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Math Help - Identities.

  1. #1
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    Identities.

    [LaTeX ERROR: Convert failed]

    \displaystyle{(a^2+b^2)(q^2+p^2) = (aq+bp)^2+(ap-bq)^2}

    [LaTeX ERROR: Convert failed]

    Where can I find more elementary identities like these?

    Ps. If you know any, I will be interested in buying it. lol.
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  2. #2
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    Quote Originally Posted by Captcha View Post
    \displaystyle{\dfrac{a^2+b^2}{p^2+q^2} = \left(\dfrac{ap+bq}{p^2+q^2}\right)^2}+\left(\frac  {aq-bp}{p^2+q^2}\right)^2

    \displaystyle{(a^2+b^2)(q^2+p^2) = (aq+bp)^2+(ap-bq)^2}

    \displaystyle{\frac{u}{v} = \frac{1}{4}\left[\left(u+\frac{1}{v}\right)^2-\left(u-\frac{1}{v}\right)^2\right]}

    Where can I find more elementary identities like these?

    Ps. If you know any, I will be interested in buying it. lol.
    What are you trying to do? Prove these?


    \left(\frac{ap + bq}{p^2 + q^2}\right)^2 + \left(\frac{aq - bp}{p^2 + q^2}\right)^2 = \frac{(ap + bq)^2}{(p^2 + q^2)^2} + \frac{(aq - bp)^2}{(p^2 + q^2)^2}

     = \frac{(ap + bq)^2 + (aq - bp)^2}{(p^2 + q^2)^2}

     = \frac{a^2p^2 + 2abpq + b^2q^2 + a^2q^2 - 2abpq + b^2p^2}{(p^2 + q^2)^2}

     = \frac{a^2p^2 + b^2q^2 + a^2q^2 + b^2p^2}{(p^2 + q^2)^2}

     = \frac{a^2p^2 + a^2q^2 + b^2p^2 + b^2q^2}{(p^2 + q^2)^2}

     = \frac{a^2(p^2 + q^2) + b^2(p^2 + q^2)}{(p^2 + q^2)^2}

     = \frac{(a^2 + b^2)(p^2 + q^2)}{(p^2 + q^2)^2}

     = \frac{a^2 + b^2}{p^2 + q^2}.
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  3. #3
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    Quote Originally Posted by Prove It View Post
    What are you trying to do? Prove these?
    I wanted to collect few useful identities like these.

    Thanks for proving, though. Goes with your name, eh?
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