What is the angle between the North Celestial Pole and the Galactic Equator?

Question

I’m trying to back-calculate a value in astronomy using spherical trigonometry. Using the following formula, where $δ$ = the Declination of the Galactic North Pole in degrees (i.e., the angle between the North Celestial Pole and the Galactic Equator).

Given that

$$\arccos(\sin(δ)\cos(23.43928°)-\sin(192.8583°)\cos(δ)\sin(23.439289°)) = 60.18894°$$

where

• $23.43928° =$ obliquity of the Ecliptic (tilt of Earth's axis relative to Ecliptic North Pole)

• $192.8583° =$ right ascension of Galactic North Pole in degrees

• $60.18894° =$ angle between the Galactic and Ecliptic North Poles (or the angle between the Galactic and Ecliptic Planes)

• $δ =$ declination of Galactic North Pole using Equatorial Coordinates

Can you solve for $δ$?

I've tried doing the calculation for δ in Excel, Wolfram Alpha, and "3.2. Galactic and equatorial coordinates" at https://aas.aanda.org/articles/aas/full/1998/01/ds1449/node3.html, but $δ$ is always given as a known value, i.e., 27.41 degrees, but I've had no luck in finding out how $δ$ is calculated in the first place.

You may ask, why do I want to calculate $δ$ if I already know its value? Call it intellectual curiosity, or the satisfaction of knowing the answer to something I've been struggling with for the past 5 years.

Answer 1

Here is how to solve for $\delta$ the following trig expression involving $\sin \delta$ and $\cos \delta$.

$$ A \sin \delta+B \cos \delta = C \tag{1}$$

Presenting below are two different methods for the same problem. In your case, use one of the two to find $\delta$.

1. Trig expansion

   Consider $\delta$ having two parts $\delta= \varphi + \psi$ and expand the trig function to

   $$ \cos \psi \left( B \cos \varphi+A \sin \varphi \right) + \sin \psi \left( A \cos \varphi - B \sin \varphi \right) = C \tag{2} $$

   Now you force $A \cos\varphi - B \sin\varphi =0$ since the choice of how to split $\delta$ is ours.

   $$ \varphi = \mathrm{atan}\left(\frac{A}{B}\right) $$

   And the equation (2) is now

   $$ \cos\psi \left( \sqrt{A^2+B^2} \right) + \sin \psi \left( 0 \right) = C $$

   which is solved for

   $$ \psi = \pm\, \mathrm{acos}\left( \frac{C}{\sqrt{A^2+B^2}} \right) $$

   Putting $\delta$ back together we have the solution

   $$ \delta = \mathrm{atan}\left(\frac{A}{B}\right) \pm\, \mathrm{acos}\left( \frac{C}{\sqrt{A^2+B^2}} \right) \tag{3}$$

2. Trig substitution

   Consider the tan half-angle substitution, $t = \tan \left( \frac{\delta}{2} \right)$ which yields the following expressions $$ \cos \delta = \frac{1-t^2}{1+t^2} \tag{4}$$ $$ \sin \delta = \frac{2 t}{1+t^2} \tag{5}$$ $$ \delta = 2 \,\mathrm{atan}(t) \tag{6}$$

   And use these in (1) to get a polynomial in terms of $t$

   $$ A \frac{2 t}{1+t^2} + B \frac{1-t^2}{1+t^2} = C $$

   $$ (B+C) t^2 -2 A t + (C-B) = 0 \tag{7}$$

   Solve the quadratic for $t$ and use it in (6)

   $$\delta= 2 \mathrm{atan}\left( \frac{A \pm \sqrt{A^2 + B^2 - C^2 }}{B+C} \right) \tag{8} $$

Even though (3) and (8) are different expressions, the numeric results are the same. But note the usual quadrant restrictions that go with trig functions.

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Consider the simple example of $2 \sin\delta+ 3 \cos \delta = 1$

If you apply any of the methods above you will find $\delta \approx -40.207818721 °$ and $\delta \approx 107.58795377 °$ as the two solutions.

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Based on the comments by @DavidHammen there is a special case when $B+C=0$. In that case the solution to (7) is $t = \frac{C}{A}$ or

$$\delta = 2\, \mathrm{atan}\left( \frac{C}{A} \right)$$

Answer 2

$$ \delta = \arcsin \left( \frac{\cos \left( \theta \right)}{\sqrt{a^{2} + b^{2}}} \right) + \arctan \left( \frac{b}{a} \right) = 27.13^{o} $$

where

$23.43928^{o} =$ obliquity of the Ecliptic (tilt of Earth's axis relative to Ecliptic North Pole)

$192.8583^{o} =$ right ascension of Galactic North Pole in degrees

$60.18894^{o} =$ angle between the Galactic and Ecliptic North Poles (or the angle between the Galactic and Ecliptic Planes)

With the substitutions:

$$ a = \cos \left( 23.43928^{o} \right) \approx 0.9174821 $$ $$ b = \sin \left( 192.8583^{o} \right) \cdot \sin \left( 23.439289^{o} \right) \approx −0.0885216 $$ $$ \theta = 60.18894^{o} $$

Calculations for deriving $\delta$ are at: https://astronomy.stackexchange.com/questions/53312/calculate-declination-of-galactic-n-pole-given-obliquity-of-ecliptic-rt-ascen/53313?noredirect=1#comment119196_53313