nth power of sine as sum of sine and cosine terms I was working on integrals of the form
$$\int_{0}^{\infty}e^{-x\cdot t}\sin^n(x) dx$$
and to solve them I tried to express $\sin^n(x)
$ in form of a sum without any powers. Interesting for me I have found a way which only depends whether the power is even or odd.
For the even ones I have got
$$\frac{1}{2^{n-1}} \left[\frac{(-1)^{\frac{n}{2}}}{2}\binom{n}{\frac{n}{2}}~+~\sum_{k=0}^{\lfloor\frac{n}{2}\rfloor-1}\binom{n}{k}(-1)^{\frac{n-2k}{2}}\cos((n-2k)x)\right] $$
and for the odd ones
$$\frac{1}{2^{n-1}}\left[\sum_{k=0}^{\lfloor\frac{n}{2}\rfloor}\binom{n}{k}(-1)^{\frac{n-2k-1}{2}}\sin((n-2k)x)\right]$$
I have got three questions concerning these series:


*

*Are they right like this, even with this kind of weird power for the minus sign and the seperated first term for the even powers?

*If it is possible, how could you simplify these sums?

*How to proof the rightness of these sums or how to show, that they are wrong?

 A: With $z=e^{ix}$,
$$\sin^nx=\left(\frac{z-z^{-1}}{2i}\right)^n=\frac1{(2i)^n}\sum_{k=0}^n\binom nkz^kz^{-(n-k)}=\frac1{(2i)^n}\sum_{k=0}^n\binom nkz^{2k-n}.$$
For even $n=2m$, the exponent runs from $-n$ to $n$ by step $2$ via $0$ and
$$\frac{(-1)^m}{2^n}\sum_{k=0}^n\binom nkz^{2k-n}=\frac{(-1)^m}{2^n}\left(\sum_{k=0}^m\binom nk\left(z^{2k-n}+z^{n-2k}\right)\right)
\\=\frac{(-1)^m}{2^n}\left(2^*\sum_{k=0}^m\binom nk\cos(2k-n)x\right).$$
We used the shorthand notation $2^*$ to express that the coefficient is $2$, unless $k=m$, which leads to a constant term $\displaystyle\binom nm$.
The development is similar for odd $m$, and all terms are paired.

It may be simpler to grasp with particular examples:
$$(2i)^4\sin^4x=\left(z-z^{-1}\right)^4=z^4-4z^2+6-4z^{-2}+z^{-4}\\=2\cos4x-2\cdot4\cos 2x+6$$
$$(2i)^5\sin^5x=\left(z-z^{-1}\right)^5=z^5-5z^3+10z-10z^{-1}+5z^{-3}-z^{-5}\\=2i\sin5x-2i\cdot5\sin3x+2i\cdot10\sin x.$$
So to a constant factor, the power of a sine is a linear combination of cosines or sines of the argument times every other integer, weighted by every other binomial coefficient and with alternating signs.
A: Just to put it all together in one post:
The right sums should be the following
For even $n$
$$\sin^{2m}(x)~=~\frac{(-1)^m}{2^n}\left[2^*\sum_{k=0}^m~\binom{n}{k}(-1)^k\cos((n-2k)x)\right]$$
where $n=2m$.
For odd $n$
$$\sin^{2m+1}(x)~=~\frac{(-1)^m}{2^{n-1}}\left[\sum_{k=0}^m~\binom{n}{k}(-1)^k\sin((n-2k)x)\right]$$
where $n=2m+1$
Am I right this time, or did I made a mistake somewhere?
Since there is still this weird constant term in the sum for even powers, which kind of annoys me.
