Find a basis of $A = (\{1, \sin(x), (\cos)^2(x), (\sin)^2(x)\})$ and determining its dimension. We consider a space F(R, R) of functions of R in R. 
Let A = ({1, $\sin(x)$, $\cos^2(x)$, $\sin^2(x)$})
Find a basis of the vector subspace of F(R,R) and determine its dimension.
So I used the identity $1 - \cos^2(x) = \sin^2(x)$, so that means that $(\cos)^2(x)$ is already a linear combination of 2 of the vectors in that space, right? So the dimension would automatically be 3. so we make the basis B = $({1, \sin(x)$, $(\sin)^2(x)$}).
But in the mark scheme, it says that we need to show that this basis B is free/linearly independent. They say let 3 scalars,: a, b, c ∈ R such that 
$a · 1 + b · \sin(x) + c · (\sin)^2(x) = 0$, so we need to show that $a = b = c = 0. $
They pose $x = 0, x = π/6$ and $x = π/2$ and they get a system of linear equations where:
$a = 0$
$a + b/2 + c/4 = 0$
$a + b + c = 0$
This is where I get confused. Why do they let $x = 0, π/2$ and $π/6$? And how did they get this system of linear equations? why over $2$ and over $4$? And lastly, when a question giving us a set of values, and they ask us to find a basis, what are the steps needed to take in order to find it? Do you first cancel all columns which are a linear combination of the others and that is a basis? 
Sorry if that's a long question. 
Thanks!
 A: Hint: Note that you can write $A$ as

$$ A = \left\{ 1, \sin(x), \frac{1}{2}-\frac{\cos(2x)}{2}, \frac{\cos(2x)}{2}+\frac{1}{2} \right\}. $$

A: When we say that $ a + b\sin x + c \sin^2 x  = 0$ we mean so for every $x \in \Bbb R$. Note that this is not an equation of $x$ but an equation of functions. So the resulting function $f(x) = a + b\sin x + c \sin^2 x  $ should be so that $f$ must be the zero function from $\Bbb R$ to $\Bbb R$ that is $f$ should be the function that is uniformly $0$ for every $x \in \Bbb R$. So setting $x = 0$ is totally justified since the equation should hold for that value. 
Then since $\sin 0 = 0 $ we have that $a = 0$. It seems the other two equations were given by taking $x = \dfrac {\pi}{ 6}$ and $x = \dfrac {\pi}{2}$. Again we can do this due to the reasons explained above. 
A: By definition, your base is linearly independent if and only if $\forall x \in \mathbb{R}$ $a\times 1 + b\times \sin x + c \times \sin ^2 x = 0 \Rightarrow a=b=c=0$.
That gives you an infinite number of equations, but you only have 3 unknowns. So the easiest thing to do is to take 3 clever values which will give you 3 equations and solve for those, in order to prove the above. As you can see, the values chosen are the ones of which you know the sine explicitly.
