More fun in 2015, Part 44

The solution I had in mind is essentially the same as Aareyan's, with some small variants.

Lemma: $\Phi_{2015}(1)=\prod(1-\omega)=1$ , where the product is taken over all primitive 2015th roots of unity $\omega$ .

Proof of lemma: Note that $\Phi_p(1)=p$ for prime $p$ . Now we show by induction that $\Phi_n(1)=1$ for square-free $n=p_1*...*p_m$ with $m>1$ . We write $x^n-1=\prod_{d|n}\Phi_d(x)$ , divide by $x-1$ , and evaluate at $x=1$ to see that $n=p_1*...*p_m*\Phi_n(1)$ so $\Phi_n(1)=1$ and, in particular, $\Phi_{2015}(1)=\boxed{1}$ .

Thus we have $\prod(1-\omega)=1$ , and taking conjugates, $\prod(1-\bar{\omega)}=1$ so $1=\prod(1-\omega)(1-\bar{\omega)}$ $=\prod_k2(1-\cos(2k\pi/2015))=2^{\phi(2015)}\prod_k(1-\cos(2\pi k/2015$ )), where the products are taken over all $1\leq k\leq 2015$ with $\gcd(1,2015)=1$ . Taking logarithms to base 2, we find that $S=\phi(2015)=\boxed{1440}$

write the sum as $-\log_2\left(\prod\left(1-\cos\left(\dfrac{2k\pi}{2015}\right)\right)\right)$ let w be the primitive 2015th roots of unity. then by eulers identity this becomes $-\log_2\left(\prod\left(1-\frac{w+w^{-1}}{2}\right)\right)=-\log_2\left(\prod\left(w^{-1}\dfrac{(1-w)^2}{2}\right)\right)$ product of all the primitive roots for all number≠2 are 1. neglect that.

a polynomial p can be written as $p(x)=\prod (x-roots)$ . what polynomial has roots w? the cyclotomic polynomial . we are just searching for $-\log_2\left(\dfrac{\Phi_{2015}(1)}{2^{1440}}\right)=1440-\log_2\left(\Phi_{2015}(1)\right)$ the formula for $\Phi_n(1)=e^{\Lambda(n)}$ for n≠1, and $\Lambda(n)$ being the von magoldt function .proof in the comments. so, $1440-\log_2\left(e^{2\Lambda(2015)}\right)=1440-\log_2\left(1\right)=\boxed{1440}$