The value of $\cos^{4} x + \sin^{4} x - 6 \cos^{2} x \sin^{2} x$ is
(a) cos 2x
(b) sin 2x
(c) cos 4x
(d) none of these

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Solution

(c) cos 4x

$\cos^{4} x + \sin^{4} x - 6 \cos^{2} x \sin^{2} x = \cos^{4} x + \sin^{4} x - 2 \cos^{2} x \sin^{2} x - 4 \cos^{2} x \sin^{2} x = {(\cos^{2} x - \sin^{2} x)}^{2} - {(2 \sin x \cos x)}^{2} = \cos^{2} 2 x - \sin^{2} 2 x = \cos 4 x$

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The minimum value of $f (x) = s i n^{4} x + c o s^{4} x, 0 \leq x \leq \frac{π}{2}$ is

\int \frac{d x}{s i n^{4} x + c o s^{4} x}

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∣ ∣ ∣ s i n x + c o s x + \frac{c o s x - s i n x}{c o s^{4} x - s i n^{4} x} ∣ ∣ ∣

sin x - cos x = 0

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