1. ## prove that

$\displaystyle \cos(\pi/14)\cos(3\pi/14)\cos(5\pi/14)=\sqrt7/8$

2. Use the Product-to-Sum Formula:

$\displaystyle \cos(A)\cos(B) = \frac{1}{2}(\cos(A-B) + \cos(A+B))$.

Use it first on the first two, then use that result with the third.

3. Henderson’s suggestion will lead to $\displaystyle \cos{\frac{\pi}{14}}+\cos{\frac{3\pi}{14}}-\cos{\frac{5\pi}{14}}=\frac{\sqrt{7}}{2}$.

Also, note that

$\displaystyle \color{white}.\quad.$$\displaystyle \cos{\frac{5\pi}{14}}\left(\cos{\frac{3\pi}{14}}+\ cos{\frac{\pi}{14}}\right) \displaystyle =\ \cos{\frac{5\pi}{14}}\left(2\cos{\frac{\pi}{14}}\c os{\frac{2\pi}{14}}\right) \displaystyle =\ \cos{\frac{\pi}{14}}\left(2\cos{\frac{5\pi}{14}}\c os{\frac{2\pi}{14}}\right) \displaystyle =\ \cos{\frac{\pi}{14}}\left(\cos{\frac{3\pi}{14}}+\c os{\frac{\pi}{2}}\right) \displaystyle =\ \cos{\frac{\pi}{14}}\cos{\frac{3\pi}{14}} i.e. \displaystyle \cos{\frac{\pi}{14}}\cos{\frac{3\pi}{14}}-\cos{\frac{3\pi}{14}}\cos{\frac{5\pi}{14}}-\cos{\frac{5\pi}{14}}\cos{\frac{\pi}{14}}=0. So if you can prove that \displaystyle \cos{\frac{\pi}{14}},\cos{\frac{3\pi}{14}},-\!\cos{\frac{5\pi}{14}} are roots of the equation \displaystyle 8x^3-4\sqrt{7}x^2+\sqrt{7}=0, you’re done. 4. I got it! I got it! Use the identity \displaystyle \cos{7\theta}\ =\ 64\cos^7{\theta}-112\cos^5{\theta}+56\cos^3{\theta}-7\cos{\theta} Substitute \displaystyle \theta=\frac{\pi}{7},\frac{3\pi}{7},\frac{5\pi}{7} into that. Thus you’ll find that \displaystyle \cos{\frac{\pi}{7}},\cos{\frac{3\pi}{7}},\cos{\fra c{5\pi}{7}} are the roots of the equation \displaystyle 0\ =\ 64x^7-112x^5+56x^3-7x+1\ =\ (x+1)(8x^3-4x^2-4x+1)^2 Obviously \displaystyle \cos{\frac{\pi}{7}},\cos{\frac{3\pi}{7}},\cos{\fra c{5\pi}{7}}\ne-1; therefore they are the roots of \displaystyle 8x^3-4x^2-4x+1=0. Now \displaystyle \cos{\frac{\pi}{7}}=2\cos^2{\frac{\pi}{14}}-1, etc. Hence the roots of the equation \displaystyle 8(2x^2-1)^3-4(2x^2-1)^2-4(2x^2-1)+1=0 are \displaystyle \pm\cos{\frac{\pi}{14}},\pm\cos{\frac{3\pi}{14}},\ pm\cos{\frac{5\pi}{14}}. All you need from the sextic equation are the leading coefficient, which is 64, and the constant term, which is \displaystyle -7. Now multiply all the six roots together and you have your answer! Oh, I am so happy! 5. Consider the equation, \displaystyle z^{13}+z^{12}+...+z+1 = 0. The solutions are \displaystyle \zeta, \zeta^2, .... , \zeta^{13} where \displaystyle \zeta = \cos \frac{\pi }{14}+i\sin \frac{\pi}{14}. Thus, we can factorize, \displaystyle 1+z+z^2+...+z^{13} = \prod_{n=1}^{13}(z - \zeta^n). Let \displaystyle z=1 and we get, \displaystyle 14 = \prod_{n=1}^{13} = (1 - \zeta^n). This implies, \displaystyle 14 = |14| = \left| \prod_{n=1}^{13} (1-\zeta^n) \right| = \prod_{n=1}^{13} \left| 1 - \zeta^n \right|. Note, \displaystyle \left| 1 - \zeta^n\right|^2 = \left| 1 - \cos \frac{2\pi n}{14} - i\sin \frac{2\pi n}{14} \right|^2 = \left( 1-\cos \frac{2\pi n}{14} \right)^2 + \sin^2 \frac{2\pi n}{14} =$$\displaystyle 1 - 2\cos \frac{2\pi n}{14} + \cos^2 \frac{2\pi n}{14} + \sin^2 \frac{2\pi n}{14} = 2\left(1 - \cos \frac{2\pi n}{14}\right) = 4\sin ^2 \frac{\pi n}{14}$.
Which means, $\displaystyle |1-\zeta^n | = \sqrt{4\sin^2 \frac{\pi n}{14} } = 2\sin \frac{\pi n}{14}$.

