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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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 $\theta \in R$):-

Similar Questions

The minimum value of $5 \sin^2 \theta + 4 \cos^2 \theta$ is

The maximum value of $(\cos \alpha_1) \cdot (\cos \alpha_2) \ldots (\cos \alpha_n)$ under the constraints $0 \leq \alpha_1, \alpha_2, \ldots, \alpha_n \leq \frac{\pi}{2}$ and $(\cot \alpha_1) \cdot (\cot \alpha_2) \ldots (\cot \alpha_n) = 1$ is

If $A+B+C=2S$,then $\sin(2S-A)+\sin(2S-B)+\sin(2S-C)-\sin(2S) = $

Assertion $(A)$: If $A=10^{\circ}, B=16^{\circ}, C=19^{\circ}$,then $\tan 2A \tan 2B + \tan 2B \tan 2C + \tan 2C \tan 2A = 1$.
Reason $(R)$: If $A+B+C=90^{\circ}$,then $\tan A \tan B + \tan B \tan C + \tan C \tan A = 1$.
Which of the following is correct?

Let $x, y, z \in [0, 1]$. Then the maximum value of $\sqrt{|x-y|} + \sqrt{|y-z|} + \sqrt{|z-x|}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module