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A^2 + b^2 = 1 proof

  1. Oct 29, 2013 #1
    given a*(1-b^2)^1/2 +b(1-a^2)^1\2 =1 prove a^2 + b^2 =1

    I tried squaring both sides and then squaring again to get
    a^4 + b^4 -2b^2 -2a^2 +2a^2b^2 +1 =0

    and that could be (a^2 + b^2)(a^2 + b^2) - 2(a^2 + b^2) = -1

    I don't know where to go from there and not sure this is even correct.

    Can someone help???
     
  2. jcsd
  3. Oct 29, 2013 #2

    mfb

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    With c=a^2+b^2, your equation becomes c^2-2c=-1. This has just one solution for c...

    I moved your thread to our homework section, as it looks like a homework question (at least it is very similar to them).
     
  4. Oct 29, 2013 #3

    WWGD

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    Try completing the square, i.e., try to express in the form $$(x+y)^2$$ . If you use

    double $'s


    ( the signs, to do Tex--tho having general $'s



    will always help in daily life too -- you can easily Tex your expressions, e.g:

    (a+b)^2 , using

    Double$'s

    (wrapping around) will get you

    $$(x+y)^2$$ .
     
  5. Oct 29, 2013 #4
    If you don't mind using some trig...

    Since [itex] -1 \leq b \leq 1 [/itex] and [itex] -1 \leq a \leq 1 [/itex] we can make the substitutions [itex] a=sin\alpha [/itex] and [itex] b=sin\beta [/itex]. You should get that if [itex] a=sin\alpha [/itex], then [itex] b=sin(\pi/2-\alpha)= cos\alpha [/itex] which solves the problem. This avoids a lot of algebra.
     
  6. Oct 29, 2013 #5


    OK so $$c^2 -2c +1 =0$$
    then (c-1)(c-1)=0 implies c=1 so
    $$a^2 + b^2 =1$$

    Right??
     
    Last edited: Oct 29, 2013
  7. Oct 29, 2013 #6

    WWGD

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    Right, but I meant to say that you need to choose both a,b here to fit your formula.

    Sorry first-of-all for my poor choice of letter a,b here. I meant you could rewrite your formula

    as a sum $$(x+y)^2 $$, with just the right choice of x,y.
     
  8. Oct 29, 2013 #7

    mfb

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    Right.

    Please use the edit button for code tests.
     
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