# Deriving Mollweide's Formula

• Aug 17th 2006, 09:23 AM
spiritualfields
Deriving Mollweide's Formula
For this post, A is alpha, B is beta, and Y is gamma, and represent the angles of the triangle. Little a, b, and c represent the sides opposite them, respectively. The form of the formula is:
Code:

a - b        sin(1/2[A - B] -----  =    -------------   c            cos([1/2]Y) There is a step in the derivation that I don't understand. Here is the derivation from my solutions book, and so this post doesn't get too protracted, I'll stop at the point that I don't understand: a - b        a    b -----  =    -- - --   c          c    c =  sin A    sin B   ----  -  ----   sin Y    sin Y = sin A - sin B   -------------       sin Y = 2sin( (A-B) / 2) cos( (A+B) / 2)   ------------------------------    makes use of sum to product formula           sin (2Y/2) = 2sin( (A-B) / 2) cos( (A+B) / 2)   --------------------------------           2sin(Y/2)cos(Y/2) = sin( (A-B) / 2) cos(PI/2 - Y/2)   --------------------------------           sin(Y/2)cos(Y/2) How does cos((A+B)/2) get turned into cos(PI/2 - Y/2)?
• Aug 17th 2006, 09:37 AM
Plato
Well you do know that $\alpha + \beta + \gamma = \pi \quad \Rightarrow \quad \alpha + \beta = \pi - \gamma.$

Now divide by 2.
• Aug 17th 2006, 09:42 AM
spiritualfields
I didn't see that for some reason. Thanks.