# Does this identity belong with the circle functions or with the conic sections?

#### steveupson

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

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

To understand the angles, itâ€™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 doesnâ€™t seem to belong to either class of functions.

#### mathman

Forum Staff
Your question is confusing. I don't see an identity.

#### steveupson

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.

#### greg1313

Forum Staff
What does $\phi$ represent?

#### steveupson

It's the small circle arc length from the North pole to the tangent point.

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

Last edited:

#### steveupson

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.

#### Attachments

• 9.9 KB Views: 1
• 20.5 KB Views: 2
Last edited:

#### steveupson

I know I'm doing something wrong, I just don't know what it is.
I think this is correct:

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