The problem asks to convert two points from spherical coordinates to Cartesian coordinates. (a) The first point is given in spherical coordinates as $(8, \frac{\pi}{4}, \frac{\pi}{6})$. (b) The second point is given in spherical coordinates as $(4, \frac{\pi}{3}, \frac{3\pi}{4})$.

1. Problem Description

The problem asks to convert two points from spherical coordinates to Cartesian coordinates.

(a) The first point is given in spherical coordinates as $(8, \frac{\pi}{4}, \frac{\pi}{6})$.

(b) The second point is given in spherical coordinates as $(4, \frac{\pi}{3}, \frac{3\pi}{4})$.

2. Solution Steps

Spherical coordinates are given by $(\rho, \theta, \phi)$, where $\rho$ is the radial distance, $\theta$ is the azimuthal angle, and $\phi$ is the polar angle.

Cartesian coordinates are given by $(x, y, z)$.

The conversion formulas are as follows:

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

$y = \rho \sin\phi \sin\theta$

$z = \rho \cos\phi$

(a) For the point $(8, \frac{\pi}{4}, \frac{\pi}{6})$, we have $\rho = 8$, $\theta = \frac{\pi}{4}$, and $\phi = \frac{\pi}{6}$.

$x = 8 \sin(\frac{\pi}{6}) \cos(\frac{\pi}{4}) = 8 (\frac{1}{2}) (\frac{\sqrt{2}}{2}) = 8(\frac{\sqrt{2}}{4}) = 2\sqrt{2}$

$y = 8 \sin(\frac{\pi}{6}) \sin(\frac{\pi}{4}) = 8 (\frac{1}{2}) (\frac{\sqrt{2}}{2}) = 8(\frac{\sqrt{2}}{4}) = 2\sqrt{2}$

$z = 8 \cos(\frac{\pi}{6}) = 8 (\frac{\sqrt{3}}{2}) = 4\sqrt{3}$

So the Cartesian coordinates are $(2\sqrt{2}, 2\sqrt{2}, 4\sqrt{3})$.

(b) For the point $(4, \frac{\pi}{3}, \frac{3\pi}{4})$, we have $\rho = 4$, $\theta = \frac{\pi}{3}$, and $\phi = \frac{3\pi}{4}$.

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

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

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

So the Cartesian coordinates are $(\sqrt{2}, \sqrt{6}, -2\sqrt{2})$.

3. Final Answer

(a) $(2\sqrt{2}, 2\sqrt{2}, 4\sqrt{3})$

(b) $(\sqrt{2}, \sqrt{6}, -2\sqrt{2})$