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Home » MARVEL RIVALS Season 3: The Rivalry Blazes Hotter Than Ever—Everyone’s on Edge! - Dyverse

MARVEL RIVALS Season 3: The Rivalry Blazes Hotter Than Ever—Everyone’s on Edge! - Dyverse

📅 March 11, 2026 👤 scraface

Mar 11, 2026

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📌 Let $ s = \sin x + \cos x $, $ p = \sin x \cos x $. Then:

📌 s^2 = 1 + 2p \Rightarrow p = \frac{s^2 - 1}{2}

📌 We minimize this expression for $ s \in (\sqrt{2}, \sqrt{2}] $, since $ \sin x + \cos x = \sqrt{2} \sin\left(x + \frac{\pi}{4}\right) $, and maximum is $ \sqrt{2} $. Wait — maximum is $ \sqrt{2} $, minimum is 1 (approaching at endpoints), but in the open interval $ (0, \frac{\pi}{2}) $, $ s \in (1, \sqrt{2}) $.

📌 f(s) = \left( \frac{2s}{s^2 - 1} \right)^2

📌 We minimize $ f(s) $ on $ (1, \sqrt{2}) $.

📌 Let’s test $ s = \sqrt{2} $:

📌 f(\sqrt{2}) = \left( \frac{2\sqrt{2}}{2 - 1} \right)^2 = (2\sqrt{2})^2 = 8

📌 Now check derivative or behavior: as $ s \to 1^+ $, $ f(s) \to \infty $. So minimum occurs at $ s = \sqrt{2} $, i.e., $ x = \frac{\pi}{4} $.

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