Is there a closed form for $\arccos\left(\dfrac{2 \pi}{2^N}\right)$ in terms of $N \in \mathbb{Z}, N \ge 3$?
I'm not super optimistic, but I'm not sure how to really start exploring the problem, either.
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2Why would you want to find an inverse trig function of a rational multiple of $\pi$. Usually a rational multiple of $\pi$ is the output of an inverse trig function, not an input. – Keshav Srinivasan Jan 01 '14 at 04:10
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It has to do with a sinusoid evaluated at Chebyshev points. – Jay Lemmon Jan 01 '14 at 06:23
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$$\arccos z= \frac {\pi} {2} - \left( z + \left( \frac {1} {2} \right) \frac {z^3} {3} + \left( \frac {1 \cdot 3} {2 \cdot 4} \right) \frac {z^5} {5} + \cdots\ \right) = $$ $$=\frac {\pi} {2} - \sum_{n=0}^\infty \frac {\binom{2n} n z^{2n+1}} {4^n (2n+1)}; \qquad | z | \le 1 $$
Adi Dani
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1Okay, but can you pull the $2^N$ term out from the infinite series? As it is, it's a bit of a tautology, isn't it? – Jay Lemmon Jan 01 '14 at 20:20
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@Adi -- where did this series come from? I considered that the Taylor's series is a "closed form", but as per Jay above, it is kind of a tautology. – Betty Mock Jan 03 '14 at 21:49
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These are kind of trivial, but maybe they are closed enough for you:
$$\arccos(z) = i \cdot \log\left(z - i\sqrt{1-z^2}\right)$$
or
$$\arccos(z) = \int_z^1 \dfrac{dz}{\sqrt{1-z^2}}$$
Ian Mateus
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Betty Mock
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