Thus $ \phi = k\pi $. Now use $ \cos^2(\phi) = \frac12 $, but $ \cos(k\pi) = \pm 1 $, so \( \cos - AMAZONAWS

📅 March 6, 2026 👤 scraface

Mar 06, 2026

$Thus $ \phi = k\pi $. Now use $ \cos^2(\phi) = \frac{1}{2} $, but $ \cos(k\pi) = \pm 1 $, so \( \cos$