$n$-th derivative of $\sin^k(x)$ I would like to know if there is a general formula to calculate the $n$-th derivative of $\sin^k(x)$ evaluated at $x=0$, that is, 
$$\left.\frac{d^n}{d x^n} (\sin^k(x))\right|_{x=0}$$
with $0\leq k \leq n$.
 A: You have a composition of two functions. To take the $n$th derivative, apply the Faa di Bruno formula outlined here, with $f(x)=x^k$ and $g(x)=\sin(x)$, and the divided $n$th derivative $f^{[n]}=\frac{1}{n!}f^{(n)}$.
$$\begin{align}
\left(f\circ g\right)^{[n]}(x)&=\sum_{P\in\mathcal{P}_n}f^{[\left\lvert P\right\rvert]}(g(x))g^{[P]}(x)\\
&=\sum_{P\in\mathcal{P}_n}f^{[\left\lvert P\right\rvert]}(\sin(x))\sin^{[P]}(x)\\
&=\sum_{P\in\mathcal{P}_n}\binom{k}{\left\lvert P\right\rvert}\sin^{k-\left\lvert P\right\rvert}(x)\sin^{[P]}(x)\\
\end{align}$$

Since you are only interested at evaluation at $x=0$, then since $\sin(0)=0$, you will only need to consider the terms where $|P|=k$.
$$\begin{align}
\left.\left(f\circ g\right)^{[n]}(x)\right|_{x=0}&=\left.\sum_{\left\lvert P\right\rvert=k}\sin^{[P]}(x)\right|_{x=0}\\
\left.\left(f\circ g\right)^{(n)}(x)\right|_{x=0}&=n!\left.\sum_{\left\lvert P\right\rvert=k}\sin^{[P]}(x)\right|_{x=0}\\
\end{align}$$
And actually the only terms that give a nonzero contribution are those that come from a partition consisting of only odd numbers (since otherwise $\left.\sin^{[P]}(x)\right|_{x=0}$ will have a factor of $\left.\sin(x)\right|_{x=0}$). For example, with $k=8$, and $n=12$, the only ordered partitions of $12$ that use $8$ odd numbers are $$(1+1+1+1+1+1+1+5)\times 8\quad(1+1+1+1+1+1+3+3)\times \binom{8}{2}$$ so $$\left.\frac{d^{12}}{dx^{12}}\sin^8(x)\right|_{x=0}=12!\cdot\left(8\cdot\frac{1}{5!}+28\cdot\left(\frac{-1}{3!}\right)^2\right)=404490240$$

Here is a summary formula:
$$\begin{align}
\left.\frac{d^n}{dx^n}\sin^k(x)\right|_{x=0}
&=n!\sum_{\begin{array}{c}j_1+j_2+\cdots j_k=n\\j_i\text{ odd},\; j_i\geq1\end{array}}\prod_{i=1}^k\frac{(-1)^{(j_i-1)/2}}{j_i!}\\
&=(-1)^{(n-k)/2}\sum_{\begin{array}{c}j_1+j_2+\cdots j_k=n\\j_i\text{ odd},\; j_i\geq1\end{array}}\binom{n}{j_1\;j_2\;\cdots\;j_k}
\end{align}$$
Some corollaries:
$$\left.\frac{d^n}{dx^n}\sin^k(x)\right|_{x=0}=\begin{cases}0&n<k\\
n!&n=k\\
0&n\not\equiv k\mod{2}\\
(-1)^{(n-1)/2}&k=1\\
-n!k/6&n=k+2
\end{cases}$$
A: Probably the easiest way to do this is to use the exponential form of $\sin$ and then the binomial theorem:
$$
\sin^k(x) = \left(\frac{e^{ix}-e^{-ix}}{2i}\right)^k=\frac{1}{(2i)^k}\sum_{j=0}^k \binom{k}{j} e^{jix}(-1)^{k-j}e^{-(k-j)ix}=\frac{1}{(2i)^k}\sum_{j=0}^k \binom{k}{j} (-1)^{k-j}e^{(2j-k)ix}
$$
The $n$th derivative of $e^{ax}$ evaluated at zero is $a^n$, so the formula you're looking for is
$$
\frac{1}{(2i)^k}\sum_{j=0}^k \binom{k}{j}(-1)^{k-j}[i(2j-k)]^n
$$
You could turn this into a purely real formula, but it'd require a certain amount of case analysis; this is probably the most compact form unless there's some simplifying binomial identity I'm missing.
(The other possible approach would be to examine the Taylor series of $\sin^k x$, but that requires the multinomial theorem instead of the binomial theorem, so it's likely to be a lot messier...)
A: In Wikipedia you find the formulas for the transformation of a power of a sine to a sum of sines of multiples. Taking the derivatives at $x=0$ is then straightforward.
For instance, for odd $k$ and odd $n$
$$2^{1-k}(-1)^{(k-1)/2}\sum_{j=0}^{(k-1)/2}(-1)^{j+(n-1)/2}\binom kj(k-2j)^n.$$

Due to the renaming of the indexes "on the fly", typos are likely.
A: First using the identity,
$$
\sin x=\frac{1}{2 i}\left(e^{x i}-e^{-x i}\right)
$$
to expand
$$\begin{aligned}
\sin ^{k} x &=\frac{1}{(2i)^ k } \sum_{j=0}^{k}(-1)^j\left(\begin{array}{l}
k \\
j
\end{array}\right) e^{x(k-j) i} e^{-x i j} \\
&=\frac{1}{2^{k} i^{k}} \sum_{j=0}^{k}(-1)^j\left(\begin{array}{l}
k \\
j
\end{array}\right) e^{x(k-2 j) i}
\end{aligned} $$
Differentiating it $n$ times w.r.t. $x$ yields
$$
\frac{d^{n}}{d x^{n}}\left(\sin ^{k} x\right)=\frac{1}{2^{k} i ^k} \sum_{j=0}^{k}(-1)^{j} \left(\begin{array}{l}
k \\
j
\end{array}\right)[(k-2 j) i]^{n} e^{x(k-2 j) i}= \boxed{\frac{i^{n-k}}{2 ^ k} \sum_{j=0}^{k}(-1)^{j} \left(\begin{array}{l}
k \\
j
\end{array}\right)(k-2 j)^{n} e^{x(k-2 j)i}
}$$
Putting $x=0$ yields $$
\left.\frac{d^{n}}{d x ^{n}}\left(\sin ^{k} x\right)\right|_{x=0}=\frac{i^{n-k}}{2^{k}} \sum_{j=0}^{k}(-1)^{j}\left(\begin{array}{c}
k \\
j
\end{array}\right)(k-2 j)^{n}
$$
Consequently, assuming $x$ is real, we have
A. When $n$ and $k$ are of different parity,
$$
\left.\frac{d^{n}}{d x^{n}}\left(\sin ^{k} x\right) \right|_{x=0} =0
$$
B. When $n$ and $k$ are of same parity,
$$
\left.\frac{d^{n}}{d x^{n}}\left(\sin ^{k} x\right) \right|_{x=0} =\frac{(-1)^{\frac{n-k}{2}}}{2^{k}} \sum_{j=0}^{k}(-1)^{j}\left(\begin{array}{c}
k \\
j
\end{array}\right)(k-2 j)^{n} 
$$
