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    Re: Spherical Law of Cosines
    From: Herbert Prinz
    Date: 2002 Oct 27, 18:31 +0000

    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
    Herbert Prinz

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