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0|0|0|||||Trigonometric proof of the Pythagorean Theorem|jdawson||22:05:35|04/02/2011|While I find Jason Zimba%27s proof interesting%2C I am puzzled why it should have taken so long for someone to cast a proof of the Pythagorean Theorem in trigonometric form. To see that the geometric definitions of the trigonometric functions are well-defined requires similarity theory. But granting that%2C it is easy to modify Euclid%27s similarity proof %28VI%2C31%29 without resort to the sum or difference identities. For let c be the hypotenuse of right triangle ABC%2C and draw the altitude from the opposite angle%2C C. Then %0D%0Ac %3D a cos B %2B b cos A%2C and similar triangles yield that%0D%0Aa%2F%28a cos B%29 %3D c%2Fa  and b%2F%28b cos A%29 %3D c%2Fb. Replacing c in the last two equations by the right member of the first equation and then cross-multiplying gives%0D%0A%0D%0Aa%5E2 %3D a%5E2 %28cos B%29%5E2 %2B ab cos A cos B  and  %0D%0Ab%5E2 %3D ab cos A cos B %2B b%5E2 %28cos A%29%5E2 . Hence%0D%0A%0D%0Aa%5E2 %2B b%5E2 %3D a%5E2 %28cos B%29%5E2 %2B2ab cos A cos B %2B b%5E2 %28cos A%29%5E2 %0D%0A  %3D %28a cos B %2B b cos A%29%5E2 %3D c%5E2 .%0D%0A%0D%0AThis would seem to me to be a %22purely trigonometric proof%22. Am I missing something%3F%0D%0A%0D%0A%0D%0A
1|1|0|||||RE%3A Trigonometric proof of the Pythagorean Theorem|alexb||22:48:22|04/03/2011|The problem is actually reverse.%0D%0A%0D%0AAlmost every proof may have been cast in a trigonometric form. I would not care about that. The question is%2C given a trigonometric proof%2C how easy it would be to recast it into the proportionality terms. That may be a subjective call. I judged that%2C for Zimba%27s proof%2C such conversion would be cumbersome. For this reason I acknowledged the proof as trigonometric.