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Let $n$ be a nonnegative integer. Can you help me prove the following ?

There exists a unique polynomial $P_{n}$ such that for all $t \in [0,\frac{\pi}{2}]$, $P_{n}(\operatorname{cotan}^2t)=\frac{\sin((2n+1)t}{(\sin t)^{2n+1}}$ with $\operatorname{cotan}(x)=\frac{\cos x}{\sin x}$.

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Where are you stuck? For the existence or the uniqueness? – Davide Giraudo Jan 11 '12 at 9:19
For the moment, both, but I guess the hardest part is proving the uniqueness. – user20010 Jan 11 '12 at 10:02

There may be an easier way but this is what I came up with:

Start with Euler's formula: \begin{align} e^{i(2n+1)t} &= \cos((2n+1)t) + i \sin((2n+1)t) = (\cos t + i \sin t)^{2n+1} \\ &= \sum_{k=0}^{2n+1} \binom{2n+1}{k}i^k \sin^k t \cos^{2n+1-k}t \\ \end{align} Now we separate the sum for even and odd values of $k$: \begin{align} &\cos((2n+1)t) + i \sin((2n+1)t) \\ &= \sum_{l=0}^{n} \binom{2n+1}{2l} i^{2l} \sin^{2l} t \cos^{2n+1-2l}t + \sum_{l=0}^{n} \binom{2n+1}{2l+1} i^{2l+1} \sin^{2l+1} t \cos^{2n+1-2l-1}t \\ &= \sum_{l=0}^{n} \binom{2n+1}{2l} (-1)^l \sin^{2l} t \cos^{2n+1-2l}t+ i \sum_{l=0}^{n} \binom{2n+1}{2l+1} (-1)^l \sin^{2l+1} t (\cos^2 t)^{n-l} \\ \end{align}

Since the imaginary parts must coincide, we get: \begin{align} \sin((2n+1)t) &= \sum_{l=0}^{n} \binom{2n+1}{2l+1} (-1)^l \sin^{2l+1} t\ (\cos^2 t)^{n-l} \\ &= \sin^{2n+1} t \quad \sum_{l=0}^{n} \binom{2n+1}{2l+1} (-1)^l (\sin^2 t)^{l-n} (\cos^2 t)^{n-l} \\ &= \sin^{2n+1} t \quad\sum_{l=0}^{n} \binom{2n+1}{2l+1} (-1)^l \Big(\frac{\cos^2t}{\sin^{2} t}\Big)^{n-l} \\ \end{align}

Hence $$\frac{\sin((2n+1)t)}{\sin^{2n+1}t} = \sum_{l=0}^{n} \binom{2n+1}{2l+1} (-1)^l \Big(\frac{\cos^2t}{\sin^{2} t}\Big)^{n-l} = P_n(\cot^2 t), $$ where $P_n$ is the (unique) $n$-degree polynomial given by $P_n(x) = \sum_{l=0}^{n} \binom{2n+1}{2l+1} (-1)^l x^{n-l}$.

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