Official solution
Using AM \[ \ge \] GM
\[ \Rightarrow {{{2^{\sin \,x}} + {2^{\cos \,x}}} \over 2} \ge \sqrt {{2^{\sin x}}{{.2}^{\cos x}}} \]
\[ \Rightarrow {2^{\sin x}} + {2^{\cos x}} \ge {2^{1 + \left( {{{\sin x + \cos x} \over 2}} \right)}}\]
\[ \Rightarrow \min ({2^{\sin x}} + {2^{\cos x}}) = {2^{1 - {1 \over {\sqrt 2 }}}}\]
As we know range of sin x + cos x is :
\[ - \sqrt 2 \] \[ \le \] sin x + cos x \[ \le \] \[\sqrt 2 \].
So Minimum value of sin x + cos x = \[ - \sqrt 2 \]