I need to prove that $$ \newcommand{\C}{\mathbb{C}} \newcommand{\span}{\text{span}_\C} \span\{\cos(kx), \sin(kx)\}=\span\{e^{ikx}, e^{-ikx}\} $$ This is my proof but I am not 100 % sure it's correct, mainly because I'm not sure I can just define the span of sine and cosine like I do..
My Proof
Using the fact that $$ e^{ikx}=\cos(kx)+i\sin(kx) \quad \text{and} \quad e^{-ikx}=\cos(kx)-i\sin(kx) $$ We can write sine and cosine as follows $$ \cos(kx)=\frac{e^{ikx}+e^{-ikx}}{2} \quad \text{and} \quad \sin(kx)=\frac{e^{ikx}-e^{-ikx}}{2i} $$ And then we can rewrite the span of cosine and sine using the definition (is this even correct?) $$ \span\{\cos(kx), \sin(kx)\}=\{a\cos(kx)+b\sin(kx): k\in\mathbb{Z}, a,b\in\C\} $$ And then substituting the formulas found above and doing some manipulations $$ \span\{\cos(kx), \sin(kx)\}=\{\left(\frac{a}{2}+\frac{b}{2i}\right)e^{ikx} + \left(\frac{a}{2}-\frac{b}{2i}\right)e^{-ikx}: k\in\mathbb{Z}, a,b\in\C\} $$ but now we can rewrite $$ a':=\frac{a}{2}+\frac{b}{2i} \quad \text{and} \quad b':=\frac{a}{2}-\frac{b}{2i} $$ and notice that $a', b'\in\C$. Therefore we obtain that $$ \span\{\cos(kx), \sin(kx)\}=\{a'e^{ikx}+b'e^{-ikx}: k\in\mathbb{Z},a',b'\in\C\}=:\span\{e^{ikx}, e^{-ikx}\} $$
Is this correct? I'm not sure this argument works
I think your proof is basically correct though, perhaps, too cumbersome. I'd go as follows:$\newcommand{\C}{\mathbb{C}}\newcommand{\span}{\text{span}_\C}$
$$ \begin{cases}\cos(kx)=\frac{e^{ikx}+e^{-ikx}}{2} \\{}\\ \sin(kx)=\frac{e^{ikx}-e^{-ikx}}{2i}\end{cases}\;\;\implies \span\{\cos(kx), \sin(kx)\}\subset\span\{e^{ikx}, e^{-ikx}\} $$
and on the other hand:
$$e^{\pm ikx}=\cos kx\pm i\sin kx\implies\span\{\cos(kx), \sin(kx)\}\supset\span\{e^{ikx}, e^{-ikx}\}$$
You can remark that all the involved coefficients are complex ones, too...