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October 17th, 2017, 06:21 PM   #1
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Math Focus: non-euclidean geometry
Does this identity belong with the circle functions or with the conic sections?

$$\alpha={\cot}^{-1 }(\cos\upsilon\tan{\sin}^{-1}(\frac{\sin\frac{\lambda}{ 2}}{ \sin\upsilon}))$$

The equation describes the relationship between 3 dihedral angles in 3D. It is expressed as: $$\alpha=f(\lambda)$$

To understand the angles, it’s easiest to construct a unit sphere with North and South poles and an axis that contains both poles. If a small circle is placed on the sphere such that it passes through the North pole, then $$\alpha$$ is the slope of the tangent to a point on the small circle (relative to a line of longitude) and $$\lambda$$ is the dihedral angle between the North pole and the tangent point. In order to define a family of functions, $$\upsilon$$ is a third variable that expresses the dihedral angle between the North pole and the center of the small circle.


With no spherical excess $$\upsilon = 0$$ the equation produces a sine curve. With maximum spherical excess $$\upsilon = \frac{\pi}{2}$$ the equation produces a hyperbola.


Even though the identity can produce either a sine or a hyperbola, it doesn’t seem to belong to either class of functions.
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October 18th, 2017, 02:33 PM   #2
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Your question is confusing. I don't see an identity.
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October 18th, 2017, 04:44 PM   #3
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Quote:
Originally Posted by mathman View Post
Your question is confusing. I don't see an identity.
I'm sort of confused, hence a confusing question. I think it is a sin vs. cos type of thing:


$$\sin{\upsilon = \frac{\sin\frac{\lambda}{2 }}{\sin\frac{\phi}{2 }}}$$

$$\cos\upsilon = \frac{\cot\alpha}{\tan\frac{\phi}{2 }}$$


is a parametrization of the equation.

I thought that since $$\cos^2(\upsilon)+\sin^2(\upsilon)=1$$ that this would mean that

$$(\frac{\sin\frac{\lambda}{2 }}{\sin\frac{\phi}{2 }})^2+(\frac{\cot\alpha}{\tan\frac{\phi}{2 }})^2 = 1$$

I know I'm doing something wrong, I just don't know what it is.
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October 18th, 2017, 08:01 PM   #4
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What does $\phi$ represent?
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October 18th, 2017, 09:02 PM   #5
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Math Focus: non-euclidean geometry
It's the small circle arc length from the North pole to the tangent point.

$$0\leq\phi\leq\pi$$

Last edited by steveupson; October 18th, 2017 at 09:28 PM.
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October 19th, 2017, 04:55 AM   #6
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Math Focus: non-euclidean geometry
Using the range and domain given, only a quarter of the full wave is present, as shown in the attachment. If the limits are changed to $$0\leq\upsilon\leq\pi$$ and $$0\leq\phi\leq2\pi$$ then this should produce the full wave function.
Attached Images
File Type: jpg Fig 5.jpg (9.9 KB, 1 views)
File Type: jpg alpha=f(lambda)x4.JPG (20.5 KB, 2 views)

Last edited by steveupson; October 19th, 2017 at 05:27 AM.
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October 25th, 2017, 10:08 AM   #7
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Math Focus: non-euclidean geometry
Quote:
Originally Posted by steveupson View Post
I know I'm doing something wrong, I just don't know what it is.
I think this is correct:

$$\left(\begin{array}{c}{\cos\frac{\phi}{2 }}\:{\sin\frac{\lambda}{2 }}\end{array}\right)^2+\left(\begin{array}{c}{\cot \frac{\phi}{2 }}\:{\cot\alpha}\end{array}\right)^2 = 1$$
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