$\begin{array}{l} {\sin ^4}x = {\left( {\dfrac{{1 - \cos 2x}}{2}} \right)^2}\\ = \dfrac{{{{\cos }^2}2x - 2\cos 2x + 1}}{4} = \dfrac{{\dfrac{{1 + \cos 4x}}{2} - 2\cos 2x + 1}}{4}\\ = \dfrac{{1 + \cos 4x - 4\cos 2x + 2}}{8} = \dfrac{3}{8} - \dfrac{1}{2}\cos 2x + \dfrac{1}{8}\cos 4x\\ \to a = 3,b = 1 \to a + b = 4 \end{array}$