Using $$A.M.\ge\ G.M.$$

$$\frac{\left(4^{\sin x}+4^{\cos x}\right)}{2}\ge\ \sqrt{\ 4^{\sin x+\cos x}}$$

or,  $$\left(4^{\sin x}+4^{\cos x}\right)\ge2\ \sqrt{\ 4^{\sin x+\cos x}}$$.

$$\left(4^{\sin x}+4^{\cos x}\right)\ge2\ \sqrt{\ 2^{2\left(\sin x+\cos x\right)}}=2.2^{\sin x+\cos x}=2^{1+\sin x+\cos x}$$

The minimum value of sinx +cosx when x lies in R is $$-\sqrt{\ 2}$$

Thus, the minimum value of the expression is $$2^{1-\sqrt{\ 2}}$$