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So, out of boredom and curiosity, today I came up with a series representation for $\sin(nu)$ when $n$ is an even integer: $$\sin(nu) = \sum_{k=1}^\frac n2 \left(\left(-1\right)^{k-1}\binom{n}{-\left|2k-n\right|+n-1}\sin\left(u\right)^{2k-1}\cos\left(u\right)^{n-2k+1}\right)\;\mathtt {if}\;n\in 2\Bbb Z$$ I was working on a similar representation for when $n$ is an odd integer, but I'm having some difficulties. There doesn't seem to be much of a pattern. If it exists, could someone please point me in its direction? If it's impossible, could you provide me with the proof?

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3 Answers 3

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Using complex methods and the binomial theorem, $$\eqalign{\sin(nu) &={\rm Im}(\cos u+i\sin u)^n\cr &={\rm Im}\sum_{m=0}^n \binom nm (\cos u)^{n-m}(i\sin u)^m\ .\cr}$$ As only the terms for odd $m$ contribute to the imaginary part we can take $m=2k-1$ to give $$\sin(nu)=\sum_{k=1}^{(n+1)/2}(-1)^{k-1}\binom n{2k-1}\cos^{n-2k+1}u\sin^{2k-1}u\ .$$

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PS If you haven't seen this kind of thing before, here is a bit of information. –  David Feb 16 at 5:32

Notice that, using Euler's Formula, for general $n$ a positive integer, $$\begin{align}\sin(nx) &= \frac{e^{inx}-e^{-inx}}{2i}\\ &=\frac{(\cos{x} + i\sin{x})^n - (\cos{x} - i\sin{x})^n}{2i}\\ &=\sum_{k=0}^{n}{\binom{n}{k}\frac{\cos^k{x}(i\sin{x})^{n-k} - \cos^k{x}(-i\sin{x})^{n-k}}{2i}}\\ &=\sum_{k=0}^{n}{\binom{n}{k}\cos^k{x}\sin^{n-k}{x}\frac{i^{n-k} - (-i)^{n-k}}{2i}}\\ &=\sum_{k=0}^{n}{\binom{n}{k}\cos^k{x}\sin^{n-k}{x}\sin{\frac{1}{2}(n-k)\pi}}\end{align}$$

I am ashamed to confess blatantly that this was taken (word for word) from here, the first link returned using the Google search query "Multiple Angle Formula".

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There are very nice and simple formulas using Chebyshev polynomials of the first kind $T_n$ (if $n$ is odd) and of the second kind $U_n$ (if $n$ is even) $$sin(nx)=(-1)^{\frac{n-1}{2}} T_n(\sin (x))$$ $$sin(nx)=(-1)^{\frac{n}{2}-1} \cos (x) U_{n-1}(\sin (x))$$

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