Subject: Why does applying trigsimp aid solvability?
From: Paul Richards
Date: Thu, 11 Jun 2009 21:20:27 +0100
I have an equation, which I have been solving using Maxima. What I
don't understand is that "solve" on its own doesn't seem to solve the
whole problem, but "solve(trigsimp(..))" does.
Why is this?
(%i1) c^2 = (cos(a/r) - 1)^2 + sin(a/r)^2;
2 2 a a 2
(%o1) c = sin (-) + (cos(-) - 1)
r r
(%i2) solve(%i1, a);
a 2 a a 2
(%o2) [sin(-) = - sqrt(- cos (-) + 2 cos(-) + c - 1),
r r r
a 2 a a 2
sin(-) = sqrt(- cos (-) + 2 cos(-) + c - 1)]
r r r
(%i3) solve(trigsimp(%i1), a);
solve: using arc-trig functions to get a solution.
Some solutions will be lost.
2
c
(%o3) [a = (%pi - acos(-- - 1)) r]
2
PS. How do I better copy/paste maxima output into email? I have a
feeling this is only viewable with fixed width fonts.
--
Paul Richards