Stuck when tackling the computation of $\Phi_n(\zeta_8)$ My current way of calculation of $\Phi_n(\zeta_8)$ where $\Phi_n(x)$ is the $n$-th cyclotomic polynomial and $\zeta_8=\cos(\frac{2\pi}{8})+i\sin(\frac{2\pi}{8})$ leave me now stuck at the problem of calculating $$\prod_{d|n}\sin(\frac{d2\pi}{16})^{\mu(\frac{n}{d})}$$ the product running over all integer divisors 0f $n$. $\mu(n)$ is the Möbius function.
My tests with WolframAlpha Online resulted for e.g. prime $n$ and $n$ squarefree with maximally $2$ prime divisors the for me astonishing results of $\pm 1$,$\pm \cot(\frac{\pi}{8})$,$\pm \cot^2(\frac{\pi}{8})$,$\pm \tan(\frac{\pi}{8})$,$\pm \tan^2(\frac{\pi}{8})$ so far.
I see no way how I could find a closed form for the product which the CAS calculations lead me to conjecture to exist.   
 A: It's a classic and elementary fact that since $x^n-1=\prod_{d\mid n}\Phi_d(x)$, by applying $\log$s to both sides followed by Mobius inversion and then exponentiating back we must have 
$$\Phi_n(x)=\prod_{d\mid n}(x^d-1)^{\mu(n/d)}.$$
This is mentioned in almost any source that covers cyclotomic polynomials. Furthermore,
$$\begin{array}{ll} \displaystyle \prod_{d\mid n}(x^d-1)^{\mu(n/d)} & \displaystyle =\prod_{d\mid n}x^{d\mu(n/d)/2}(x^{d/2}-x^{-d/2})^{\mu(n/d)} \\ & \displaystyle =x^{\large\frac{1}{2}\left[\sum\limits_{d\mid n}d\mu(n/d)\right]}\prod_{d\mid n}(x^{d/2}-x^{-d/2})^{\mu(n/d)}. \end{array}$$
Here is a well-known technique utilized in the number theory of arithmetic functions:  since $\sum_{d\mid n}d\mu(n/d)$ is a convolution of multiplicative functions it is itself multiplicative, hence equals 
$$\prod_{p^e\|n}\left(\sum_{r=0}^e p^r\mu(p^e/p^r)\right)=\prod_{p^e\|n}(p^e-p^{e-1})= \varphi(n).$$
The notation $p^e\|n$ when $p$ is a prime means that $p^e\mid n$ but $p^{e+1}\nmid n$, or in other words that $p^e$ is the precise power of $p$ present in $n$'s prime factorization. Thus we have
$$\Phi_n(x)=x^{\varphi(n)/2}\prod_{d\mid n}(x^{d/2}-x^{-d/2})^{\mu(n/d)}.$$
If $x=e^{2\pi i\frac{k}{m}}$ then $x^{d/2}-x^{-d/2}=2i\sin(\pi\frac{kd}{m})$ by Euler's formula, and so by factoring out all of the terms $(2i)^{\mu(n/d)}$ and then using the fact $\sum_{d\mid n}\mu(n/d)=0$ if $n>1$ we get
$$\Phi_n(e^{2\pi i\frac{k}{m}})=e^{\pi i\frac{\varphi(n)k}{m}}\prod_{d\mid n}\sin\left(\pi\frac{kd}{m}\right)^{\mu(n/d)}.$$
Now let's specialize to the case $k/m=1/8$ and $n$ odd. It is geometrically "obvious" that
$$\begin{array}{|c|rrrrrrrr|}\hline d\bmod 16 & 1 & 3 & 5 & 7 & 9 & 11 & 13 & 15 \\ \hline
