# Math Help - prove that

1. ## prove that

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

2. Use the Product-to-Sum Formula:

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

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

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

Also, note that

$\color{white}.\quad.$ $\cos{\frac{5\pi}{14}}\left(\cos{\frac{3\pi}{14}}+\ cos{\frac{\pi}{14}}\right)$

$=\ \cos{\frac{5\pi}{14}}\left(2\cos{\frac{\pi}{14}}\c os{\frac{2\pi}{14}}\right)$

$=\ \cos{\frac{\pi}{14}}\left(2\cos{\frac{5\pi}{14}}\c os{\frac{2\pi}{14}}\right)$

$=\ \cos{\frac{\pi}{14}}\left(\cos{\frac{3\pi}{14}}+\c os{\frac{\pi}{2}}\right)$

$=\ \cos{\frac{\pi}{14}}\cos{\frac{3\pi}{14}}$

i.e. $\cos{\frac{\pi}{14}}\cos{\frac{3\pi}{14}}-\cos{\frac{3\pi}{14}}\cos{\frac{5\pi}{14}}-\cos{\frac{5\pi}{14}}\cos{\frac{\pi}{14}}=0$.

So if you can prove that $\cos{\frac{\pi}{14}},\cos{\frac{3\pi}{14}},-\!\cos{\frac{5\pi}{14}}$ are roots of the equation $8x^3-4\sqrt{7}x^2+\sqrt{7}=0$, you’re done.

4. I got it! I got it!

Use the identity

$\cos{7\theta}\ =\ 64\cos^7{\theta}-112\cos^5{\theta}+56\cos^3{\theta}-7\cos{\theta}$

Substitute $\theta=\frac{\pi}{7},\frac{3\pi}{7},\frac{5\pi}{7}$ into that. Thus you’ll find that $\cos{\frac{\pi}{7}},\cos{\frac{3\pi}{7}},\cos{\fra c{5\pi}{7}}$ are the roots of the equation

$0\ =\ 64x^7-112x^5+56x^3-7x+1\ =\ (x+1)(8x^3-4x^2-4x+1)^2$

Obviously $\cos{\frac{\pi}{7}},\cos{\frac{3\pi}{7}},\cos{\fra c{5\pi}{7}}\ne-1$; therefore they are the roots of $8x^3-4x^2-4x+1=0$.

Now $\cos{\frac{\pi}{7}}=2\cos^2{\frac{\pi}{14}}-1$, etc. Hence the roots of the equation

$8(2x^2-1)^3-4(2x^2-1)^2-4(2x^2-1)+1=0$

are $\pm\cos{\frac{\pi}{14}},\pm\cos{\frac{3\pi}{14}},\ pm\cos{\frac{5\pi}{14}}$. All you need from the sextic equation are the leading coefficient, which is 64, and the constant term, which is $-7$. Now multiply all the six roots together and you have your answer!

Oh, I am so happy!

5. Consider the equation, $z^{13}+z^{12}+...+z+1 = 0$. The solutions are $\zeta, \zeta^2, .... , \zeta^{13}$ where $\zeta = \cos \frac{\pi }{14}+i\sin \frac{\pi}{14}$.
Thus, we can factorize, $1+z+z^2+...+z^{13} = \prod_{n=1}^{13}(z - \zeta^n)$.

Let $z=1$ and we get, $14 = \prod_{n=1}^{13} = (1 - \zeta^n)$. This implies, $14 = |14| = \left| \prod_{n=1}^{13} (1-\zeta^n) \right| = \prod_{n=1}^{13} \left| 1 - \zeta^n \right|$.

Note, $\left| 1 - \zeta^n\right|^2 = \left| 1 - \cos \frac{2\pi n}{14} - i\sin \frac{2\pi n}{14} \right|^2 = \left( 1-\cos \frac{2\pi n}{14} \right)^2 + \sin^2 \frac{2\pi n}{14} =$ $1 - 2\cos \frac{2\pi n}{14} + \cos^2 \frac{2\pi n}{14} + \sin^2 \frac{2\pi n}{14} = 2\left(1 - \cos \frac{2\pi n}{14}\right) = 4\sin ^2 \frac{\pi n}{14}$.
Which means, $|1-\zeta^n | = \sqrt{4\sin^2 \frac{\pi n}{14} } = 2\sin \frac{\pi n}{14}$.

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

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

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

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

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

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

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

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