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January 11th, 2013, 07:22 PM  #1 
Newbie Joined: Jan 2013 Posts: 3 Thanks: 0  Radical And trigonometry simplification
The following expression pertains to the fact that the sum of the distances from the foci to any point on an ellipse is 2a, where a is the radius along the major axis. I have verified this equation for various values of a, b, and theta, so I know that it is correct. I don't know how to verify it algebraically, though. Can anyone show me how to do it? I've tried squaring it, but it just got really nasty. 
January 11th, 2013, 07:28 PM  #2 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs  Re: Radical And trigonometry simplification
Did you try arranging the equation with a radical on either side, then squaring twice? I believe that's what's usually done with ellipses and hyperbolas.

January 11th, 2013, 07:42 PM  #3 
Newbie Joined: Jan 2013 Posts: 3 Thanks: 0  Re: Radical And trigonometry simplification
Sounds like you're suggesting something like I expanded both the left and right sides in maple, and it seems to just generate more complex terms that are difficult to cancel :\ The issues come from the interior terms of the expansion that do not eliminate the square roots. 
January 11th, 2013, 07:50 PM  #4 
Newbie Joined: Jan 2013 Posts: 3 Thanks: 0  Re: Radical And trigonometry simplification
Nevermind haha, I think I get what to do now. This reference seems to show exactly what I'm trying to do http://en.wikipedia.org/wiki/Proofs_inv ... he_ellipse Thanks for the help, and you're right about the double squaring! 
January 11th, 2013, 07:53 PM  #5 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs  Re: Radical And trigonometry simplification
Yes, you've got radicals under radicals...perhaps if you rewrite the expressions a different way. For convenience, let F be half the interfocal distance, where: 

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