*(This item, "The minimum value of 2sinx+2cosx", is a trigonometry/AM-GM question, not a Statistics topic item --- it appears misplaced under this heading in the source PDF. It is retained here for completeness of the source document.)*
Solutions · 1
Official solution
By AM--GM: 2sinx+2cosx≥2⋅22sinx+cosx. Since −2≤sinx+cosx≤2, the minimum of 22sinx+cosx is 2−1/2.
\[2^{\sin x}+2^{\cos x} \ge 2\cdot2^{-1/\sqrt{2}} = 2^{1-\frac{1}{\sqrt{2}}}\]