Answer
$$\left( {\sqrt 6 ,\sqrt 2 ,2\sqrt 2 } \right)$$
Work Step by Step
$$\eqalign{
& \left( {4,\frac{\pi }{6},\frac{\pi }{4}} \right) \cr
& \left( {\rho ,\theta ,\phi } \right):{\text{ }}\left( {4,\frac{\pi }{6},\frac{\pi }{4}} \right) \to \rho = 4,{\text{ }}\theta = \frac{\pi }{6},{\text{ }}\phi = \frac{\pi }{4} \cr
& {\text{Spherical to rectangular }}\left( {\rho ,\theta ,\phi } \right) \cr
& x = \rho \sin \phi \cos \theta ,{\text{ }}y = \rho \sin \phi \sin \theta ,{\text{ }}z = \rho \cos \phi \cr
& x = \left( 4 \right)\sin \left( {\frac{\pi }{4}} \right)\cos \left( {\frac{\pi }{6}} \right) = 4\left( {\frac{{\sqrt 2 }}{2}} \right)\left( {\frac{{\sqrt 3 }}{2}} \right) = \sqrt 6 \cr
& y = \left( 4 \right)\sin \left( {\frac{\pi }{4}} \right)\sin \left( {\frac{\pi }{6}} \right) = 4\left( {\frac{{\sqrt 2 }}{2}} \right)\left( {\frac{1}{2}} \right) = \sqrt 2 \cr
& z = \left( 4 \right)\cos \left( {\frac{\pi }{4}} \right) = 4\left( {\frac{{\sqrt 2 }}{2}} \right) = 2\sqrt 2 \cr
& {\text{The rectangular }}\left( {x,y,z} \right){\text{ coordinates are:}} \cr
& \left( {\sqrt 6 ,\sqrt 2 ,2\sqrt 2 } \right) \cr} $$
