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

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(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 $\si

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$2 + \sqrt{2}$

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

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