# How do you prove cos (pi/7) cos ((2pi)/7) cos ((3pi)/7) = 1/8?

Mar 31, 2016

Since this is a proof, we do not have a short answer. See the explanation below.

#### Explanation:

First you should prove the following:

$\setminus \cos \left(2 \setminus \frac{\pi}{7}\right) + \setminus \cos \left(4 \setminus \frac{\pi}{7}\right) + \setminus \cos \left(6 \setminus \frac{\pi}{7}\right) = - \frac{1}{2}$

To do that, draw a regular heptagon ABCDEFG with unit sides. Think of the sides as a vector, from A to B, then from B to C, etc. The vectors form a cycle, so from the "head to tail" rule the resultant is zero. At the same time we can define an "x-axis" from A to B and work out the components of each side's vector along that axis. Add them up and the sum must match the zero resultant:

$1 + \setminus \cos \left(2 \setminus \frac{\pi}{7}\right) + \setminus \cos \left(4 \setminus \frac{\pi}{7}\right) + \setminus \cos \left(6 \setminus \frac{\pi}{7}\right) + \setminus \cos \left(8 \setminus \frac{\pi}{7}\right) + \setminus \cos \left(10 \setminus \frac{\pi}{7}\right) + \setminus \cos \left(12 \setminus \frac{\pi}{7}\right) = 0$

Since $\setminus \cos \left(2 \setminus \pi - x\right) = \setminus \cos \left(x\right)$ the sum above becomes:

$1 + 2 \setminus \cos \left(2 \setminus \frac{\pi}{7}\right) + 2 \setminus \cos \left(4 \setminus \frac{\pi}{7}\right) + 2 \setminus \cos \left(6 \setminus \frac{\pi}{7}\right) = 0$

From that we then have the required result

$\setminus \cos \left(2 \setminus \frac{\pi}{7}\right) + \setminus \cos \left(4 \setminus \frac{\pi}{7}\right) + \setminus \cos \left(6 \setminus \frac{\pi}{7}\right) = - \frac{1}{2}$

With that result proven, we are now ready to tackle the given product. First use the sum-product relation for cosines

$\setminus \cos \left(a\right) \setminus \cos \left(b\right) = \left(\frac{1}{2}\right) \left(\setminus \cos \left(a + b\right) + \setminus \cos \left(a - b\right)\right)$

to get

$\setminus \cos \left(\setminus \frac{\pi}{7}\right) \setminus \cos \left(2 \setminus \frac{\pi}{7}\right) = \left(\frac{1}{2}\right) \left(\setminus \cos \left(3 \setminus \frac{\pi}{7}\right) + \setminus \cos \left(- \setminus \frac{\pi}{7}\right)\right)$

Then:

\cos(\pi/7)\cos(2\pi/7)\cos(3\pi/7) =(1/2)(\cos(3\pi/7)(\cos(3\pi/7)+(1/2)(\cos(-\pi/7)(\cos(3\pi/7))

Apply the sum-product relation once more to each of the above terms to get

$\left(\frac{1}{4}\right) \left(\setminus \cos \left(6 \setminus \frac{\pi}{7}\right) + \setminus \cos \left(0\right) + \setminus \cos \left(2 \setminus \frac{\pi}{7}\right) + \setminus \cos \left(\left(- 4\right) \setminus \frac{\pi}{7}\right)\right)$

Since

$\setminus \cos \left(0\right) = 1$ and $\setminus \cos \left(- 4 \setminus \frac{\pi}{7}\right) = \setminus \cos \left(4 \setminus \frac{\pi}{7}\right)$
this can be rearranged to

$\left(\frac{1}{4}\right) \left(1 + \setminus \cos \left(2 \setminus \frac{\pi}{7}\right) + \setminus \cos \left(4 \setminus \frac{\pi}{7}\right) + \setminus \cos \left(6 \setminus \frac{\pi}{7}\right)\right) = \left(\frac{1}{4}\right) \left(1 - \frac{1}{2}\right) = \frac{1}{8}$

Oct 1, 2016

we shall use the formula

$2 \sin \theta \cos \theta = \sin 2 \theta$

$L H S = \cos \left(\frac{\pi}{7}\right) \cos \left(\frac{2 \pi}{7}\right) \cos \left(\frac{3 \pi}{7}\right)$

$= \frac{4}{8 \sin \left(\frac{\pi}{7}\right)} \times 2 \sin \left(\frac{\pi}{7}\right) \cos \left(\frac{\pi}{7}\right) \cos \left(\frac{2 \pi}{7}\right) \cos \left(\frac{3 \pi}{7}\right)$

$= \frac{2}{8 \sin \left(\frac{\pi}{7}\right)} \times 2 \sin \left(\frac{2 \pi}{7}\right) \cos \left(\frac{2 \pi}{7}\right) \cos \left(\frac{3 \pi}{7}\right)$

$= \frac{1}{8 \sin \left(\frac{\pi}{7}\right)} \times 2 \sin \left(\frac{4 \pi}{7}\right) \cos \left(\frac{3 \pi}{7}\right)$

$= \frac{1}{8 \sin \left(\frac{\pi}{7}\right)} \times 2 \sin \left(\pi - \frac{3 \pi}{7}\right) \cos \left(\frac{3 \pi}{7}\right)$

$= \frac{1}{8 \sin \left(\frac{\pi}{7}\right)} \times 2 \sin \left(\frac{3 \pi}{7}\right) \cos \left(\frac{3 \pi}{7}\right)$

$= \frac{1}{8 \sin \left(\frac{\pi}{7}\right)} \times \sin \left(\frac{6 \pi}{7}\right)$

$= \frac{1}{8 \sin \left(\frac{\pi}{7}\right)} \times \sin \left(\pi - \frac{6 \pi}{7}\right)$

$= \frac{1}{8 \sin \left(\frac{\pi}{7}\right)} \times \sin \left(\frac{\pi}{7}\right)$

$= \frac{1}{8}$

proved