Phương pháp giải:

+) Tính \(\cos a,\,\,\cos b\).

+) Sử dụng các công thức \(\sin 2x=2\sin x\cos x,\,\,\sin \left( a+b \right)=\sin a\cos b+\cos a\sin b,\) \(\cos \left( a+b \right)=\cos a\cos b-\sin a\sin b\).

Lời giải chi tiết:

Ta có:

\(\begin{align} {{\cos }^{2}}a=1-{{\sin }^{2}}a=1-\frac{1}{9}=\frac{8}{9}\Leftrightarrow \cos a=\frac{2\sqrt{2}}{3}\,\,\left( Do\,\,0<a<{{90}^{0}} \right) \\ {{\cos }^{2}}b=1-{{\sin }^{2}}b=1-\frac{1}{4}=\frac{3}{4}\Leftrightarrow \cos b=\frac{\sqrt{3}}{2}\,\,\left( Do\,\,{{0}^{0}}<b<{{90}^{0}} \right) \\\end{align}\)

\(\begin{array}{l}\sin 2\left( {a + b} \right) = 2\sin \left( {a + b} \right)\cos \left( {a + b} \right)\\\sin 2\left( {a + b} \right) = 2\left( {\sin a\cos b + \cos a\sin b} \right)\left( {\cos a\cos b - \sin a\sin b} \right)\\\sin 2\left( {a + b} \right) = 2\left( {\dfrac{1}{3}.\dfrac{{\sqrt 3 }}{2} + \dfrac{{2\sqrt 2 }}{3}.\dfrac{1}{2}} \right)\left( {\dfrac{{2\sqrt 2 }}{3}.\dfrac{{\sqrt 3 }}{2} - \dfrac{1}{3}.\dfrac{1}{2}} \right)\\\sin 2\left( {a + b} \right) = 2\dfrac{{\sqrt 3  + 2\sqrt 2 }}{6}.\dfrac{{2\sqrt 6  - 1}}{6}\\\sin 2\left( {a + b} \right) = \dfrac{{7\sqrt 3  + 4\sqrt 2 }}{{18}}\end{array}\).

Chọn C.