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Geometric theorems

  1. Apr 20, 2005 #1
    Can someone tell me (or help me find) the derivation of the pythagorean theorem, and the laws of sin,cos, and tangent. I know the first is a derivation of the low of cosins, but I'd like to know if theres a writeout as to how he actually came up with those results.
  2. jcsd
  3. Apr 20, 2005 #2
    There are many proofs to the phytagorean theorem but this is one of my favourite. It was told to me as the applied mathematicians proof of phythagorous. Take a triangle with hypotinuse h, width x, and height y. By a dimension argument the area is [tex]ch^2[/tex] where c is a dimensionless constant. Now draw a line from the right angle that meets the hypotinous at a right angle. Now you have two triangles similar to the original with hypotinuses x and y. So the are of each of these is [tex]cx^2[/tex] and [tex]cy^2[/tex]. And they sum up to the total area so [tex] cx^2 + cy^2 = ch^2 [/tex]

    This surely wasn't the original proof but it's very much how an applied mathematician thinks.
  4. Apr 20, 2005 #3
    The law of cos is just a generalization of the Pythagorean theorem, let theta = 90. The derivation of the law of cos I don’t remember off hand but it’s just about every trig and pre-cal book.
  5. Apr 20, 2005 #4
    Isnt that backwards, jonf?
  6. Apr 20, 2005 #5


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    The Pythagorean theorem long predates the Law of Cosines.
  7. Apr 20, 2005 #6
    I'm not following this part, can you elaborate? What 'dimension argument'? Are we drawing any triangle or a right triangle?
  8. Apr 20, 2005 #7
    The law of cos is just a generalization of the Pythagorean theorem, let theta = 90


    a^2+b^2 = c^2, or is it

    a^2 + b^2 -2abcos(t) = c^2?
  9. Apr 20, 2005 #8

    Yeah that is definitely the big jump and this is the sort of stuff some applied mathematicians (especially russian ones) tend to just sweep under the rug. You can do this with any triangle or polygon. Just take any side length and say it is x cm (centimetres). The area will be in cm's squared. So the function between sidelength to area is some multiple of the square of the length since the units have to match and you can't seperate the length from the unit. So the constant changes for different triangles, even among right triangles.

    Again this is not the sort of proof a Euclidean geometer would come up with. The typical proof is sort of a jig saw puzzle.
  10. Apr 20, 2005 #9


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    It don't believe one can say HOW Pythagoras himself proved the "Pythagorean theorem" (he certainly did NOT derive it from the cosine law since cosines handn't been invented then!). There are probably more different proofs of the Pythagorean theorem than any other single theorem. Even a president of the United States (James Garfield) developed an original proof- given here: http://jwilson.coe.uga.edu/emt669/Student.Folders/Huberty.Greg/Pythagorean.html
  11. Apr 20, 2005 #10
    Thanks ivy thats exactly what I was looking for.
  12. Apr 20, 2005 #11
    Here's a diagram for another proof. I'll leave it to you to show the angles work out. Just remember the sum of interior angles is 180degrees.

    Attached Files:

  13. Apr 21, 2005 #12
    Remeber that back in the days of the Greeks there were no sturctured algerbra like we have now. There proofs were all based of geometry.
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