TITLE: Proving functions are linearly independent
QUESTION [9 upvotes]: I'm currently going through Harvard's Abstract Algebra using Michael Artin's book, and have no real way of verifying my proofs, and was hoping to make sure that my proof was right.
The question reads:

Let $V$ be the vector space of functions on the interval $[0, 1]$. Prove that the functions $x^{3}$, $\sin(x)$, and $\cos(x)$ are linearly independent.

My proof goes as follows:
For these to be linearly dependent there must exist an $a_{i} \neq0$, where $ i = 1, 2, 3$ such that
$$a_{1}x^{3} + a_{2}\sin(x) + a_{3}\cos(x) = 0. $$
So, we'll do this in 3 cases:
Case 1: $x = 0$
In this case, $x^{3} = 0$, $\sin(x) = 0$ but $\cos(x) = 1$.
So, we have 
$$0\times a_{1} + 0\times a_{2} + 1\times a_{3} = 0.$$
So, $a_{1}$ and $a_{2}$ could be anything but $a_{3}$ must be 0.
Case 2: $x \in (0,1)$
In this case, $x^{3} \neq 0$, $\sin(x) \neq 0$ and $\cos(x) \neq 0$.
So, for this to be true, $a_{1}$, $a_{2}$ and $a_{3}$ all must be $0$.
Case 3: $x = 1$
In this case, $x^{3} = 1$, $\sin(x) = .8...$ and $\cos(x) = .5...$.
So, we have 
$$1\times a_{1} +.8\times a_{2} + .5\times a_{3} = 0.$$
So, $a_{3}$ could be any value, while $a_{1}$ and $a_{2}$ must be $0$.
So, if $a_{1} \neq 0$ then we have a problem in Case 3. If $a_{2} \neq 0$ we have a problem in Case 3. If $a_{3} \neq 0$ we have a problem in Case 1. So, we know that all of the $a$ values must be $0$ and we complete the proof.

REPLY [6 votes]: Differentiation can be powerful in cases like this.
BobaFret's suggestion of using the Wronskian is one possibility, but you can do it more directly.
Your errors have been discussed well in Jonas Meyer's answer, and he also discussed differentiation.
I want to add why differentiation is a good idea.
As you noticed, the condition is simplest to work with at $x=0$.
You can of course take three points in $[0,1]$ and evaluate the condition at each one, but it is more convenient to work at the origin if at all possible.
Since looking at only $x=0$ gives you just one condition, you should next look near the origin.
This process of "looking at a condition near a point" is often means differentiating the condition there.
It is often easier to differentiate a given condition at an easy point than to study the condition elsewhere.
Even if there are several easy points you could pick, it can be beneficial to choose only one but look at different derivatives at the same point.
This is not just a little calculus trick; this idea is not at all unusual in my research.

Here's a full solution with this approach:
You have a function
$$
f(x)=a_{1}x^{3} + a_{2}\sin(x) + a_{3}\cos(x)
$$
and you know that $f(x)=0$ for all $x\in[0,1]$.
Evaluating at $x=0$ gives you $f(0)=a_3=0$, so in fact
$$
f(x)=a_{1}x^{3} + a_{2}\sin(x).
$$
Since $f(x)=0$ for all $x\in[0,1]$ and all functions are differentiable (on the whole real line), you also have
$$
f'(x)=3a_1x^2+a_2\cos(x)=0
$$
for all $x\in[0,1]$.
Evaluating this at zero leaves you $0=f'(0)=a_2$.
Therefore
$$
f(x)=a_{1}x^{3}.
$$
Now you can study $f(1)$ to find that $a_1=0$.
If you want to keep on differentiating, you can also inspect $f'''(0)$ and draw the same conclusion.