<s>[INST] Proving functions are linearly independent 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.
 [/INST] Your case 2 and case 3 are wrong.
For case 2,
just because
$a_{1}x^{3} + a_{2}sin(x) + a_{3}cos(x) = 0$
does not imply that
the $a_i$ are all zero.
Actually,
for any $x \ne 0$,
and any $a_2, a_3$,
$a_1$ can always be chosen
so that
$a_{1}x^{3} + a_{2}sin(x) + a_{3}cos(x) = 0$
by setting
$a_{1}=-\dfrac{a_{2}sin(x) + a_{3}cos(x)}{x^{3}}$.
For your case 3,
$\sin(1) \ne 1$
and
$\cos(1) \ne 0$.
If you use
$\pi/2$,
then this works.
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