The maximum value of the expression frac{1}{s | vedclass

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(A) Let $f(	heta) = \sin^2 	heta + 3 \sin 	heta \cos 	heta + 5 \cos^2 	heta$. To find the maximum value of $\frac{1}{f(	heta)}$,we need to find

Vedclass · Accepted Answer

$2$

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(A) Let $f(	heta) = \sin^2 	heta + 3 \sin 	heta \cos 	heta + 5 \cos^2 	heta$. To find the maximum value of $\frac{1}{f(	heta)}$,we need to find the minimum value of $f(	heta)$. Using trigonometric identities: $f(	heta) = \frac{1 - \cos 2	heta}{2} + \frac{3}{2} \sin 2	heta + 5 \frac{1 + \cos 2	heta}{2}$ $f(	heta) = \frac{1}{2} - \frac{1}{2} \cos 2	heta + \frac{3}{2} \sin 2	heta + \frac{5}{2} + \frac{5}{2} \cos 2	heta$ $f(	heta) = 3 + 2 \cos 2	heta + \frac{3}{2} \sin 2	heta$. The expression $a \cos x + b \sin x$ ranges between $-\sqrt{a^2 + b^2}$ and $\sqrt{a^2 + b^2}$. Here,$2 \cos 2	heta + \frac{3}{2} \sin 2	heta$ ranges between $-\sqrt{2^2 + (\frac{3}{2})^2} = -\sqrt{4 + \frac{9}{4}} = -\sqrt{\frac{25}{4}} = -\frac{5}{2}$ and $\frac{5}{2}$. Thus,the minimum value of $f(	heta) = 3 - \frac{5}{2} = \frac{1}{2}$. The maximum value of $\frac{1}{f(	heta)}$ is $\frac{1}{1/2} = 2$.

The maximum value of the expression $\frac{1}{\sin^2 \theta + 3 \sin \theta \cos \theta + 5 \cos^2 \theta}$ is

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