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 $~\{0\}~$ 
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.
 A: 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.
A: Let$$a\sin^2(x)+b\cos^2(x) = 0 $$
$$x=0\implies b=0$$
and $$x=\pi/2\implies a=0$$
Thus the two functions are linearly independent.
A: Let $C[-\pi,\pi]$ be the vector space of all continuous functions. Let us consider a subset of the vector space $C[-\pi,\pi]$ as $V = \{\sin^2(x),\cos^2(x)\}$. We want to analyze whether set $V$ is linearly independent?
Let us consider the linear combination set to zero, like $a \sin^2(x) + b \cos^2(x) =0$ and if that implies $a=0,b=0$, then we have the set $V$ as the linearly independent set.
We must observe that this is true for all $x \in [-\pi,\pi]$.
So we put the values of $x$ (we utilize this freedom) such that we get the value of $a,b$ easily. you can put $x = \frac{\pi}{4}$ and obtain an equation in $a$ and $b$ and then you can also put say $x = \frac{\pi}{3}$ and then obtain another equation in $a$ and $b$. Now having two equations in two unknowns $a$ and $b$, you can solve for $a$ and $b$.
But an easy way to go is $x=0$ where $\sin(x) =0, \cos(x) = 1$ giving us $b=0$ and next considering $x=\frac{\pi}{2}$, we have $\sin(x) = 1, \cos(x)  =0$, giving us $a=0$. thus since we have $a=0,b=0$. This implies that the set $V$ is linearly independent.
