TITLE: Combinatorics and trigonometry identity
QUESTION [5 upvotes]: Prove the following:
$\displaystyle\prod_{n=1}^{180}\left(\cos{\left(\dfrac{n\pi}{180}\right)}+2\right)=\displaystyle\sum_{n=0}^{89}\binom{180}{2n+1}\left(\dfrac{3}{4}\right)^n$.
We can state this as $\displaystyle \binom{180}{1}\left(\dfrac{3}{4}\right)^0+ \binom{180}{3}\left(\dfrac{3}{4}\right)^1+\cdots+\binom{180}{179}\left(\dfrac{3}{4}\right)^{89} = \left(\cos\left(\dfrac{\pi}{180}\right)+2\right)\left(\cos\left(\dfrac{2\pi}{180}\right)+2\right) \cdots \left(\cos\left(\dfrac{180\pi}{180}\right)+2\right)$ 
The problem I am encountering is that there is no general result we can prove about this for $n$ it seems. I tried using induction on $n$ to prove this but it fails for the base case. Also, combinations and cosines seem unrelated, so how could I prove the equation is true?

REPLY [3 votes]: First rewrite $\cos(x)$ as $\cos(x) = \dfrac{e^{ix}+e^{-ix}}{2}$. Then we also have $\dfrac{1}{2}(4+e^{ix}+e^{-ix}) = \cos(x) + 2$. We can factorize the left-hand side of the last equality as $\dfrac{1}{2}(4+e^{ix}+e^{-ix}) = \dfrac{1}{2}\left(a+\dfrac{e^{ix}}{a}\right)\left(a+\dfrac{e^{-ix}}{a} \right)$ where $a^2+\dfrac{1}{a^2} = 4$. This gives four solutions for $a$, so I will pick the positive one, which is $\dfrac{1+\sqrt{3}}{\sqrt{2}}$. Now notice that this allows us to evaluate the left-hand side of $(1)$. We have that $\dfrac{1}{2a^2}(a^2+e^{ix})(a^2+e^{-ix}) = \cos(x) + 2$. Then make the substitution $x = \dfrac{n\pi}{180}$ to get that $$\dfrac{1}{2a^2}(a^2+e^{\frac{in\pi}{180}})(a^2+e^{-\frac{in\pi}{180}}) = \cos \left ( \dfrac{n\pi}{180} \right) + 2,$$ which we can use to rewrite the left-hand side of $(1)$. Thus, $$\displaystyle \prod_{n=1}^{180} \left (\cos \left( \dfrac{n\pi}{180} \right) + 2 \right) = \prod_{n=1}^{180} \left(\dfrac{1}{2a^2}(a^2+e^{\frac{in\pi}{180}})(a^2+e^{-\frac{in\pi}{180}}) \right).$$ Then see that the numbers $e^{\frac{in\pi}{180}}$ for $n = -180,-179,\ldots,-1,0,1,\ldots,179$ are the roots of the equation $x^{360} = 1$. Now, we see that $$\prod_{n=1}^{180} \left(\dfrac{1}{2a^2}(a^2+e^{\frac{in\pi}{180}})(a^2+e^{-\frac{in\pi}{180}}) \right) = \dfrac{1}{2^{180}a^{360}}\dfrac{(a^{720}-1)(a^2-1)}{a^{2}+1}.$$ To calculate the RHS, we see that $\displaystyle \sum_{n = 0}^{89} \binom{180}{2n+1} \left( \dfrac{3}{4} \right)^n = \dfrac{\left(1+\dfrac{\sqrt{3}}{2}\right)^{180} - \left(1-\dfrac{\sqrt{3}}{2}\right)^{180}}{\sqrt{3}}.$ It remains to show that $$\dfrac{1}{2^{180}a^{360}}\dfrac{(a^{720}-1)(a^2-1)}{a^{2}+1} = \dfrac{\left(1+\dfrac{\sqrt{3}}{2}\right)^{180} - \left(1-\dfrac{\sqrt{3}}{2}\right)^{180}}{\sqrt{3}}.$$ To see this, first observe that $1-\dfrac{\sqrt{3}}{2} = \dfrac{1}{2a^2}$, $a^2 = 1+\dfrac{\sqrt{3}}{2}$, and $$\dfrac{1}{2^{180}a^{360}}\dfrac{(a^{720}-1)(a^2-1)}{a^{2}+1}=\left[\left(\dfrac{a^2}2\right)^{180}-\left(\dfrac1{2a^2}\right)^{180}\right]\cdot\dfrac{a^2-1}{a^2+1}.$$ Then finally we see that $\dfrac{a^2-1}{a^2+1} = \dfrac{1}{\sqrt{3}}$, proving the desired result.