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Home » The Ultimate Guide to Finding Treasure-D Lavish Vintage Dresses You Need! - American Beagle Club

The Ultimate Guide to Finding Treasure-D Lavish Vintage Dresses You Need! - American Beagle Club

📅 February 27, 2026 👤 scraface

Feb 27, 2026

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📌 $ \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.

📌 $ x = rac{\pi}{2} $: $ \cos 2x = \cos \pi = -1 $, $ \cos 4x = \cos 2\pi = 1 $, so $ f(x) = 1 - rac{1}{2}(-1 - 1) = 1 - rac{1}{2}(-2) = 1 + 1 = 2 $. Correct.

📌 But earlier we said $ f(x) = 1 - rac{1}{2}(\cos 2x - \cos 4x) $. When $ \cos 2x = -1 $, $ \cos 4x = 1 $, so $ -1 - 1 = -2 $, multiplied by $ -rac{1}{2} $ gives $ +1 $, so $ 1 + 1 = 2 $. Yes.

📌 But we want **minimum**, not maximum. Can $ f(x) $ be less than 1?

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