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Prove cos2(x)+sin2(x) = 1

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    (Original post by ben-smith)
    -Euler Polyhedron formula
    Any proof of this is going to use Gauss-Bonnet somewhere, I'm willing to bet.
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    (Original post by around)
    Any proof of this is going to use Gauss-Bonnet somewhere, I'm willing to bet.
    hmmm. Are we talking about the same thing?:
    any convex polyhedron's surface has an euler characteristic 2?
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    (Original post by ben-smith)
    hmmm. Are we talking about the same thing?:
    any convex polyhedron's surface has an euler characteristic 2?
    http://en.wikipedia.org/wiki/Gauss%E...Bonnet_theorem (under special cases: polyhedra)

    The Euler characteristic (stop me if I'm boring you) is a topological invariant of surfaces, and Gauss-Bonnet relates the topological properties of a surface with its geometry.
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    (Original post by around)
    http://en.wikipedia.org/wiki/Gauss%E...Bonnet_theorem (under special cases: polyhedra)

    The Euler characteristic (stop me if I'm boring you) is a topological invariant of surfaces, and Gauss-Bonnet relates the topological properties of a surface with its geometry.
    I asked because I'd seen proofs* of the polyhedron theorem that didn't require gauss-bonnet.

    *Perhaps these "proofs" weren't rigorous...
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    (Original post by ben-smith)
    I asked because I'd seen proofs* of the polyhedron theorem that didn't require gauss-bonnet.

    *Perhaps these "proofs" weren't rigorous...
    These?

    There are also given in "Proof and Refutations".
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    (Original post by ukdragon37)
    These?

    There are also given in "Proof and Refutations".
    yes!
    It was a while ago so I couldn't remember where I had seen them.
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    Some of the proofs below use only the topology of the planar graph, some use the geometry of its embedding, and some use the three-dimensional geometry of the original polyhedron
    So basically they all use Gauss-Bonnet in some way.

    (you have to understand that Gauss-Bonnet is pretty deep)
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    (Original post by around)
    So basically they all use Gauss-Bonnet in some way.

    (you have to understand that Gauss-Bonnet is pretty deep)
    I see. Well, you learn something every day
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    (Original post by ben-smith)
    Any ideas for a similar thread? Some possible ones:
    prove: -the derivative of e^x is e^x
    -Euler Polyhedron formula
    The idea for other threads sounds really nice.

    However, that was a desired result, wasn't it?
    I mean, the exponential function was defined to have this property. It will be interesting to see an independent proof of this.

    How about, prove that \sqrt{2} is irrational. To be honest, I haven't seen more than two proofs of this.
    It is kind of embarrassing to know that it will take me, like, forever, just to draw the hypotenuse of a right angled triangle.
    We can never draw this triangle with absolute accuracy, if we start with the hypotenuse.

    *
    A funny anecdote:

    "Prove or disprove that, given a line and an independent point, there exists at most one straight line, parallel to the given one, which can be drawn through that point."
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    (Original post by gff)
    "Prove or disprove that, given a line and an independent point, there exists at most one straight line, parallel to the given one, which can be drawn through that point."
    ...in euclidean space?
    This postulate is not true in hyperbolic geometry which is equiconsistent with euclidean geometry. I think.
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    This is an anecdote.

    As far as I know, this is an axiom derived from Euclid's parallel postulate, which you can't prove, but neither you can disprove the existence of such a line.
    I may be wrong, it will be good if somebody clarifies this. Thanks.

    (maybe it didn't have "straight" in the original, I don't remember, but I don't really know if this matters )
    I don't know much about Bolyai's ideas, neither what Gauss and Riemann did with all this. I have a long time until Riemannian geometry.

    I think they've just replaced the axiomatic system used by Euclid and created this new geometry, but this does not "disprove" Euclidean's geometry?
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    (Original post by gff)
    This is an anecdote.

    As far as I know, this is an axiom derived from Euclid's parallel postulate, which you can't prove, but neither you can disprove the existence of such a line.
    I may be wrong, it will be good if somebody clarifies this. Thanks.

    (maybe it didn't have "straight" in the original, I don't remember, but I don't really know if this matters )
    I don't know much about Bolyai's ideas, neither what Gauss and Riemann did with all this. I have a long time until Riemannian geometry.

