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Consider the metric space $C_{per}[-\pi,\pi]$ (with the supremum norm and metric).

Can $sin(x)$ be approximated by a trigonometric polynomial that does not contain an appearance of $sin(x)$ itself?

Meaning, is there an approximation $p$ such that $$p\in Span \{sin(nx) | n\ne1\}\cup\{cos(nx)\}$$ for all $n\in\Bbb{N}$?

And if not, how would you show that?

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    What permissible trig polynomials can you use? – 2017-01-06
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    Did you mean $p$ is a *linear combination* of...? – 2017-01-06
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    Martin - Yes I did mean that. I will edit shortly. Paul - See my answer to Martin. – 2017-01-06

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Idea: let be $p$ an $\epsilon$- approximation of $\sin$. The integral $\int_{-\pi}^{\pi}p\sin$ must be $\approx\int_{-\pi}^{\pi}\sin^2 = \pi$, but by the orthogonality of the trigonometrical polynomials, $\int_{-\pi}^{\pi}p\sin = 0$.

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    I understand the idea, and frankly it's quite aesthetic, but orthogonality assumes that an inner product is defined, and in the context of my question it is not defined (the only connection to it is that the metric is defined by the norm). – 2017-01-06
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    @NadavSchweiger, the orthogonality is an elementary fact based in trigonometric formulas like $2\sin(u)\sin(v) = \cos(u − v) − \cos(u + v))$ and the periodicity. – 2017-01-06
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As pointed out in Martín-Blas Pérez Pinilla's answer, there does not exist such approximations on $[-\pi,\pi]$. However, we can prove that there is one on $[0, \pi ]$.

Consider the sequence of polynomials $(P_n)$ defined by $$P_0 = 0 \mbox{ and } \ \forall n,\ P_{n+1} = P_n + \frac{1}{2} \Big( X-P_n^2 \Big)$$

We want to prove that $(P_n)$ converges uniformly to $x \mapsto \sqrt{x}$ on $[0,1]$.

You can show, using induction, that $\forall n \ge 0,\ \forall x \in \mathbb{R}^+,\ 0 \le P_n(x) \le P_{n+1}(x) \le \sqrt{x}$.

Now we define $(\varepsilon_n)_{n \ge 0}$ by : $\varepsilon_0=1$ and $\forall n \ge 0,\ \varepsilon_{n+1} = \varepsilon_n-\frac{\varepsilon_n^2}{4}$. Note that $(\varepsilon_n) \underset{n \to +\infty}{\longrightarrow} 0$.

Firstly $\forall x \in [0,1],\ x-P_n(x)^2 \le \varepsilon_0$. Moreover, you can prove easily that $\forall n \ge 0,\ \forall x \in [0,1],\ x-P_{n+1}(x)^2 \le x-P_n(x)^2 - \frac{\big( x-P_n(x)^2 \big)^2}{4} $. It follows by induction that $$\forall n\ge 0, \forall x \in [0,1],\ x-P_n(x)^2 \le \varepsilon_n$$

Thus, we can conclude that $(P_n)$ converges uniformly on $[0,1]$ to $x\mapsto \sqrt{x}$.

Now we come back to your problem. Note that $\forall x \in [0,\pi],\ \sin(x) = \sqrt{1-\cos(x)^2}$.

Thus $x \mapsto P_n \big( 1 - \cos(x)^2 \big)$ converges uniformly on $[0, \pi]$ to $\sin$.

Finally, it is a well-known fact that these functions can be written as linear combination of the functions $x \mapsto \cos(px)$ for $p \ge 0$.

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    Big problem: $\sqrt{1-\cos(x)^2} = |\sin(x)|\ne\sin(x)$. – 2017-01-06
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    Even more, $P_n(1−\cos^2)$ is obviously even while $\sin$ is odd. – 2017-01-06
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    Of course, sorry, I was approximating on $[0, \pi]$ – 2017-01-06