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Subject: Re: Spherical Law of Cosines
From: Herbert Prinz (hprinz@XXX.XXX)
Date: Sun Oct 27 2002 - 13:31:23 EST
While removing Dan Allen's confusion about the cosine theorem, Trevor
Kenchington adds some of his own:
>
> 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.
>
Trevor is meaning to say
sin(a) = cos(b)
which isn't absurd at all (because it is true if a and b are complementary),
but certainly imposes a limitation on the applicability of the formula that
we cannot accept. At any rate, this goes to shows how easy it is to mix up
one's sin and cos.
A remark about notation. To avoid any potential misunderstanding about what
kind of a product "ab" might be, it's probably better to remember the
formula in the form
(I) cos(c) = cos(a)*cos(b) + sin(a)*sin(b)*cos(C)
where lower case letter denote "sides" and upper case letters denote their
opposite "angles". (Of course, we know that the "sides" are, in fact also
angles, measured at the center of the sphere).
Finally, a mnemotechnical crutch for the mathematically oriented. For C = 0,
the above formula degenerates into the well known addition theorem for
cosines, which is frequently required to simplify messy terms or to
re-produce these obscure formulas in manuals of olden times.
If C equals 0, sides a and b come to lie on one line, therefore c = a - b.
We get
(II) cos(c) = cos(a-b) = cos(a)*cos(b) + sin(a)*sin(b)
So, you have two important formulas for the price of one.
If you have fun with this sort of thing, you can show with the above formula
that the ordinary noon sight is but a special case of the nautical triangle,
where the hour angle (C in formula I) becomes 0 or, in other words, a and b
are on the same meridian. The cosine of zenith distance of the noon Sun is
the cosine of (your latitude minus the Sun's declination). If you ever
program the nautical triangle into your calculator, you can re-use that same
formula for your noon sights, too. It automatically takes care of the whole
same side/contrary side - business that presents so many nightmares for
programmers.
Herbert Prinz
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