<s>[INST] How do I find the exact value of $\cos\frac{\pi}{12}\cos\frac{5\pi}{12}\cos\frac{7\pi}{12}\cos\frac{11\pi}{12}$? I know that $\cos(6\phi)\equiv32c^6-48c^4+18c^2-1$ where $c=\cos\phi$.
I also know that when $\cos(6\phi)=0$, then $\phi=\frac{k\pi}{12}$ ($k = 1,3,5,7,9,11$).
How do I find the exact value of:
$$\cos\left(\frac{\pi}{12}\right) \cos\left(\frac{5\pi}{12}\right) \cos\left(\frac{7\pi}{12}\right) \cos\left(\frac{11\pi}{12}\right)$$
 [/INST] Mosquito-nuking linear algebraic solution: it can be shown that the tridiagonal matrix
$$\mathbf T=\frac12\begin{pmatrix}-1&1&&&&\\1&0&1&&&\\&1&0&1&&\\&&1&0&1&\\&&&1&0&1\\&&&&1&1\end{pmatrix}$$
has the eigenvalues $\cos\dfrac{(2k-1)\pi}{12},\quad k=1,\dots,6$. Since the product of the eigenvalues is equal to the determinant of the matrix,
$$\cos\frac{\pi}{12}\cos\frac{5\pi}{12}\cos\frac{7\pi}{12}\cos\frac{11\pi}{12}=\frac{\det\mathbf T}{\cos\tfrac{\pi}{4}\cos\tfrac{3\pi}{4}}=\frac{-\tfrac1{32}}{-\tfrac12}=\frac1{16}$$
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