The minimum value of 2^{sin x}+2^{cos x} is | vedclass

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(A) We know that by the $AM-GM$ inequality, for positive real numbers $a$ and $b$, $\frac{a+b}{2} \geq \sqrt{ab}$.Applying this to $2^{\sin x}$ and $2

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

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(A) We know that by the $AM-GM$ inequality, for positive real numbers $a$ and $b$, $\frac{a+b}{2} \geq \sqrt{ab}$.Applying this to $2^{\sin x}$ and $2^{\cos x}$, we get:$\frac{2^{\sin x}+2^{\cos x}}{2} \geq \sqrt{2^{\sin x} \cdot 2^{\cos x}}$$\Rightarrow 2^{\sin x}+2^{\cos x} \geq 2 \cdot 2^{\frac{\sin x+\cos x}{2}} = 2^{1+\frac{\sin x+\cos x}{2}}$.We know that the range of $\sin x+\cos x$ is $[-\sqrt{2}, \sqrt{2}]$.To minimize the expression $2^{1+\frac{\sin x+\cos x}{2}}$, we must minimize the exponent $1+\frac{\sin x+\cos x}{2}$.The minimum value of $\sin x+\cos x$ is $-\sqrt{2}$.Substituting this into the expression, the minimum value is $2^{1+\frac{-\sqrt{2}}{2}} = 2^{1-\frac{1}{\sqrt{2}}}$.

The minimum value of $2^{\sin x}+2^{\cos x}$ is

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