Ytukyg

Is there any way to express the trigonometric functions as infinitely long polinomials? If so, how? If not why?

Obviously doing $x(x-pi)(x-2pi)(x-3pi)...$ does not work as it doesn't match for values in between the zeroes.

I have heard that there are ways of making your approach work, but I don't know anything about it personally.

The standard approach would be Taylor series. The idea of Taylor series is to make an "infinite degree" polynomial for which the function value and all derivatives agree with those of your function.

Specifically, for $sin$, we have
$$
sin(x)=x-frac{x^3}{3!}+frac{x^5}{5!}-frac{x^7}{7!}+cdots
$$
For any given $x$, the factorial $n!$ eventually grows much faster than the power $x^n$, so the series will converge.

A Taylor series is an expression like the above (we do allow even degree terms, it's just that $sin$ happens to not have them), possibly with $x$ replaced by $(x-a)$ for some real number $a$. We say that such a Taylor series is centered around $a$. The above series for $sin$ is centered at $0$.

In general, in order to have a Taylor series, a function must be infinitely differentiable. And even then, there is no guarantee that the Taylor series converges, or that it converges to the function it is derived from. And even if it does, the series may not converge everywhere.

A function which for each point $a$ in its domain has a Taylor series centered at $a$ which converges on some open neighborhood of $a$ is called analytic. Many (most?) named functions that one learn about in school are analytic. Trigonometric functions like $sin$ are analytic, as are polynomials, $sqrt{phantom x}$, $e^x$, and rational functions, to name a few.

Another answer has mentioned the infinite sum Taylor series for sine. Here is the infinite product Weierstrass factorization for sine:
$$ sin(x) = x left(1 - left(frac{x}{pi}right)^2 right)
left(1 - left(frac{x}{2pi}right)^2 right)
left(1 - left(frac{x}{3pi}right)^2 right) cdots. $$
There is a similar factorization for cosine.

You already received good answers to the question.

In this area, we can do other things which are amazing. If you look here, being myself fascinated by the approximation $$sin(x) simeq frac{16 (pi -x) x}{5 pi ^2-4 (pi -x) x}qquad (0leq xleqpi)$$ which wasproposed, more than $1400$ years ago by Mahabhaskariya of Bhaskara I (a seventh-century Indian mathematician), I wrote
$$sin(x)=sum_{k=1}^n a_k [(pi-x)x]^k$$ which respects the value of $0$ at $x=0$ and $x=pi$. The first coefficients are
$$left(
begin{array}{cc}
k & a_k \
1 & frac{1}{pi } \
2 & frac{1}{pi ^3} \
3 & frac{12-pi ^2}{6 pi ^5} \
4 & frac{10-pi ^2}{2 pi ^7} \
5 & frac{1680-180 pi ^2+pi ^4}{120 pi ^9} \
6 & frac{1008-112 pi ^2+pi ^4}{24 pi ^{11}} \
7 & frac{665280-75600 pi ^2+840 pi ^4-pi ^6}{5040 pi ^{13}} \
8 & frac{308880-35640 pi ^2+450 pi ^4-pi ^6}{720 pi ^{15}}
end{array}
right)$$ The maximum error is at $x=frac pi 2$; for example, using $n=8$, the error is just $4.99 times 10^{-13}$.