Thus, we established that, $\displaystyle 14 = \prod_{n=1}^{13} 2\sin \frac{\pi n}{14} \implies \prod_{n=1}^{13}\sin \frac{\pi n}{14} = \frac{14}{2^{13}}$.

In the product, $\displaystyle \sin \frac{\pi }{14} \cdot ... \cdot \sin \frac{5\pi}{14}\cdot\sin \frac{6\pi }{14}\cdot \sin \frac{7\pi}{14}\cdot ... \cdot \sin \frac{13\pi}{14} = \frac{14}{2^{13}}$
Look at the sines from $\displaystyle 8\pi/14,...,13\pi/14$.
Use the fact that $\displaystyle \sin \frac{8\pi}{14} = \sin \left( \pi - \frac{8\pi}{14} \right) = \sin \frac{6\pi}{14}$, $\displaystyle \sin \frac{9\pi}{14} = \sin \left( \pi - \frac{9\pi}{14} \right) = \sin \frac{5\pi}{14}$, ... , $\displaystyle \sin \frac{13\pi}{14} = \sin \frac{\pi}{14}$.

This means, we can write this product as,
$\displaystyle \sin^2 \frac{\pi}{14}\cdot \sin^2 \frac{2\pi}{14} \cdot ... \cdot \sin ^2\frac{6\pi}{14} \sin \frac{7\pi}{14} = \frac{14}{2^{13}}=\frac{7}{2^{12}}$.

This means, (note $\displaystyle \sin \frac{7\pi}{14} = 1$),
$\displaystyle \sin \frac{\pi}{14} \sin \frac{2\pi}{14} \cdot ... \cdot \sin \frac{6\pi}{14} = \frac{\sqrt{7}}{2^6}$.

Now, use fact that, $\displaystyle \sin A\sin B = \frac{1}{2}\left( \cos (A-B) - \cos (A+B) \right)$.

Apply the above identity to the opposite ends and work inside-out.
$\displaystyle \sin \frac{\pi}{14} \sin \frac{6\pi}{14} = \frac{1}{2}\cos \frac{5\pi}{14}$
$\displaystyle \sin \frac{2\pi}{14} \sin\frac{5\pi}{14} = \frac{1}{2}\cos \frac{3\pi}{14}$
$\displaystyle \sin \frac{3\pi}{14}\sin \frac{4\pi}{13} = \frac{1}{2}\cos \frac{\pi}{14}$.

Thus,
$\displaystyle \cos \frac{\pi}{14} \cos \frac{3\pi}{14} \cos \frac{5\pi}{14} = \frac{\sqrt{7}}{2^3}$.

Generalize! If $\displaystyle n$ is odd then: $\displaystyle \boxed{ \cos \frac{\pi}{2n} \cos \frac{3\pi}{2n} \cdot ... \cdot \cos \frac{(n-2)\pi}{2n} = \frac{\sqrt{n}}{2^{(n-1)/2}} }$.