Re: Spherical geometry

In article <11e9jmtj7t0su02@xxxxxxxxxxxxxxxxxx>,
"Jim Heckman" <wnzrfeurpxzna@xxxxxxxxxxxxxxxxx> wrote:
>On 23-Jul-2005, gregegan@xxxxxxxxxxxxxxxxxxxxx (Greg Egan)
>wrote in message
><gregegan-2307052119030001@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx>:
>
>[...]
>
>> Suppose we have a triangle on the unit sphere, with A, B, C the angles at
>> the three vertices, and a, b, c the angles subtended at the centre of the
>> sphere by the sides opposite those vertices.
>>
>> We can always choose a coordinate system such that the vertices are the
>> following unit vectors in R^3:
>>
>> vA = (0, 0, 1)
>> vB = (sin c, 0, cos c)
>> vC = (cos A sin b, sin A sin b, cos b)
>>
>> In terms of spherical polar coordinates, we put vertex A at the north
>> pole, vertex B on the prime meridian, and vertex C on the meridian with
>> longitude A. The co-latitudes (angles from the pole) of vertices B and C
>> are c and b respectively.
>>
>> Cosine Laws
>> ===========
>
>[...]
>
>> Substituting the geometry of the second triangle into the First Cosine Law
>> gives the Second Cosine Law:
>>
>> cos A = cos a sin B sin C - cos B cos C
>
>I haven't checked the steps in your derivation, but there must be a
>sign error somewhere. It should be:
>
>cos A = cos a sin B sin C + cos B cos C
>
>[...]

Nope. As Greg said, for every spherical triangle, there is a dual
triangle where each side is replaced by a supplementary angle and each
angle is replaced by a supplementary side. Since cos(pi-x) = -cos(x),
and sin(pi-x) = sin(x), the dual of the formula

cos(a) = cos(A) sin(b) sin(c) + cos(b) cos(c)

is

-cos(A) = -cos(a) sin(B) sin(C) + cos(B) cos(C)

which is the same as

cos(A) = cos(a) sin(B) sin(C) - cos(B) cos(C)

Rob Johnson <rob@xxxxxxxxxxxxxx>
take out the trash before replying
.

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