HMMT-Feb 2015 guts, Problem 20

What is the largest real number $\theta$ less than $\pi$ (i.e. $\theta<\pi$ ) such that \[ \prod_{k=0}^{10} \cos \left(2^{k} \theta\right) \neq 0 \] and \[ \prod_{k=0}^{10}\left(1+\frac{1}{\cos \left(2^{k} \theta\right)}\right)=1 ? \]

Solution: For equality to hold, note that $\theta$ cannot be an integer multiple of $\pi$ (or else $\sin =0$ and $\cos = \pm 1)$. Let $z=e^{i \theta / 2} \neq \pm 1$. Then in terms of complex numbers, we want \[ \prod_{k=0}^{10}\left(1+\frac{2}{z^{2^{k+1}}+z^{-2^{k+1}}}\right)=\prod_{k=0}^{10} \frac{\left(z^{2^{k}}+z^{-2^{k}}\right)^{2}}{z^{2^{k+1}}+z^{-2^{k+1}}} \] which partially telescopes to \[ \frac{z+z^{-1}}{z^{2^{11}}+z^{-2^{11}}} \prod_{k=0}^{10}\left(z^{2^{k}}+z^{-2^{k}}\right) \] Using a classical telescoping argument (or looking at binary representation; if you wish we may note that $z-z^{-1} \neq 0$, so the ultimate telescoping identity hold\&2), this simplifies to \[ \frac{z+z^{-1}}{z^{2^{11}}+z^{-2^{11}}} \frac{z^{2^{11}}-z^{-2^{11}}}{z-z^{-1}}=\frac{\tan \left(2^{10} \theta\right)}{\tan (\theta / 2)} \] Since $\tan x$ is injective modulo $\pi$ (i.e. $\pi$-periodic and injective on any given period), $\theta$ works if and only if $\frac{\theta}{2}+\ell \pi=1024 \theta$ for some integer $\ell$, so $\theta=\frac{2 \ell \pi}{2047}$. The largest value for $\ell$ such that $\theta<\pi$ is at $\ell=1023$, which gives $\theta=\frac{2046 \pi}{2047} 3$ Remark. It's also possible to do this without complex numbers, but it's less systematic. The steps are the same, though, first note that $1+\sec 2^{k} \theta=\frac{1+\cos 2^{k} \theta}{\cos 2^{k} \theta}=\frac{2 \cos ^{2} 2^{k-1} \theta}{\cos 2^{k} \theta}$ using the identity $\cos 2 x=$ $2 \cos ^{2} x-1$ (what does this correspond to in complex numbers?). hen we telescope using the identity $2 \cos x=\frac{\sin 2 x}{\sin x}$ (again, what does this correspond to in complex numbers?). $\fbox{\frac{2046 \pi}{2047}}$.

Answer: \frac{2046 \pi}{2047}