# Evaluating $(1+\cos\frac{\pi}{8})(1+\cos\frac{3\pi}{8}) (1+\cos\frac{5\pi}{8})(1+\cos\frac{7\pi}{8})$

How to find the value of $$\left(1+\cos\frac{\pi}{8}\right)\left(1+\cos\frac{3\pi}{8}\right) \left(1+\cos\frac{5\pi}{8}\right)\left(1+\cos\frac{7\pi}{8}\right)$$

I used: $1+ \cos \theta=2\sin^2(\theta/2)$ to get $$2^4 \cdot\prod_{n=0}^3 \sin\left(\frac{(2n+1)\pi}{16}\right)$$

After that I tried using the formula: $$\sin\left(1 \frac{\pi}n\right)\sin\left(2 \frac{\pi}n\right) \cdots \sin\left((n-1) \frac{\pi}n\right)= \frac n {2^{n-1}}$$

to find the value of the expression in the parenthesis in the first equation, but currently I'm stuck. I don't think this method will lead me to the correct answer.

Any help would be appreciated!

## 5 Answers

First notice that $\dfrac{7\pi }{8}=\pi-\dfrac{\pi}{8}$ and also $\dfrac{5\pi }{8}=\pi-\dfrac{3\pi}{8}$. Substituting these values in the original expression, you'll get:

$$\left(1+\cos {\pi \over 8}\right)\left(1+\cos {3\pi \over 8}\right)\left(1-\cos {3\pi \over 8}\right)\left(1-\cos {\pi \over 8}\right)$$

Using the obvious algebraic identity, you get $$\left(1-\cos^2 {\pi \over 8}\right)\left(1-\cos^2 {3\pi \over 8}\right)$$ Which is nothing but:

$$\sin^2{\pi \over 8}.sin^2{3\pi \over 8}$$ Now finally notice that $\dfrac{3\pi}{8}=\dfrac{\pi}{2}-\dfrac{\pi}{8}$ and use the identity $\sin2\theta=2\sin\theta \cos\theta$ to get the simple form $$\frac{1}{4}\sin^2{\pi \over 4}$$

Here is a solution using complex numbers: Let $\omega = \exp(i\pi/2n)$. Then

\begin{align*} \prod_{k=1}^{n} \left(1 + \cos \left( \tfrac{2k-1}{2n}\pi \right) \right) &= \frac{1}{2^{n}} \prod_{k=1}^{n} \left(1 + 2\Re(\omega^{2k-1})+ 1 \right) \\ &= \frac{1}{2^{n}} \prod_{k=1}^{n} (1 + \omega^{2k-1})(1 + \omega^{-(2k-1)}). \end{align*}

But notice that $\omega^{-(2n-1)}, \cdots, \omega^{2n-1}$ are $2n$ distinct zeros of the polynomial $z^{2n} + 1$. Consequently we have

\begin{align*} \prod_{k=1}^{n} \left(1 + \cos \left( \tfrac{2k-1}{2n}\pi \right) \right) &= \frac{1}{2^{n}} \prod_{k=1}^{n} (-1 - \omega^{2k-1})(-1 - \omega^{-(2k-1)}) \\ &= \frac{1}{2^{n}} (z^{2n}+1)|_{z=-1} \\ &= \frac{1}{2^{n-1}}. \end{align*}

Hint: $1+\cos \theta = 2\cos^2\frac{\theta}{2}$, and use:

$\cos \frac{\theta}{2} = \dfrac{\sin \theta}{2\sin \frac{\theta}{2}}$, or

try to create a quartic equation and use Viete's formula for the product of zeroes.

Use the linearisation formulae. The expression can be rewritten as: $$2^2\Bigl(2\sin\frac{\pi}{16}\,\sin\frac{7\pi}{16}\Bigr)\Bigl(2\sin\frac{3\pi}{16}\,\sin\frac{5\pi}{16}\Bigr)=2^2\cos\frac{3\pi}{8}\cos\frac{\pi}{8}=2\cos\frac{\pi}{4}=\sqrt2.$$

If $\cos4x=0,4x=(2n+1)\dfrac\pi2$ where $n$ is any integer

$x=(2n+1)\dfrac\pi8$ where $n\equiv0,1,2,3\pmod4$

Now $\cos4x=2\cos^22x-1=\cdots=8\cos^4x-8\cos^2x+1$

So, the roots of $8c^4-8c^2+1=0$ are $\cos(2n+1)\dfrac\pi8$ where $n\equiv0,1,2,3\pmod4$

So, the equation whose roots are $1+\cos(2n+1)\dfrac\pi8=y\iff\cos(2n+1)\dfrac\pi8=y-1$ is

$8(y-1)^4-8(y-1)^2+1=0\iff8y^4+\cdots+1=0$

Using Vieta's formula, $$\prod_{n=0}^3\left[1+\cos(2n+1)\dfrac\pi8\right]=\dfrac18$$