Show that $1, x, x^2 ,\sin(x), \cos(x)$ are linearly independent I need to show that ${1, x, x^2 ,\sin(x), \cos(x)}$ is a basis of $E =<1, x, x^2 ,\sin(x), \cos(x)>$. As they already span $E$, I just need to show that they are linearly independent. But I can't figure out if there is a numerical reason, or it is just because the functions themselves.
 A: Hint. Let $c_1,c_2,c_3,c_4, c_5$ be real numbers such that FOR ALL $x\in \mathbb{R}$,
$$c_1\cdot 1+c_2\cdot x+c_3\cdot x^2+c_4\cdot \sin(x)+c_5\cdot\cos(x)=0.$$
Then, after dividing by $x^2$ and by taking the limit as $x\to +\infty$, we get $c_3=0$. Similarly, after dividing by $x$, we get that also $c_2=0$.
So $$c_1\cdot 1+c_4\cdot \sin(x)+c_5\cdot\cos(x)=0.$$
Now by letting  $x=0$, $x=\pi/2$, and $x=\pi$,  we will obtain more conditions about $c_1$, $c_4$, and $c_5$. 
Are you able to conclude that $c_1=c_4=c_5=0$?
A: Write
$$a+bx+cx^2+d\sin x+e\cos x=0$$
and choose $5$ values of $x$ (or more) such that you obtain a system of rank $5$.
You can ease the task by processing the even and odd parts separately, as they are certainly linearly independent:
$$a+cx^2+e\cos x=0,\\by+d\sin y=0.$$
It is convenient to cancel the trigonometric terms by choosing some appropriate multiples of $\pi$.

Even better, you can split in periodic and non periodic contributions and work with
$$a+cx^2=0,\\e\cos y=0,\\bz=0,\\d\sin t=0.$$
This makes the computation trivial.
A: $1,x,x^2$ are independent as solutions to $f'''(x)=0$.
$1,\sin(x),\cos(x)$ are independent as solutions to $f'''(x)+f'(x)=0$.
$\{\sin(x),\cos(x)\}$ and $\{x,x^2\}$ are independent since $a\cos(x)+b\sin(x)$ is always a bounded function (by the Cauchy-Schwarz inequality the absolute value of $a\cos x+b\sin x$ is $\leq\sqrt{a^2+b^2}$) while $cx+dx^2$ is never a bounded function (unless $(c,d)=(0,0)$).
