Back to the Roots

This is my overkill solution, I hope you all guys have a shorter one. Let $w=e^{2\pi i/n}$ . Now, with the identity $\sin^2 \theta=\frac{1-\cos 2\theta}{2}$ the product $P$ is:

$P=\displaystyle \prod_{k=1}^{(n+1)/2}\left(5-4\cos\left(\dfrac{(2k-1) \pi}{n}\right)\right)$

We can evaluate the last term of the product, which is just $9$ , and also with the identity $\cos(\pi-\theta)=-\cos\theta$ , we can work with the product backwards to make it simpler:

$P=9\displaystyle \prod_{k=1}^{(n-1)/2} \left(5+4\cos\left(\dfrac{2\pi k}{n}\right)\right)$

Now, with the identity $\cos(2\pi-\theta)=\cos\theta$ , we can say that:

$P=9\sqrt{\displaystyle \prod_{k=1}^{n-1} \left(5+4\cos\left(\dfrac{2\pi k}{n}\right)\right)}$

The next step is consider the polynomial $P(x)=\dfrac{x^n-1}{x-1}$ where $x\neq 1$ . It clearly contains the roots $w,w^2,\ldots,w^{n-1}$ , and since $n$ is odd, we can factorize it like this: $P(x)=(w-x)(w^2-x)\cdots(w^{n-1}-x)$ .

Next, we know that $w^k+w^{-k}=2\cos\left(\dfrac{2\pi k}{n}\right)$ , so the product is:

$P=9\sqrt{\displaystyle \prod_{k=1}^{n-1} \left(5+2w+2w^{-1}\right)}=9\sqrt{\displaystyle \prod_{k=1}^{n-1} \dfrac{2(w^k+1/2)(w^k+2)}{w^k}}$

Note that we can distribute the product, so it becomes to:

$P=9\sqrt{2^{n-1} \displaystyle \prod_{k=1}^{n-1} (w^k+1/2) \displaystyle \prod_{k=1}^{n-1} (w^k+2)}=9\sqrt{2^{n-1} \cdot P(-1/2)\cdot P(-2)}$

Finally:

$P=9\sqrt{2^{n-1}\cdot\dfrac{(-1/2)^n-1}{-1/2-1}\cdot\dfrac{(-2)^n-1}{-2-1}}$

Since $n$ is odd, we can simplify the powers:

$P=9\sqrt{2^{n-1}\cdot\dfrac{2^{-n}+1}{3/2}\cdot\dfrac{2^n+1}{3}}=9\sqrt{\dfrac{2^n(2+2^{-n}+2^n)}{9}}$

$P=3\sqrt{2^{2n}+2^{n+1}+1}=3\sqrt{(2^n+1)^2}$

$P=3(2^n+1)=3\cdot 2^n+3$

On comparing we get $a=3,b=2,c=3$ , hence $a+b+c=\boxed{8}$ .

Like Alan, I use $\sin^2{t}=\frac{1-\cos2t}{2}$ and I evaluate the last term to find $P=9\prod_{k=1}^{(n-1)/2}\left(5-4\cos\left(\frac{(2k-1)\pi}{n}\right)\right)$ Now I consider the $n$ th roots of $-1$ , which are the odd powers of $e^{\pi{i}/n}$ , to find $x^n+1$ $=(x+1)\prod_{k=1}^{(n-1)/2}(x-e^{(2k-1)\pi{i}/n})(x-e^{-(2k-1)\pi{i}/n})=(x+1)\prod_{k=1}^{(n-1)/2}\left(x^2+1-2x\cos\left(\frac{(2k-1)\pi}{n}\right)\right)$ Plugging in $x=2$ gives $2^n+1=3\prod_{k=1}^{(n-1)/2}\left(5-4\cos\left(\frac{(2k-1)\pi}{n}\right)\right)$ so $P=9\left(\frac{2^n+1}{3}\right)=3(2^n+1)=3*2^n+3$ with $a+b+c=3+2+3=\boxed{8}$