But earlier algebra:

February 22, 2026

But earlier algebra:

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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 $.
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.

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