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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.

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