Relating E to Nu
Relating E to \nu is a problem of trigonometry. We will be referring to the diagram
which we have extracted (with slight modifications) from one of our previous diagrams wherein we juxtaposed an ellipse and a circle. Recall that P was a point on the ellipse, P’ a point on the circle, S the sun (one focus of the ellipse), and C the center of both the circle and the ellipse. The point Q is the foot of the perpendicular line through P and P’ to the major axis.
In this section, we will use the trigonometric identity
In the diagram, QF=-r\cos \nu. Recalling our formula for r from a previous section, this becomes
We may also observe CQ=a \cos E and CF=ae, so noting that CF=CQ+QF, we get
Solving this equation for \cos E gives us
From basic trigonometry, we may calculate that QP’=a\sin E and QP=r \sin \nu. Let us be mindful that P is on the ellipse and P’ is directly above it on the circle, and that the first can be obtained from the second by shrinking by a factor of
So
Solving for \sin \nu and substituting for r will give us
It is an easy series of calculations (he wrote with a malicious grin) to see
and
and
Using our formula for the tangent of the half-angle, we may combine our formulas for 1-\cos \nu and \sin \nu to obtain
Some algebra and another application of the tangent of the half-angle formula will reveal
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