# Find a Maclaurin series for sin(2+x)? Trouble finding a representation for it?

Here's the work i've done to find the Maclaurin series. However, I'm having a very hard time finding a representation for the series using sum, n for the nth term, and x from g(x).

• It is absolutely fine. Where do you have a problem? – Rohan Dec 12 '16 at 4:17
• @Rohan I need a Maclaurin series in the form of a summation representation: for example, $$\sum_{n=0}^ ∞ sin(x) = \frac{(-1)^n*(x)^{2n+1}}{(2n+1)!}$$ and I wasn't sure if it was sufficient to just alter the x into (x+2) in this notation – gticecream8 Dec 12 '16 at 4:27
• @gticecream8 I believe you got that backwards : p. – YoTengoUnLCD Dec 12 '16 at 4:52
• Why wouldn't you approximate it at x0 = -2 ? – Itay.V Dec 12 '16 at 5:56
• Or you just could change variable such as t = x + 2 and approximate at t=0 which will be maclaurin – Itay.V Dec 12 '16 at 5:58

One way to get the "${+}{+}{-}{-}$" pattern is with $(-1)^{n(n-1)/2}$.

To alternate between $\sin(2)$ and $\cos(2)$: $$\sin(2)\frac{(-1)^n+1}{2}+\cos(2)\frac{(-1)^{n-1}+1}{2}$$

So you could write $$\sum_{n=0}^{\infty}(-1)^{n(n-1)/2}\left(\sin(2)\frac{(-1)^n+1}{2}+\cos(2)\frac{(-1)^{n-1}+1}{2}\right)\frac{x^n}{n!}$$