We're Only On Coprime Numbers?

We will find the product for $1\leq k\leq 2016$ and $\gcd(k,2016)=1$ . Let $n=2016$ , an even number.

We will work with moduli so that we don't have to keep track of messy factors $i$ and $e^{it}$ . Now

$2^{\phi(n)}P^2=\prod_{k}2\sin(k\pi/n)=\prod_{k}|e^{ik\pi/n}-e^{-ik\pi/n}|=\prod_{k}|1-e^{2ik\pi/n}|=|\Phi_n(1)|,$

where $\Phi_n(x)$ is the cyclotomic polynomial. We can use induction on $n$ (or fancier methods) to show that $\Phi_n(1)=p$ if $n=p^k$ is a prime power, and $\Phi_n(1)=1$ otherwise.

Thus $P=\sqrt{\frac{\Phi_n(1)}{2^{\phi(n)}}}=\frac{1}{2^{576/2}}=\frac{1}{2^{288}}$

and the answer is $\boxed{289}$

Relevant wiki: Dirichlet Convolution

My approach to this problem is less elegant.

Lemma: $\prod _{k=1}^n\sin \left(\frac{k\pi }{2n}\right)=\frac{\sqrt{n}}{2^{n-1}}$

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Proof:

$\left(2^n\prod _{k=1}^n\sin \left(\frac{k\pi }{2n}\right)\right)^2=\left(\prod _{k=1}^n2\sin \left(\frac{k\pi }{2n}\right)\right)^2=2\left(\prod _{k=1}^{2n-1}2\sin \left(\frac{k\pi }{2n}\right)\right)$

Using a method similar to Otto's, the product is

$2\left(\prod _{k=1}^{2n-1}\left|1-e^{\frac{ik\pi }{n}}\right|\right)$

Since though roots of unity,

$2\left(\prod _{k=1}^{2n-1}\left|x-e^{\frac{ik\pi }{n}}\right|\right)=\frac{2}{\left|x-e^{2ik\pi}\right|}\prod _{k=1}^{2n}\left|x-e^{\frac{ik\pi }{n}}\right|=2\left|\frac{x^{2n}-1}{x-1}\right|=2\left|\sum _{k=0}^{2n-1}x^k\right|$

$2\left(\prod _{k=1}^n\left|1-e^{\frac{ik\pi }{n}}\right|\right)\left(\prod _{k=n+1}^{2n-1}\left|1-e^{\frac{ik\pi }{n}}\right|\right)=4n$

Putting it all together gives

$\prod _{k=1}^n\sin \left(\frac{k\pi }{2n}\right)=\frac{\sqrt{n}}{2^{n-1}}$

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Now this is where mobius inversion comes in. Let $\prod _{k=1}^n\sin \left(\frac{k\pi }{2n}\right)=F\left(n\right)$ and $\prod _{1\le k\le n, (n,k)=1}\sin \left(\frac{k\pi }{2n}\right)=f(n)$

Notice that

$\ln \left(F\left(n\right)\right)=1 * \ln \left(f\left(n\right)\right)$

Where $*$ is the dirichlet convolution. Through mobius inversion,

$\ln \left(f\left(n\right)\right)= \mu*\ln \left(F\left(n\right)\right)$

Substituting $F(n)=\frac{\sqrt{n}}{2^{n-1}}$ ,

$\ln \left(f\left(n\right)\right)=\frac{1}{2}\sum _{d \mid n}^{ }\mu\left(\frac{n}{d}\right)\ln \left(d\right)-\ln \left(2\right)\left(\sum _{d \mid n}^{ }\mu\left(\frac{n}{d}\right)d-\sum _{d \mid n}^{ }\mu\left(\frac{n}{d}\right)\right)$

$=\frac{1}{2}\Lambda \left(n\right)-\ln \left(2\right)\left(\varphi \left(n\right)-\varepsilon \right)$

Where $\Lambda$ is the von Mangoldt function, $\varphi$ is euler's totient function and $\varepsilon$ is the multiplicative identity.

Substituting n=1008 gives

$f\left(1008\right)=\boxed{2^{-288}}$

At the beginning the mention may be allowed that the following solution is obviously not so elegant - but "The end justifies the means!":

First calculate the value for the product via EXCEL: P=2.01076E-87

Now adapt the result to the necessary form: $\frac{ln(\frac{P}{a})}{ln(2)}=-b$

Already for a=1 obtain b=288 and therefore solution $\boxed{289}$ .