Ist it possible to "write out" f(x) = sin(x) or other trigonomical functions Today I had a discussion with a friend about trigonometric functions. We were stuck at a function where we couldn't extract $x$ from a trigonometric function, like (oversimplifying)
$$
(x-4)\cdot(x-2)=\sin\left(x\cdot\frac{\pi}{2}\right) 
$$
I know this question can't be solved by analytic methods, but we asked us why. If $\sin$ is just another function (like our professor told us) why can't we get the instructions of the function and apply them to $x$?
To make it clearer: if we have the function $g(x) = x + 2$ we could write 
$52 + g(2) = 56$ or $52 + (2 + 2)= 56$, both equations are identical. But if we can do this why can we only write $52 + sin(90) = 52,89\dots$ and not
$$ 
52 + \text{insert some instructions based on a pattern here} = 52,89\dots
$$
Thanks to everybody in advance.
 A: You can write 
$$\sin(x) = \sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n+1}}{(2n+1)!}$$
or also 
$$\sin(x) = x \prod_{n=1}^{\infty}(1 - \frac{x^2}{n^2 \pi^2}) $$
and in some other ways. 
Note though that the sum and the product is infinite so it might not be apparent to you what this means precisely. 
But for example for say $|x| \le 1/2$, if you compute 
$$\sin(x) = \sum_{n=0}^{10}\frac{(-1)^nx^{2n+1}}{(2n+1)!}$$
that'll be already very close to $\sin(x)$. 
But you never can write $\sin(x)$ as  polynomial. For one thing $\sin(x)$ has infinitely many zeroes/roots while a non-zero polynomial only has finitely many. Note how in the product description you can see the roots, it'll be $0$ when $1 - \frac{x^2}{n^2 \pi^2}=0$ for some $n$
 that is $x= \pm n\pi$ (and the factor $x$ gives the zero at $0$).
A: 
If sin is just another function [...] why cant we get the instructions of the function and apply them

Sure you can. Unfortunately, there are infinitely many of these "instructions". Its MacLaurin series is:
$$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \frac{x^9}{9!} ...$$ and on and on towards infinity (for $x$ close to $0$).
There are a myriad of other ways to rewrite $\sin x$. For example if you know complex numbers, as:
$$\frac{e^{\pi i} - e^{-\pi i}}{2i}$$
Although this probably doesn't help you in your quest anyway.
Note: Even without such "complicated" functions like Sine and Cosine, general polynomials of degree six or higher, where you probably consider that all "instructions" are present, (like $x^6+x^5+x^4+x^3+x^2+x+1=0$) are impossible to solve without guessing at least one root.
A: You can't "get the instructions of the function" because trigonometric functions are non-linear: this means that you can't express $\sin(x)$ as a combination of $x$ and its powers — in other words, $\sin(x)$ cannot be exactly expressed as a polynomial. Why?
Look at the definition of sine on Wikipedia:

The sine of an angle $x$ is the ratio of the length of the opposite side to the length of the hypotenuse.

The magnitude of $x$ is not involved in the mentioned ratio! Given $x$ only, you can't reconstruct the length of opposite side and hypotenuse.

However, not everything is lost. You can still get an approximation of the sine function by a polynomial using an interpolation method, for example Lagrange polynomial.
