If the spherical coordinates of a point $P$ are $\left( 4,\dfrac{\pi }{6}, \dfrac{2\pi }{3}\right)$, find the rectangular coordinates of $P$.

Solution We use the equations (1) to convert spherical coordinates to rectangular coordinates. Then $$\begin{eqnarray*} x& =&\rho \sin \phi \cos \theta =4\sin \dfrac{2\pi }{3}\cos \dfrac{\pi }{6} =4\!\left( \dfrac{\sqrt{3}}{2}\right) \left( \dfrac{\sqrt{3}}{2}\right) =3 \qquad\!\! {\color {#0066A7}{\hbox{\(\rho =4; \theta =\dfrac{\pi}{6}, \phi =\dfrac{2\pi }{3}\)}}}\\ y& =&\rho \sin \phi \sin \theta =4\sin \dfrac{2\pi }{3}\sin \dfrac{\pi }{6} =4\!\left( \dfrac{\sqrt{3}}{2}\right) \left( \dfrac{1}{2}\right) =\sqrt{3} \\[4pt] z& =&\rho \cos \phi =4\cos \dfrac{2\pi }{3}=4\!\left( -\dfrac{1}{2}\right) =-2 \end{eqnarray*}$$

The rectangular coordinates of $P$ are $( 3,\sqrt{3},-2)$.