$ \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}

Sat, 09 May 2026 20:38:39 +0000

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