# NavList:

## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding

**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 programmers. Herbert Prinz