TITLE: How do I find the exact value of $\cos\frac{\pi}{12}\cos\frac{5\pi}{12}\cos\frac{7\pi}{12}\cos\frac{11\pi}{12}$?
QUESTION [5 upvotes]: 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:
$$\cos\left(\frac{\pi}{12}\right) \cos\left(\frac{5\pi}{12}\right) \cos\left(\frac{7\pi}{12}\right) \cos\left(\frac{11\pi}{12}\right)$$

REPLY [3 votes]: 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 $\sin2x=2\sin x\cos x,$
$$\cos\left(\frac{\pi}{12}\right) \cos\left(\frac{5\pi}{12}\right)=\cos\left(\frac{\pi}{12}\right) \sin\left(\frac{\pi}{12}\right)=\frac{\sin \frac\pi6}2=\frac{\frac12}2=\frac14$$