How to expand $\sin (2z+1)$ in powers of $(z+1)$? I'm trying expand $\sin (2z+1)$ in powers of $(z+1)$ without search the coeficients of the Taylor Series.
I'm trying using the Mclaurin expansion for $\sin$ and $\cos$, so:
\begin{align}
\sin(2z+1) &= \sin(2(z+1)-1) \\
&= \sin(2(z+1))\cdot \cos(1)-\cos(2(z+1))\cdot \sin(1) \\
&= \sum_{n=0}^{\infty}\frac{(-1)^n\cdot2^{2n+1} \cdot (z+1)^{2n+1}\cdot cos(1)}{(2n+1)!}
\\  &-\sum_{n=0}^{\infty}\frac{(-1)^n\cdot2^{2n} \cdot (z+1)^{2n}\cdot \sin(1)}{(2n)!}
\end{align}
I'm stuck here, i've tried put in the same sum the terms but i can't get a "friendly" result.
 A: First, you can move the subtraction into the power of $-1$ in the second set of terms to get
$$-\sum_{n=0}^{\infty}\frac{(-1)^n\cdot2^{2n} \cdot (z+1)^{2n}\cdot \sin(1)}{(2n)!} = \sum_{n=0}^{\infty}\frac{(-1)^{n+1}\cdot2^{2n} \cdot (z+1)^{2n}\cdot \sin(1)}{(2n)!} \tag{1}\label{eq1A}$$
Next, note the first summation uses only all of the odd powers of $z + 1$ while the second one uses only all of the even powers of $z + 1$, so there's no overlap of terms, making it easier to combine the $2$ of them. However, you need to deal with the powers of $-1$ being $-1$ or $1$, and the $\cos(1)$ or $\sin(1)$ factors, in a way to distinguish which ones are used, such as in a special function I show below. You could thus use something like
$$\sin(2z + 1) = \sum_{n=0}^{\infty}\frac{2^{n} \cdot (z+1)^{n} \cdot f(n)}{n!} \tag{2}\label{eq2A}$$
where
$$
f(n) = 
\begin{cases}
-\sin(1), & \text{if $n \equiv 0 \pmod 4$} \\
\cos(1), & \text{if $n \equiv 1 \pmod 4$} \\
\sin(1), & \text{if $n \equiv 2 \pmod 4$} \\
-\cos(1), & \text{if $n \equiv 3 \pmod 4$}
\end{cases}
\tag{3}\label{eq3A}$$
Update: As stated in The_Sympathizer's comment, you can use the phase-shifting aspect between $\sin$ and $\cos$ (e.g., $\sin\left(x + \frac{\pi}{2}\right) = \cos(x)$) to get a simpler, non-case form for $f(n)$ such as
$$f(n) = -\sin\left(1 - \frac{\pi n}{2}\right) \tag{4}\label{eq4A}$$
which you could just use directly instead in \eqref{eq2A}.
