Show that $\sin(kx)$ and $\cos(kx)$ are polynomial uniform limits Since linear combinations of $\sin(kx)$ and $\cos(kx)$ for $n\in \mathbb{N}$ form a dense subset in $(C([-\pi , \pi]), \|\cdot \|_{\infty})$. We want to show that $\sin(kx)$ and $\cos(kx)$ are polynomial uniform limits in term of $x$ on the interval $[-\pi , \pi]$, and conclude that the set of polynomial functions is dense in $(C([-\pi , \pi]), \|\cdot \|_{\infty})$.
I am not sure how to prove this. I have proven that for $A, B\subset (X, d)$, if $B\subset \overline{A}$ and $B$ is $d$-dense in $X$ then $A$ is $d$-dense in $X$. I think this should help me but I don't know how to apply this my problem. If anyone has an idea, I would appreciate it.
 A: Hints:
Use the Taylor expansion of $\sin(kx)$ and $\cos(kx)$ to find polynomials that approximate
these functions.
Then use some of the formulas for the remainder.
Then you can estimate the reminder for a bound that does not depend on $x$.
This bound should then go to $0$ as you increase the degree of the polynomial.
As for the overall question: For a given continuous function,
first approximate it using a linear combination of $\sin(kx),\cos(kx)$,
then use polynomials to approximate this linear combination.
Using a triangle inequality should yield the result.
A: It is well-known that $\sin y$ and $\cos y$ can be represented by their Taylor series
$$\sum_{n=0}^\infty (-1)^n\frac{y^{2n+1}}{(2n+1)!} \text{ and } \sum_{n=0}^\infty (-1)^n\frac{y^{2n}}{(2n)!} .$$
Both series have radius of convergence $= \infty$, thus convergence is uniform on each compact subset of $\mathbb R$, especially on $[-k\pi,k\pi]$. This means that for each $\epsilon > 0$ we find $M$ such that for all $m \ge N$ and all $y \in [-k\pi,k\pi]$
$$\left\lvert \sin(y) - \sum_{n=0}^m (-1)^n\frac{y^{2n+1}}{(2n+1)!} \right\rvert < \epsilon \text{ and } \left\lvert \cos(y) - \sum_{n=0}^m (-1)^n\frac{y^{2n}}{(2n)!} \right\rvert < \epsilon .$$
Substituting $y = kx$ shows that for each $\epsilon > 0$ we find $M$ such that for all $m \ge N$ and all $x \in [-\pi,\pi]$
$$\left\lvert \sin(kx) - \sum_{n=0}^m (-1)^nk^{2n+1}\frac{x^{2n+1}}{(2n+1)!} \right\rvert < \epsilon \text{ and } \left\lvert \cos(kx) - \sum_{n=0}^m (-1)^nk^{2n}\frac{x^{2n}}{(2n)!} \right\rvert < \epsilon .$$
This shows that the functions $\sin(kx)$ and $\cos(kx)$ are uniform limits of polynomial functions.
In other words, if $P([-\pi,\pi])$ denotes the set of all polynomial functions, then for each $k \in \mathbb N$
$$\sin(kx), \cos(kx) \in \overline{P([-\pi,\pi])} , \tag{1}$$
where $\overline{\phantom{X}}$ denotes closure in $(C([-\pi , \pi]), \|\cdot \|_{\infty})$.
If $T([-\pi,\pi])$ denotes the set of all linear combinations of elements in $\{ \sin(kx) \mid k \in \mathbb N \} \cup \{ \cos(kx) \mid k \in \mathbb N \} $, then of course
$$T([-\pi,\pi]) \subset \overline{P([-\pi,\pi])} . \tag{2}$$
Since $\overline{P([-\pi,\pi])} $ is closed, we have
$$\overline{T([-\pi,\pi])} \subset \overline{P([-\pi,\pi])} . \tag{3}$$
But we know that $\overline{T([-\pi,\pi])} = C([-\pi,\pi])$, thus
$$\overline{P([-\pi,\pi])} =  C([-\pi,\pi]) .$$.
