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Pythagoras's theorem proof

  1. Aug 13, 2013 #1
    I found this proof of Pythagoras theorem, I need to find if it's "legitimate" and if it's correct. I mean by that: does the trigonometric relations I used need Pythagoras to be proved , or they were found by another method. Thank you in advance.
     

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  3. Aug 13, 2013 #2
    They are found in much simpler methods, but it looks right.
     
  4. Aug 13, 2013 #3

    symbolipoint

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  5. Aug 13, 2013 #4

    Ray Vickson

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    Your proof is invalid: it uses ##\sin(\theta) = \sqrt{1-\cos^2(\theta)},## but this property requires Pythagoras in its proof!
     
  6. Aug 13, 2013 #5

    vanhees71

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    It depends on how you define the trigonometric functions. If you use the definition over their series
    [tex]\cos x=\sum_{k=0}^{\infty} \frac{(-1)^k}{(2k)!} x^{2k}, \quad \sin x=\sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)!} x^{2k+1},[/tex]
    which defines these functions for all [itex]\mathbb{R}[/itex], you can easily prove that
    [tex]\cos' x=-\sin x, \quad \sin' x=\cos x[/tex]
    and from this
    [tex]\frac{\mathrm{d}}{\mathrm{d} x}[\cos^2 x+\sin^2 x]=0 \; \Rightarrow \cos^2 x+\sin^2 x=\text{const},[/tex]
    and from that, because of [itex]\cos 0=1, \quad \sin 0=0[/itex] you get [itex]\cos^2 x + \sin^2 x=1[/itex].
    Then you can define the number [itex]\pi/2[/itex] as the smallest positive solution of [itex]\cos(\pi/2)=0[/itex].

    In this way you can derive step by step all the properties of sin and cos without using any geometric definitions. Then you can prove the geometrical properties of these functions from the unit circle's standard parametrization
    [tex]x=\cos \varphi, \quad y=\sin \varphi, \quad \varphi \in [0,2 \pi).[/tex]
    This is, however, obviously not precalculus anymore.
     
  7. Aug 13, 2013 #6
    Well about this equality I saw on Wikipedia that (sin(x))^2=(1-cos(2x))/2 is obtained from the cosine double-angle formula which can be proved without Pythagoras. And that's how I can find this equality without Pythagoras.
     
  8. Aug 13, 2013 #7
    Thank you very much, I should have thought of that earlier. Now I can confirm that the proof is right.
     
  9. Aug 13, 2013 #8

    Ray Vickson

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    Of course---Rudin does it this way. But that is hardly pre-calculus, as you say.
     
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