<p>As discussed <a href="https://en.wikipedia.org/wiki/Position_of_the_Sun#Overview" rel="nofollow noreferrer">here</a>, the path of the Sun <em>resembles</em> a sine wave (see the image below), with a (semi-)amplitude equal to the <a href="https://en.wikipedia.org/wiki/Axial_tilt#Earth" rel="nofollow noreferrer">axial tilt of the Earth</a>, usually denoted $\varepsilon$, about 23.4 degrees. A simple formula, which might be accurate for your purposes, is the following: if $a$ denotes the right ascension (in hours), the declination $\delta$ is given by the following formula:</p>

<p>$$\delta = \varepsilon \sin(2\pi\frac{a}{24h})$$</p>

<p>The actual formula is more complicated. From the <a href="https://en.wikipedia.org/wiki/Position_of_the_Sun#Equatorial_coordinates" rel="nofollow noreferrer">formulas linked to by @barrycarter</a>:</p>

<p>$$\alpha = \arctan(\cos\varepsilon \tan\lambda)$$</p>

<p>$$\delta = \arcsin(\sin\varepsilon \sin\lambda)$$</p>

<p>we obtain</p>

<p>$$\tan\alpha = \cos\varepsilon \tan\lambda$$</p>

<p>$$\frac{\tan\alpha}{\cos\varepsilon} = \tan\lambda$$</p>

<p>so</p>

<p>$$\lambda = \arctan\frac{\tan\alpha}{\cos\varepsilon}$$</p>

<p>Plugging this into the declination formula, and using the fact that $\sin(\arctan(x)) = \frac{x}{\pm\sqrt{x^2+1}}$, we get</p>

<p>$$\delta = \arcsin(\sin\varepsilon \frac{\tan\alpha}{\pm\sqrt{\tan^2\alpha+\cos^2\varepsilon}})$$</p>

<p>(here the right ascension $\alpha$ is measured in radians or degrees, not hours).</p>

<p><a href="https://i.sstatic.net/3HptH.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/3HptH.jpg" alt="enter image description here"></a></p>

<p><sub>(picture taken from <a href="https://en.wikipedia.org/wiki/Ecliptic#/media/File:Constellations_ecliptic_equirectangular_plot.svg" rel="nofollow noreferrer">Wikipedia</a>, originally by Cmglee, Timwi)</sub></p>

CC BY-SA 3.0
2018-03-24T13:51:14.310Z
@barrycarter it is, the [conversion between ecliptic and equatorial coordinates](https://en.wikipedia.org/wiki/Ecliptic_coordinate_system#Conversion_between_celestial_coordinate_systems) should look familiar.
Glorfindel
CC BY-SA 3.0
2018-03-24T14:20:14.870Z
you're right. Trigonometry is hard :(
Glorfindel
CC BY-SA 3.0
2018-03-24T09:40:28.547Z
2018-03-24T14:18:13.237Z
2018-03-24T14:18:13.237Z
Glorfindel
<p>How to calculate declination of the Sun if its right ascension is known?<br>
I am BSc (NS) student.</p>

CC BY-SA 3.0
2018-03-25T10:45:12.367Z
For basic homework questions it would be far better to show your current thinking/working. Otherwise you will not learn...
Rory Alsop
<the-sun>
How to calculate declination of the Sun if its right ascension is known?
Chetan Pandey
# How to calculate declination of the Sun if its right ascension is known?

(Asked by: username_0 on 2018-03-24)

How to calculate declination of the Sun if its right ascension is known?  
I am BSc (NS) student.

**Tags:** the-sun

> **Comments:**
> **username_1**: For basic homework questions it would be far better to show your current thinking/working. Otherwise you will not learn...  

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## 1 answers

### Answer by username_2 on 2018-03-24. Score: 2

As discussed [here](https://en.wikipedia.org/wiki/Position_of_the_Sun#Overview), the path of the Sun *resembles* a sine wave (see the image below), with a (semi-)amplitude equal to the [axial tilt of the Earth](https://en.wikipedia.org/wiki/Axial_tilt#Earth), usually denoted $\varepsilon$, about 23.4 degrees. A simple formula, which might be accurate for your purposes, is the following: if $a$ denotes the right ascension (in hours), the declination $\delta$ is given by the following formula:

$$\delta = \varepsilon \sin(2\pi\frac{a}{24h})$$

The actual formula is more complicated. From the [formulas linked to by @barrycarter](https://en.wikipedia.org/wiki/Position_of_the_Sun#Equatorial_coordinates):

$$\alpha = \arctan(\cos\varepsilon \tan\lambda)$$

$$\delta = \arcsin(\sin\varepsilon \sin\lambda)$$

we obtain

$$\tan\alpha = \cos\varepsilon \tan\lambda$$

$$\frac{\tan\alpha}{\cos\varepsilon} = \tan\lambda$$

so

$$\lambda = \arctan\frac{\tan\alpha}{\cos\varepsilon}$$

Plugging this into the declination formula, and using the fact that $\sin(\arctan(x)) = \frac{x}{\pm\sqrt{x^2+1}}$, we get

$$\delta = \arcsin(\sin\varepsilon \frac{\tan\alpha}{\pm\sqrt{\tan^2\alpha+\cos^2\varepsilon}})$$

(here the right ascension $\alpha$ is measured in radians or degrees, not hours).

[![enter image description here](https://i.sstatic.net/3HptH.jpg)](https://i.sstatic.net/3HptH.jpg)

(picture taken from [Wikipedia](https://en.wikipedia.org/wiki/Ecliptic#/media/File:Constellations_ecliptic_equirectangular_plot.svg), originally by Cmglee, Timwi)

> **Comments:**
> **username_2**: @barrycarter it is, the [conversion between ecliptic and equatorial coordinates](https://en.wikipedia.org/wiki/Ecliptic_coordinate_system#Conversion_between_celestial_coordinate_systems) should look familiar.  
> **username_2**: you're right. Trigonometry is hard :(  

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