Equation of a straight line in spherical coordinates I'm trying to prove the angle sum formula for a triangle on the surface of a sphere. In order to do this I wanted to create a general triangle on the sphere, with one vertex at $\theta = 0$  and one side along the angle $\phi = 0$ for simplicity. (I think this can be done without loss of generality?)
In order to do this I need to be able to express one of the other sides as a straight line in spherical coordinates. Intuition tells me that it is probably just $\theta = k\phi + \alpha$ and $\phi$ as a free parameter. I've tried proving this to myself by showing the the resulting curve is a great circle. I then tried to compute a path integral around this curve so I could check that it is equal to the circumference of my general sphere, however wolfram alpha tells me that I must upgrade to pro to get more computation time to do my integral. ($\int_{0}^{2\pi}\sqrt{k^2 + \sin{(kt + \alpha)}}$ dt)
I've also tried to show that in cartesian coordinates that any points resulting from this parameterization must be coplanar. I have had little success with this approach. 
Am I completely wrong here? Is there some good way to see whether or not this is true?
 A: I suppose you assumed the $r$-coordinate of your path is constant.
Since you're looking for a path on the surface of a sphere, that's appropriate.
When you're trying to describe an object in a coordinate system, it helps to have very clear notions about the properties of that object as expressed in
that coordinate system.
There are a few important facts to keep in mind about a great circle.
One is that every point on the great circle is exactly opposite another
point on the great circle, that is, if $(r,\phi,\theta)$ is on the
great circle then so is $(r,\pi-\phi,\theta + pi)$.
Another fact is that as you travel around the great circle, after you cross
a "line of latitude" going "upward", you must later cross that same
line of latitude at a different point while going "downward".
A third fact is that unless the great circle happens to be a
"line of longitude", it will not pass through the poles. That is,
each great circle has a minimum latitude angle $\phi_0$ and a maximum
latitude angle (which by the first fact above must be $\pi - \phi_0$),
where $0 < \phi_0 < \pi/2$.
The great circle only attains its minimum latitude at one point; likewise the maximum latitude.
So if you have an equation of a great circle in the form
$$f(\phi, \theta) = 0,$$
this equation must have exactly one solution with $\phi = \phi_0$
and one with $\phi = \pi - \phi_0$;
but for each $\phi$ such that $\phi_0 < \phi_0 < \pi - \phi_0$,
there must be exactly two solutions of the equation.
On further examination of a great circle drawn on a sphere,
it should be obvious that at the point on the great circle where $\phi = \phi_0$, the circle is tangent to a line of latitude,
and at that point $\frac{d}{d\theta}\phi = 0$.
Your proposed equation, $\theta = k\phi + \alpha$,
is equivalent to $\theta - k\phi - \alpha = 0$.
Now we can see several flaws.
First, there is nothing in this equation to require that $\phi$ have
a maximum or minimum value anywhere except possibly at $\phi=0$ and $\phi=\pi$.
Second, there is only one solution to the equation for any given $\phi$.
(There should be two solutions for most values of $\phi$.)
Third, there is no point at which $\frac{d}{d\theta}\phi = 0$.
Let's try a different approach. The great circle is the intersection of
the surface of the sphere and a plane through the center of the sphere.
The equation of that plane is
$$ax + by + cz = 0.$$
Now convert the Cartesian coordinates to spherical coordinates:
\begin{align}
x &= r \sin \phi \cos \theta, \\
y &= r \sin \phi \sin \theta, \\
z &= r \cos \phi.
\end{align}
Making these substitutions directly into the equation of the plane, we have
$$ar \sin \phi \cos \theta + br \sin \phi \sin \theta + cr \cos \phi = 0.$$
Take out the factor of $r$ and combine some terms and we have
$$(a \cos \theta + b \sin \theta)\sin \phi + c \cos \phi = 0.$$
A: I think i have finally found how to go about this: Any circle on the surface of the circle must satisfy two equations. 1) $x^2 + y^2 + z^2 = r_0^2$ and 2) $ax + by + cz = d$ (The equation of a plane). Thus, converting to spherical coordinates, we have that $r = r_0$ and $a r sin(\phi)cos(\theta) + b r sin(\phi) sin(\theta) + c cos(\phi) = d$. You can then take the $sin(\phi)$ common to get $sin(\phi)(acos(\theta) + bsin(\theta))$ and multiply and divide by $\sqrt{a^2 + b^2}$ to enable you to have that portion in the form $sin(\theta + \theta_0)$. Now you can see that a parametric equation can be given in a form $(r, \theta, \phi) = (1, sin^{-1}( \frac{1}{\sqrt{a^2 + b^2}} (\frac{(d - c cos(\phi)}{sin(\phi)})) - \theta_0, u), u = 0...2π$.
What do you think?   
A: To create a line in circular coordinates, you can just do it like this: $sin(\theta) = m cos(\theta) + c/r$, where $m$ is the gradient and $c$ is the y-intercept. This comes from the $y = mx + c$ definition.
The $y = mx$ part is straightforward, but it is not as obvious. To see how this works explicitly, consider that to your base equation $sin(\theta) = m cos(\theta)$, you are adding the equation for a horizontal line: $sin(\theta) = c/r$ which comes straight from geometry.  
You can use similar principles to extend this to spherical coordinates.
