[tex]\text{Given that,} \sin x = \dfrac{\sqrt 5}3,~~ \text{and}~~~ \sin y =\dfrac{\sqrt 2}2\\\\\\~~~~~~~\sin x = \dfrac{\sqrt 5}3\\\\\implies \sin^2 x = \dfrac 59\\\\\implies 1- \cos^2 x = \dfrac 59\\\\\implies \cos^2 x =1- \dfrac59 \\\\\implies \cos^2 x = \dfrac 49\\\\\implies \cos x = \dfrac 23~~~~~~~;[\text{In quadrant I, all ratios are positive.}]\\\\\\\\[/tex]
[tex]~~~~~~\sin y =\dfrac{\sqrt{2}}2\\ \\\implies \sin y = \dfrac 1{\sqrt 2}\\\\\implies \sin^2 y = \dfrac 12\\\\\implies 1-\cos^2 y = \dfrac 12\\\\\implies \cos^2 y = 1-\dfrac 12=\dfrac 12\\\\\implies \cos y = \dfrac 1{\sqrt 2}\\\\\\[/tex]
[tex]\text{Now,}\\\\\cos(x+y) = \cos x \cos y - \sin x \sin y\\\\\\~~~~~~~~~~~~~~=\dfrac 23 \cdot \dfrac 1{\sqrt 2}- \dfrac{\sqrt 5}3 \cdot \dfrac 1{\sqrt 2}\\\\\\~~~~~~~~~~~~~~=\dfrac{2- \sqrt 5}{3\sqrt 2}\\\\\\~~~~~~~~~~~~~~=\dfrac{\sqrt2\left(2 -\sqrt 5 \right)}6[/tex]
