Variation of Weierstrass Approximation Theorem Let $f:[-1,1] \to \mathbb R$ be a continuous even function. Show that for any $\epsilon >0$ there exists an even polynomial $p(x)= \sum _{k=0} ^{n} a_k x^{2k}$ such that $|f(x)-p(x)|<\epsilon$ for any $x \in [-1,1]$. Show a similar result for a continuous odd function. 
I know that since $f(x)$ is continuous, that a polynomial exists such that $|f(x)-p(x)|<\epsilon$ by the Weierstrass Approximation Theorem. However, I do not know how to show that this polynomial is also even. The proof for the odd case should be basically the same. 

The polynomial furnished by Weierstrass Approximation Theorem doesn't need to be even/odd in this case. But you can decompose $p$ in its even and odd parts $p=p_e+p_o$ where $p_e(x)=\frac{p(x)+p(-x)}{2}$ and $p_o(x)=\frac{p(x)-p(-x)}{2}$. Then $p_e$ will still be a good enough approximation to $f$ since $p_o$ will be small.
