The minimum value of cos theta + sin theta + | vedclass

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

Question

(A) Let $f(	heta) = \cos 	heta + \sin 	heta + \frac{2}{\sin 2 	heta}$.Let $x = \sin 	heta + \cos 	heta$. Then $x^2 = 1 + \sin 2 	heta$,so $\sin

Vedclass · Accepted Answer

$2 + \sqrt{2}$

Vedclass · Answer

(A) Let $f(	heta) = \cos 	heta + \sin 	heta + \frac{2}{\sin 2 	heta}$.Let $x = \sin 	heta + \cos 	heta$. Then $x^2 = 1 + \sin 2 	heta$,so $\sin 2 	heta = x^2 - 1$.Since $	heta \in (0, \pi / 2)$,$x = \sqrt{2} \sin(	heta + \pi / 4) \in (1, \sqrt{2}]$.The expression becomes $f(x) = x + \frac{2}{x^2 - 1}$.To find the minimum,we differentiate $f(x)$ with respect to $x$:$f'(x) = 1 - \frac{2(2x)}{(x^2 - 1)^2} = 1 - \frac{4x}{(x^2 - 1)^2}$.Setting $f'(x) = 0$,we get $(x^2 - 1)^2 = 4x$.For $x = \sqrt{2}$,$f(\sqrt{2}) = \sqrt{2} + \frac{2}{2 - 1} = \sqrt{2} + 2$.Checking the interval $(1, \sqrt{2}]$,the function is decreasing as $x$ approaches $\sqrt{2}$.Thus,the minimum value is $2 + \sqrt{2}$.

The minimum value of $\cos \theta + \sin \theta + \frac{2}{\sin 2 \theta}$ for $\theta \in (0, \pi / 2)$ is

Similar Questions

The minimum value of $\cos 2\theta + \cos \theta$ for real values of $\theta$ is

If $\alpha+\beta+\gamma=\pi$,then the expression $\sin^2 \alpha+\sin^2 \beta-\sin^2 \gamma$ is equal to:

Let $x, y, z$ be real numbers such that $x \geq y \geq z \geq \frac{\pi}{12}$. If $x+y+z = \frac{\pi}{2}$,then the minimum value of $\cos x \cdot \sin y \cdot \cos z$ is

If $A = \sin^2 x + \cos^4 x$,then for all real $x :$

The maximum value of $\cos \alpha_1 \cdot \cos \alpha_2 \cdots \cos \alpha_n$ under the restrictions $0 \le \alpha_1, \alpha_2, \dots, \alpha_n \le \frac{\pi}{2}$ and $\cot \alpha_1 \cdot \cot \alpha_2 \cdots \cot \alpha_n = 1$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module