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Re: Spherical Law of Cosines
From: Trevor Kenchington
Date: 2002 Oct 26, 11:10 -0300

```Dan Allen wrote:

[snip]

> However, I went back and found support in Smart's book for the
> form that I had written, i.e.,
>
> cos(c) = sin(a)*sin(b) + cos(a)*cos(b)*cos(ab)
>
> in determining the length of twilight and other such calculations.
>
> In thinking about things I realized that both versions are
> right, but it simply is a matter of origin.  Are the angles
> measured down from the pole (co-latitudes and such) or are
> they measured from the equator up (latitudes)?
> They are equivalent.
>
> The mental picture that I work from is the canonical version,
>
> cos(c) = cos(a)*cos(b) + sin(a)*sin(b)*cos(ab)

If it were true that:

cos(c) = sin(a)*sin(b) + cos(a)*cos(b)*cos(ab)

and:

cos(c) = cos(a)*cos(b) + sin(a)*sin(b)*cos(ab)

then it would necessarily be true that:

sin(a)*sin(b) + cos(a)*cos(b)*cos(ab) = cos(a)*cos(b) +
sin(a)*sin(b)*cos(ab)

since both are equal to cos(c). And so we would have to suppose that
sin(a)=cos(a), which is obviously absurd.

The correct equation of this pair is Dan's "cononical version" (unless I
am wildly off base). His alternate should, I suspect, be written:

cos(c) = sin(A)*sin(B) + cos(A)*cos(B)*cos(ab)

where A=90-a and B=90-b. Stretching memory back to high-school
triginometry, I think it is true that sin(90-a)=cos(a), making this form
ofthe alternate identical to the canonical version. Of course, provided
one is careful over using (e.g.) latitudes rather than co-latitudes with
the alternate, you could forget about explicitly subtracting anything
from 90. But remembering the alternate version as a solution to the
spherical triangle could get you into serious confusion.

Trevor Kenchington

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Nova Scotia  B0J 2L0, CANADA                      Home  (902) 889-3555

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