The minimum value of 2 sin^2 theta + 8 csc^2 | vedclass

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Question

(A) Let $f(	heta) = 2 \sin^2 	heta + 8 \csc^2 	heta$.Using the Arithmetic Mean-Geometric Mean $(AM \ge GM)$ inequality for positive terms:$\frac{2

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(A) Let $f(	heta) = 2 \sin^2 	heta + 8 \csc^2 	heta$.Using the Arithmetic Mean-Geometric Mean $(AM \ge GM)$ inequality for positive terms:$\frac{2 \sin^2 	heta + 8 \csc^2 	heta}{2} \ge \sqrt{2 \sin^2 	heta \cdot 8 \csc^2 	heta}$$\frac{2 \sin^2 	heta + 8 \csc^2 	heta}{2} \ge \sqrt{16 \sin^2 	heta \cdot \frac{1}{\sin^2 	heta}}$$\frac{2 \sin^2 	heta + 8 \csc^2 	heta}{2} \ge \sqrt{16}$$2 \sin^2 	heta + 8 \csc^2 	heta \ge 2 	imes 4$$2 \sin^2 	heta + 8 \csc^2 	heta \ge 8$.The equality holds when $2 \sin^2 	heta = 8 \csc^2 	heta$,which implies $\sin^4 	heta = 4$,which is impossible as $\sin^2 	heta \le 1$.Since $\sin^2 	heta \in (0, 1]$,let $x = \sin^2 	heta$. Then $f(x) = 2x + \frac{8}{x}$.For $x \in (0, 1]$,the function $f(x) = 2x + \frac{8}{x}$ is a decreasing function because $f'(x) = 2 - \frac{8}{x^2} Thus,the minimum value occurs at the largest possible value of $x$,which is $x = 1$.$f(1) = 2(1) + \frac{8}{1} = 10$.

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

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