The minimum value of 2 ^ x + 2 ^ x is :

Using $A . M \ge G . M .$

$\dfrac{{2}^{\sin x} + {2}^{\cos x}}{2} \ge \sqrt{{2}^{\sin x} \cdot {2}^{\cos x}}$

$⇒ {2}^{\sin x} + {2}^{\cos x} \ge {2}^{1 + \dfrac{\sin x + \cos x}{2}}$

Now $- \sqrt{2} \le \sin x + \cos x \le \sqrt{2}$

So minimum value of

${2}^{\sin x} + {2}^{\cos x} = {2}^{1 + \dfrac{- \sqrt{2}}{2}} = {2}^{1 - \dfrac{1}{\sqrt{2}}}$