The minimum value of f(x) = a^{2} cos^{2} x + | vedclass

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(B) Given the function $f(x) = a^{2} \cos^{2} x + b^{2} \sin^{2} x$. Using the trigonometric identities $\cos^{2} x = \frac{1 + \cos 2x}{2}$ and $\sin

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$b^{2}$

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(B) Given the function $f(x) = a^{2} \cos^{2} x + b^{2} \sin^{2} x$. Using the trigonometric identities $\cos^{2} x = \frac{1 + \cos 2x}{2}$ and $\sin^{2} x = \frac{1 - \cos 2x}{2}$,we get: $f(x) = a^{2} \left( \frac{1 + \cos 2x}{2} ight) + b^{2} \left( \frac{1 - \cos 2x}{2} ight)$ $f(x) = \frac{a^{2} + a^{2} \cos 2x + b^{2} - b^{2} \cos 2x}{2}$ $f(x) = \frac{a^{2} + b^{2}}{2} + \left( \frac{a^{2} - b^{2}}{2} ight) \cos 2x$. Since $-1 \leq \cos 2x \leq 1$,the function $f(x)$ attains its minimum value when $\cos 2x = -1$ (given $a^{2} > b^{2}$,so $\frac{a^{2} - b^{2}}{2} > 0$). Substituting $\cos 2x = -1$: $f_{	ext{min}} = \frac{a^{2} + b^{2}}{2} + \left( \frac{a^{2} - b^{2}}{2} ight) (-1)$ $f_{	ext{min}} = \frac{a^{2} + b^{2} - a^{2} + b^{2}}{2} = \frac{2b^{2}}{2} = b^{2}$. Thus,the minimum value is $b^{2}$.

The minimum value of $f(x) = a^{2} \cos^{2} x + b^{2} \sin^{2} x$ if $a^{2} > b^{2}$,is

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