Can $\sum _{k=0}^n \frac {\sin(\frac {k\pi}2)}{k!}x^k$ be the $n$th Maclaurin polynomial of $\sin x$?

Question

When finding an expression for the $n$th Maclaurin polynomial for $\sin x$, I have given $\sum _{k=0}^n \frac {\sin(\frac {k\pi}2)}{k!}x^k$, as the successive derivatives of $\sin x $ at $0$, i.e. $0,1, 0, -1, 0 \ldots$ are the values of $\sin x$ at successive intervals of $\pi \over 2$. However, I am unsure as to whether my expression is correct because the Maclaurin polynomial for $\sin x$ is always given to me as $\sum _{k=0}^n (-1)^k \frac {x^{2k+1}}{(2k+1)!}$. Is this last sum merely a better representation, or is it the case that the sum I have come up with is incorrect?