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chateau de la caze lobby - Decision Point

📅 May 4, 2026 👤 scraface

May 04, 2026

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📌 But earlier we saw $ f(x) = 1 - rac{1}{2}(\cos 2x - \cos 4x) $. Letâs test $ x = rac{\pi}{2} $:

📌 $ \sin^2 rac{\pi}{2} = 1 $, $ \cos^2(2 \cdot rac{\pi}{2}) = \cos^2 \pi = (-1)^2 = 1 $, so $ f = 1 + 1 = 2 $? No â wait:

📌 Wait: $ \cos(2x) = \cos \pi = -1 $, $ \cos^2(2x) = 1 $, $ \sin^2 x = 1 $, so $ f(x) = 1 + 1 = 2 $? But that canât be â maximum of each is 1, but sum could be 2? But letâs compute $ f(x) = \sin^2 x + \cos^2(2x) \leq 1 + 1 = 2 $, but is $ f(x) = 2 $ possible? Only if $ \sin^2 x = 1 $ and $ \cos^2(2x) = 1 $.

📌 $ \sin^2 x = 1 \Rightarrow x = rac{\pi}{2} + k\pi $. At $ x = rac{\pi}{2} $, $ 2x = \pi $, $ \cos \pi = -1 $, $ \cos^2 \pi = 1 $. So yes: $ f\left(rac{\pi}{2}

📌 ight) = 1 + 1 = 2 $.

📌 But earlier algebra:

📌 f(x) = 1 - rac{1}{2}(\cos 2x - \cos 4x)

📌 Let $ x = 0 $: $ \sin^2 0 = 0 $, $ \cos^2 0 = 1 $, $ f(0) = 1 $. Formula: $ 1 - rac{1}{2}(1 - 1) = 1 $. Correct.

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