My Math Forum Radical And trigonometry simplification

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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. $2a=\sqrt{\left(a \cdot \cos{\theta}+\sqrt{a^2-b^2}\right)^2+b^2\cdot \sin^2{\theta}}+\sqrt{\left(a \cdot \cos{\theta}-\sqrt{a^2-b^2}\right)^2+b^2\cdot \sin^2{\theta}}$ 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 $\left(2a - \sqrt{\left(a \cdot \cos{\theta}+\sqrt{a^2-b^2}\right)^2+b^2\cdot \sin^2{\theta}}\right)^4= \left(\left(a \cdot \cos{\theta}-\sqrt{a^2-b^2}\right)^2+b^2\cdot \sin^2{\theta}\right)^2$ 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: $F=\sqrt{a^2-b^2}$

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