# Identities.

• Feb 2nd 2010, 06:41 PM
Identities.
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$\displaystyle{(a^2+b^2)(q^2+p^2) = (aq+bp)^2+(ap-bq)^2}$

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Where can I find more elementary identities like these?

Ps. If you know any, I will be interested in buying it. lol.
• Feb 2nd 2010, 09:51 PM
Prove It
Quote:

$\displaystyle{\dfrac{a^2+b^2}{p^2+q^2} = \left(\dfrac{ap+bq}{p^2+q^2}\right)^2}+\left(\frac {aq-bp}{p^2+q^2}\right)^2$

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

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

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?

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

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

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

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

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

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

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

$= \frac{a^2 + b^2}{p^2 + q^2}$.
• Feb 5th 2010, 10:16 AM