Rectangular to Spherical coordinate conversion

[MathsLearner]

I need to convert from rectangular to spherical coordinates $$(\sqrt3, -1, 2\sqrt3)$$. The formulae are
$$\rho = \sqrt{(x^2+y^2+z^2)} -eq1 \\ x = \rho \sin\phi \cos\theta; -eq2\\ y = \rho\sin\phi\sin\theta; -eq3\\ z = \rho\cos\phi; -eq4 \\ \rho = \sqrt{(3+1+12)} = 4; \\ \cos\phi = \frac {2\sqrt3} 4; \\ \phi = \frac \pi 6$$The confusion is with the $$\theta;$$. I can have two values using eq 2 and 3,
using eq1
$$\cos\theta = \frac x {\rho\sin\phi} = \frac {\sqrt3} {4\sin(\frac \pi 6)} \\ \theta = \frac \pi 6; -r1$$ using eq2
$$\sin\theta = \frac y {\rho\sin\phi} = -\frac {1} {4\sin(\frac \pi 6)} = -\frac {1} 2 \\ \theta = -\frac {\pi} 6$$.
Since $$\theta$$ is negative. The angle is in the anti clockwise direction and lies in the 4th quadrant, the value in terms of clockwise is
$$2\pi - \frac \pi 6 = \frac {11\pi} 6$$. It also matches since y is negative.
Does it mean the $$r1$$ is wrong?

 

[Dr.Peterson]

There is only one [MATH]\theta[/MATH] that satisfies both equations 2 and 3. Neither alone determines the quadrant.

From equation 2, [MATH]\theta = \frac{\pi}{6} \text{ or } -\frac{\pi}{6}[/MATH] or any coterminal angle.

From equation 3, [MATH]\theta = \frac{7\pi}{6} \text{ or } \frac{11\pi}{6}[/MATH] or any coterminal angle.

We conclude that [MATH]\theta = \frac{11\pi}{6}.[/MATH]

 

[MathsLearner]

Ok to avoid confusion i could probably do this satisfying equations 2 and 3, Divide eq3 with eq2 to obtain
$$\tan\theta = \frac y x = -\frac 1 {\sqrt3}; \\ \theta = -\frac \pi 6 = \frac {11\pi} 6;$$

 

[Dr.Peterson]

Unfortunately, there too we have two solutions, which differ by [MATH]\pi[/MATH] this time. No matter how you approach it, you have to take quadrants into account to decide between [MATH]\frac {5\pi}{6}[/MATH] and [MATH]\frac {11\pi}{6}[/MATH].

But this is the way I prefer, when combined with quadrant considerations.

 

[HallsofIvy]

There is only one [MATH]\theta[/MATH] that satisfies both equations 2 and 3. Neither alone determines the quadrant.

From equation 2, [MATH]\theta = \frac{\pi}{6} \text{ or } -\frac{\pi}{6}[/MATH] or any coterminal angle.

From equation 3, [MATH]\theta = \frac{7\pi}{6} \text{ or } \frac{11\pi}{6}[/MATH] or any coterminal angle.

We conclude that [MATH]\theta = \frac{11\pi}{6}.[/MATH]

Note that Dr. Peterson "concludes" that [MATH]\theta = \frac{11\pi}{6}.[/MATH] because $$\frac{11\pi}{6}= \frac{12\pi}{6}- \frac{\pi}{6}= 2\pi- \frac{\pi}{6}$$ so is a "coterminal" angle with $$-\frac{\pi}{6}$$.