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:
$$\cos\left(\frac{\pi}{12}\right) \cos\left(\frac{5\pi}{12}\right) \cos\left(\frac{7\pi}{12}\right) \cos\left(\frac{11\pi}{12}\right)$$
 A: From the formulas: $\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)\\\cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b)$
we have: $\cos(a)\cos(b)=\frac{1}{2}\big(\cos(a+b)+\cos(a-b)\big)$
Apply this formula:
$\cos\left(\frac{\pi}{12}\right)\cos\left(\frac{11\pi}{12}\right) \cos\left(\frac{5\pi}{12}\right) \cos\left(\frac{7\pi}{12}\right) \\=\frac{1}{2}\big(\cos(\pi)+\cos(\frac{10\pi}{12})\big)\frac{1}{2}\big(\cos(\pi)+\cos(\frac{2\pi}{12})\big)=\frac{1}{4}\big(-1-\cos(\frac{2\pi}{12})\big)\big(-1+\cos(\frac{2\pi}{12})\big)=\frac{1}{4}\big(1-\cos(\frac{2\pi}{12})^2\big)$
You certainly know: $\cos(\frac{2\pi}{12})=\cos(\frac{\pi}{6})=\frac{\sqrt{3}}{2}$
Result is: $\frac{1}{16}$
A: As $\frac{11\pi}{12}=\pi-\frac\pi{12}$ and $\frac{7\pi}{12}=\pi-\frac{5\pi}{12}$
and $\cos(\pi-x)=-\cos x,$
$$\cos\left(\frac{\pi}{12}\right) \cos\left(\frac{5\pi}{12}\right) \cos\left(\frac{7\pi}{12}\right) \cos\left(\frac{11\pi}{12}\right)$$
$$=\cos\left(\frac{\pi}{12}\right) \cos\left(\frac{5\pi}{12}\right)\left( - \cos\left(\frac{5\pi}{12}\right)\right)\left( -\cos\left(\frac{\pi}{12}\right)\right)$$
$$=\left(\cos\left(\frac{\pi}{12}\right) \cos\left(\frac{5\pi}{12}\right)\right)^2$$
Now as $\frac{5\pi}{12}=\frac\pi2-\frac\pi{12}$ and $\sin2x=2\sin x\cos x,$
$$\cos\left(\frac{\pi}{12}\right) \cos\left(\frac{5\pi}{12}\right)=\cos\left(\frac{\pi}{12}\right) \sin\left(\frac{\pi}{12}\right)=\frac{\sin \frac\pi6}2=\frac{\frac12}2=\frac14$$
A: HINT:
SO, the equation whose roots are $\cos\frac{r\pi}{12}$ where $r=3,9$ is
$$\left(c-\cos \frac{3\pi}{12}\right)\left(c-\cos \frac{9\pi}{12}\right)=0$$
$$\implies \left(c-\frac1{\sqrt2}\right)\left(c+\frac1{\sqrt2}\right)=0\text{ as }\cos \frac{9\pi}{12}=\cos\left(\pi- \frac{3\pi}{12}\right)=-\cos\frac{3\pi}{12}$$ 
$$\implies 2c^2-1=0$$
SO, the equation whose roots are $\cos\frac{r\pi}{12}$  where $r=1,5,7,11$ will be
$$\frac{32c^6-48c^4+18c^2-1}{2c^2-1}=0$$
Now, apply Vieta's Formulas 1,2,3
A: 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,
$$\cos\frac{\pi}{12}\cos\frac{5\pi}{12}\cos\frac{7\pi}{12}\cos\frac{11\pi}{12}=\frac{\det\mathbf T}{\cos\tfrac{\pi}{4}\cos\tfrac{3\pi}{4}}=\frac{-\tfrac1{32}}{-\tfrac12}=\frac1{16}$$
