2^{sin theta} + 2^{cos theta} is greater than | vedclass

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Question

(D) By the Arithmetic Mean-Geometric Mean inequality $(AM \ge GM)$:$\frac{2^{\sin 	heta} + 2^{\cos 	heta}}{2} \ge \sqrt{2^{\sin 	heta} \cdot 2^{\co

Vedclass · Accepted Answer

$2^{(1 - 1/\sqrt{2})}$

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(D) By the Arithmetic Mean-Geometric Mean inequality $(AM \ge GM)$:$\frac{2^{\sin 	heta} + 2^{\cos 	heta}}{2} \ge \sqrt{2^{\sin 	heta} \cdot 2^{\cos 	heta}}$$2^{\sin 	heta} + 2^{\cos 	heta} \ge 2 \cdot 2^{(\sin 	heta + \cos 	heta)/2} = 2^{1 + \frac{\sin 	heta + \cos 	heta}{2}}$Since $\sin 	heta + \cos 	heta = \sqrt{2} \sin(	heta + \pi/4)$,the minimum value of $\sin 	heta + \cos 	heta$ is $-\sqrt{2}$.Therefore,$2^{\sin 	heta} + 2^{\cos 	heta} \ge 2^{1 + \frac{-\sqrt{2}}{2}} = 2^{1 - \frac{1}{\sqrt{2}}}$.

$2^{\sin \theta} + 2^{\cos \theta}$ is greater than

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