Minimum value of 2 sin^2 theta + 8 csc^2 thet | vedclass

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(A) Let $f(	heta) = 2 \sin^2 	heta + 8 \csc^2 	heta$.Since $\sin^2 	heta$ can take values in the interval $(0, 1]$,let $x = \sin^2 	heta$,where $

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(A) Let $f(	heta) = 2 \sin^2 	heta + 8 \csc^2 	heta$.Since $\sin^2 	heta$ can take values in the interval $(0, 1]$,let $x = \sin^2 	heta$,where $x \in (0, 1]$.Then the expression becomes $f(x) = 2x + \frac{8}{x}$.To find the minimum value,we analyze the function $f(x) = 2x + \frac{8}{x}$ for $x \in (0, 1]$.The derivative is $f'(x) = 2 - \frac{8}{x^2}$.Setting $f'(x) = 0$ gives $x^2 = 4$,so $x = 2$ or $x = -2$. Neither of these values lies in the interval $(0, 1]$.Since $f'(x) Therefore,the minimum value occurs at the largest possible value of $x$,which is $x = 1$.Substituting $x = 1$ into the expression: $f(1) = 2(1) + \frac{8}{1} = 2 + 8 = 10$.Thus,the minimum value is $10$.

Minimum value of $2 \sin^2 \theta + 8 \csc^2 \theta$ is (where $\theta \in R$):-

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