Prove that $\{\cos x, \sin x, e^x, e^{-x}\}$ is a linearly independent subset of $C^\infty (\mathbb R)$ I am going to prove $\cos x, \sin x, e^x$ and $e^{-x}$ is a linearly independent subset of $C^\infty (\mathbb R)$, which is smooth functions.
first we have $a\cos x+b\sin x+ce^x+de^{-x}=0$, WTS that $a=b=c=d=0$.
Suppose $c \ne 0$, then for all $x$ in $\mathbb R$, $ce^x$ is sometimes much larger than other 3 terms which is contradiction. So $c=0$.
So we have $a\cos x+b\sin x+de^{-x}=0$, if $d \ne 0$, then by the same logic in $c$, for all $x$ in $\mathbb R$, $x$ is also sometimes much larger than other 2 terms which is contradiction. So $d=0$.
Here it becomes $a\cos x+b\sin x=0$,
when $x=0, a*0+b*1=0$
when $x=\pi/2, a*1+b*0=0$ there is no such situation that $\cos x=\sin x=0$.
So $a=b=0$.
Above is my proof, we have not learned Wronskian or det, so I could only prove it by definition. While since it is going to prove subset of $C^\infty(\mathbb R)$, is there any correction or improvement for the above? Or is there a more clear way to prove this?

You just need to say "much" larger a bit more rigorously.
If $c>0$, let $x=\ln(\frac{a+b+d+\epsilon}{c}),\epsilon>0$ Then
$$
a \cos x + b \sin x + c \exp x + d \exp (-x)\geq -a-b-d+a+b+d+\epsilon>0,
$$
So $c=0$.
Similarly $d=0$.
