Taylor/Maclaurin Series vs Taylor/Maclaurin Polynomial

For this question, I will use $$sin(x)$$ as an example. The Maclaurin series for $$\sin(x)$$ is $$\sum_{n=0}^{\infty} \frac{{(-1)}^nx^{2n+1}}{(2n+1)!}$$ This gives us a Maclaurin polynomial to represent $$\sin(x)$$, but is there another way to represent $$\sin(x)$$ which doesn't give us a polynomial? Is there any case where the Taylor/Maclaurin series gives us a non-polynomial representation for a certain function?

• If I remember correctly, the Taylor/Maclaurin series by definition have to have be a polynomial (perhaps of "infinite degree," since a polynomial is by definition finite in degree but that's another issue). Fourier series of functions can be used as a means of writing functions as a (possibly infinite) sum of sine and cosine terms, and I imagine other series based on different functions can be developed as well. (None come to mind since this isn't something I've looked too deeply into.) – Eevee Trainer Feb 9 at 2:49

There are many ways to represent $$\sin\left(x\right)$$ which doesn't use a polynomial (of any degree), but with Maclaurin & Taylor series, due to how they're defined, they will always give a "polynomial" (almost always of infinite degree, although polynomials may be defined as only allowing a finite degree, as Eevee Trainer commented to the question) representation for any function.
• @Jondue There are basic trigonometric identities like $\sin\left(x\right) = \cos\left(\frac{\pi}{2} - x\right)$ or $\sin\left(x\right) = \frac{1}{\csc\left(x\right)}$. With complex #'s, Euler's formula of $e^{ix} = cos\left(x\right) + i\sin\left(x\right)$ gives $\sin\left(x\right) = \frac{e^{ix} - cos\left(x\right)}{i}$. As Eevee Trainer stated, you could also use a Fourier type series of functions, with I believe it being possible to use something other than $\sin$ and $\cos$, although it'll likely be more complicated & less useful. – John Omielan Feb 9 at 4:30