the element zero).
Because if you plug the value $0$ in the equation,
it is equal to $a(0)+b1=0$ so $b$ must be zero, but $a$ can be any value and the equation remains true, Right?
So by the definition would be linearly dependent instead.
I'm confused. Because I'm aware that those functions are linearly independent for all reals.
Best Answer
Using $\sin^2(x) + \cos^2(x) = 1 \implies \sin^2(x) = 1 - \cos^2(x)$, you get
$$\begin{equation}\begin{aligned} a\sin^2(x) + b\cos^2(x) & = 0 \\ a(1 - \cos^2(x)) + b\cos^2(x) & = 0 \\ a - a\cos^2(x) + b\cos^2(x) & = 0 \\ a + (b - a)\cos^2(x) = 0 \end{aligned}\end{equation}\tag{1}\label{eq1}$$
Since $a$ and $b$ are constants, but $\cos^2(x)$ varies with $x$ with $0 \le \cos^2(x) \le 1$, the equation in \eqref{eq1} can only always be true only if $b - a = 0$, so then $a = 0$ also, resulting in $b = 0$. Thus, this shows $\sin^2(x)$ and $\cos^2(x)$ are linearly independent.
Update: I misinterpreted the question's original wording. As discussed in the comments below, if the domain is just a single point, you're using $2$ functions and the concept of (in)dependence can be considered to apply, then regardless of what the domain value is and what $2$ functions you're checking, you'll always determine they are linearly dependent. This is because you have one equation in $2$ unknowns of $a$ and $b$. Thus, this system is under-determined, meaning there will always be at least one free parameter among $a$ and $b$ which can take on any value. As stated in the question, for this particular case using $x = 0$ only, $b = 0$ and $a$ can take on any value.