    I think they've just replaced the axiomatic system used by Euclid and created this new geometry, but this does not "disprove" Euclidean's geometry?
    What you mentioned was playfair's axiom which is logically equivalent to euclid's parallel postulate. The point is that Hyperbolic, Euclidean geometry are equally valid and so the parallel postulate is independent of the other axioms as it is not necessarily true.
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    (Original post by ben-smith)
    Any ideas for a similar thread? Some possible ones:
    prove: -the derivative of e^x is e^x
    -Euler Polyhedron formula
    You can prove the  e^x differentiation using the definitions of sinhx and coshx and their derivatives, but they are themselves exponentials, so it's kind of moot and circular, but the only other way I know how to do it (which I've forgotten) is by exhaustion.

    I'll do it anyway:

     \frac{d}{dx}sinhx = coshx
     \frac{d}{dx}coshx = sinhx

     sinhx = \frac {e^x-e^{-x}}{2}
     coshx = \frac {e^x+e^{-x}}{2}

    Therefore,  sinhx + coshx = e^x

     \frac {d}{dx} (sinhx + coshx) = coshx + sinhx .

    Therefore,  \frac {d}{dx} e^x = e^x
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    (Original post by TwilightKnight)
    You can prove the  e^x differentiation using the definitions of sinhx and coshx and their derivatives, but they are themselves exponentials, so it's kind of moot and circular, but the only other way I know how to do it (which I've forgotten) is by exhaustion.

    I'll do it anyway:

     \frac{d}{dx}sinhx = coshx
     \frac{d}{dx}coshx = sinhx

     sinhx = \frac {e^x-e^{-x}}{2}
     coshx = \frac {e^x+e^{-x}}{2}

    Therefore,  sinhx + coshx = e^x

     \frac {d}{dx} (sinhx + coshx) = coshx + sinhx .

    Therefore,  \frac {d}{dx} e^x = e^x
    Showing my ignorance here once again, isn't the usual way to just make use of the derivative of lnx? (which you determine by the definition of e)

    (ln(x+h)-lnx)/h etc.

    Then of course y = ln(x) so dy/dx = 1/x = e^-y so dx/dy = e^y and hence
    d/dy e^y = e^y ?
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    (Original post by Jodin)
    Showing my ignorance here once again, isn't the usual way to just make use of the derivative of lnx? (which you determine by the definition of e)

    (ln(x+h)-lnx)/h etc.

    Then of course y = ln(x) so dy/dx = 1/x = e^-y so dx/dy = e^y and hence
    d/dy e^y = e^y ?
    I wouldn't know, to be honest.

    I don't think we're aiming at elementary proofs here (a lot of people used definitions of sin and cos that were in themselves based on other definitions) so I would imagine using already defined values like the natural log (which is base e) and mine are ok for the purpose we're using it for.
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    (Original post by L'art pour l'art)
    \begin{aligned}1 & = \cos(0) \\& = \cos(x-x) \\& = \cos{x}\cos{x}+\sin{x}\sin{x} \\& = \cos^2{x}+\sin^2{x}.\end{aligned  }
    thats brilliant!
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    (Original post by jsmith6131)
    thats brilliant!
    I liked it too but it requires you to expand cos(A-B) which is more tricky than proving Pythagoras in the first place!
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    If we are allowing trig identities what about using half-tangent identities?

    \sin^2 x + \cos^2 x = (\frac{2t}{1+t^2})^2 + (\frac{1-t^2}{1+t^2})^2 = \frac{4t^2 + 1 - 2t^2 + t^4}{(1+t^2)^2} = \frac{(1+t^2)^2}{(1+t^2)^2}=1
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    (Original post by ben-smith)
    Any ideas for a similar thread? Some possible ones:
    prove: -the derivative of e^x is e^x
    -Euler Polyhedron formula
    Probably not very clever, but this is what makes sense to me:

    \frac{d}{dx}(ln(e^x)) = \frac{d}{dx} (xlne)= \frac{d}{dx}x = 1

    Edit: ah, crap, I just noticed that your post is about 4 months old.
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    (Original post by Dadeyemi)
    Well this is just for fun,
    I proved it using some matrices stuff but i wanna know if any1 else has an interesting proof.

    Prove cos2(x) + sin2(x) = 1

    see how many different proofs we can get.

    my proof is added as an attachment.
    Love this!

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