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

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

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

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(D) Using 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^{\frac{\sin 	heta + \cos 	heta}{2}}$We know that $\sin 	heta + \cos 	heta = \sqrt{2} \sin(	heta + \frac{\pi}{4})$.The minimum value of $\sin(	heta + \frac{\pi}{4})$ is $-1$,so the minimum value of $\sin 	heta + \cos 	heta$ is $-\sqrt{2}$.Thus,$2^{\sin 	heta} + 2^{\cos 	heta} \ge 2 \cdot 2^{\frac{-\sqrt{2}}{2}} = 2^1 \cdot 2^{-\frac{1}{\sqrt{2}}} = 2^{(1 - \frac{1}{\sqrt{2}})}$.

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

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