# 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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