# How do you find the rectangular coordinates of the point whose spherical coordinates are (4, 3π/4, π/3)?

Feb 20, 2015

The answer is: $\left(\sqrt{2} , \sqrt{6} , - 2 \sqrt{2}\right)$.

The formulas are:

$x = r \sin \theta \cos \phi$

$y = r \sin \theta \sin \phi$

$z = r \cos \theta$

So:

$x = 4 \sin \left(\frac{3}{4} \pi\right) \cos \left(\frac{\pi}{3}\right) = 4 \cdot \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \sqrt{2}$

$y = 4 \sin \left(\frac{3}{4} \pi\right) \sin \left(\frac{\pi}{3}\right) = 4 \cdot \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} = \sqrt{6}$

$z = 4 \cos \left(\frac{3}{4} \pi\right) = 4 \cdot \left(- \frac{\sqrt{2}}{2}\right) = - 2 \sqrt{2}$.