To convert spherical coordinates $(r, \theta, \phi)$ to rectangular coordinates $(x, y, z)$, we use the following formulas:

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

The given spherical coordinates are $(3, -\frac{\pi}{6}, -\frac{\pi}{2})$, where:

• $r = 3$
• $\theta = -\frac{\pi}{6}$
• $\phi = -\frac{\pi}{2}$

Now, let's calculate the rectangular coordinates step by step.

1. For $x$: $x = 3 \sin \left( -\frac{\pi}{6} \right) \cos \left( -\frac{\pi}{2} \right)$ We know:

• $\sin \left( -\frac{\pi}{6} \right) = -\frac{1}{2}$
• $\cos \left( -\frac{\pi}{2} \right) = 0$

Thus: $x = 3 \times \left( -\frac{1}{2} \right) \times 0 = 0$

2. For $y$: $y = 3 \sin \left( -\frac{\pi}{6} \right) \sin \left( -\frac{\pi}{2} \right)$ We know:

• $\sin \left( -\frac{\pi}{6} \right) = -\frac{1}{2}$
• $\sin \left( -\frac{\pi}{2} \right) = -1$

Thus: $y = 3 \times \left( -\frac{1}{2} \right) \times (-1) = \frac{3}{2} = 1.5$

3. For $z$: $z = 3 \cos \left( -\frac{\pi}{6} \right)$ We know:

• $\cos \left( -\frac{\pi}{6} \right) = \frac{\sqrt{3}}{2}$

Thus: $z = 3 \times \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2} \approx 2.598$

So, the rectangular coordinates are approximately:

$x = 0, \quad y = 1.5, \quad z = 2.598$

Would you like further clarification or details? Here are 5 questions related to this concept:

1. How are spherical coordinates related to cylindrical coordinates?
2. What is the significance of the angles $\theta$ and $\phi$ in spherical coordinates?
3. Can you explain how the rectangular coordinate system differs from the spherical one?
4. How do we handle negative angles when converting between spherical and rectangular coordinates?
5. Can spherical coordinates be used to represent all points in 3D space?

Tip: It's helpful to visualize the spherical and rectangular coordinate systems to better understand how points transform between them.