\sin(\pi\frac{d}{8}) & s & c & c & s & -s & -c & -c & -s \\ \hline \end{array} $$
where $s=\sin(\frac{\pi}{8})~\left(=\frac{1}{2}\sqrt{2-\sqrt{2}}\,\right)$ and $c=\cos(\frac{\pi}{8})~\left(=\frac{1}{2}\sqrt{2+\sqrt{2}}\,\right)$. Therefore
$$\Phi_n(e^{2\pi i/8})=e^{i\pi \varphi(n)/8}(-1)^{v_{9,11,13,15}(n)}\sin(\frac{\pi}{8})^{v_{1,7,9,15}(n)}\cos(\frac{\pi}{8})^{v_{3,5,11,13}(n)} \tag{1}$$
using the ad hoc functions $v_S(n)=\sum_{d\mid n,d\in S\bmod16}\mu(n/d)$. Let's simply abbreviate this
$$\Phi_n(e^{2\pi i/8})=e^{i\pi\varphi(n)/8}(-1)^{\gamma(n)}\sin(\frac{\pi}{8})^{\alpha(n)}\cos(\frac{\pi}{8})^{\beta(n)}.$$
The fact $\sum_{d\mid n}\mu(n/d)=0$ (given $n>1$) tells us $\alpha(n)+\beta(n)=0$. Simplifying we get
$$\Phi_n(e^{2\pi i/8})=e^{i\pi\varphi(n)/8}(-1)^{\gamma(n)}\tan(\frac{\pi}{8})^{\alpha(n)}. $$
Notice something special: the residues $\overline{1},\overline{7},\overline{9},\overline{15}$ form an index two subgroup of $U(16)$, so they are the kernel of some homomorphism $\theta:U(16)\to\{\pm1\}$. Indeed we have
$$\theta(x)=\begin{cases}1 & x^2\equiv1 \mod{16} \\ -1 & x^2\not\equiv 1\mod{16}.\end{cases} \tag{2}$$
Therefore with some trickery we can rewrite $\alpha(n)$ via
$$\alpha(n)=\frac{\alpha(n)-\beta(n)}{2}=\frac{1}{2}\sum_{d\mid n}\mu(n/d)\theta(\overline{d})=\frac{1}{2}\prod_{p^e\|n}\left(\sum_{r=0}^e\mu(p^e/p^r)\theta(\overline{p})^r\right)$$
$$=\frac{1}{2}\prod_{p^e\|n}\left(\theta(\overline{p})^e-\theta(\overline{p})^{e-1}\right)=\frac{\theta(\overline{n})}{2}\prod_{p\mid n}(1-\theta(\overline{p})) $$
$$\alpha(n)=\begin{cases}\theta(\overline{n})2^{\omega(n)-1} & {\rm if}~p^2\not\equiv1\bmod16~{\rm for~each~prime~}p\mid n \\ 0 & {\rm if}~p^2\equiv1\bmod16~{\rm for~any~~prime~}p\mid n. \end{cases} \tag{3}$$
The function $\omega(n)$ counts the number of prime factors of $n$. And so we conclude
Theorem. $\Phi_n(e^{2\pi i/8})=e^{i\pi\varphi(n)/8}(-1)^{\gamma(n)}\tan(\frac{\pi}{8})^{\alpha(n)}$ where $\alpha,\gamma$ are defined in $(1),(2),(3)$, for any odd integer $n>1$.
In particular, the first $n$ for which the product of sines fails to have absolute value $\tan(\frac{\pi}{8})^e$ for an exponent $|e|\le2$ occurs at $n=3\cdot5\cdot11=165$, where $|\Phi_{165}(e^{2\pi i/8})|=\tan(\frac{\pi}{8})^{-4}$.
A: I think I derived a proof of the following formula for $n>1$. I did also a CAS test run for $n=1$ to $n=262$ confirming the formula. $$\Phi_n(\zeta_8)=\zeta_{16}^{\phi(n)}\prod_{d|n}\sin(d\frac{2\pi}{16})^{\mu(\frac{n}{d})}$$
