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Math Help - Calculate !

  1. #1
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    Calculate !

    Hii !

    \tan(a) and \tan(b) are the solutions of the equation :

    x^2+\pi x+\sqrt{2}=0

    calculate: sin^2(a+b)+\pi\sin(a+b)\cos(a+b)+\sqrt{2}cos^2(a+b  )
    Last edited by Perelman; February 3rd 2010 at 09:21 AM.
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  2. #2
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    Quote Originally Posted by Perelman View Post
    Hii !

    \tan(a) and \tan(b) are the solutions of the equation :

    x^2+\pi x+\sqrt{2}=0

    calculate: \sin^2(a+b)+\pi\sin(a+b)\cos(a+b)+\sqrt{2}\cos^2(a  +b)
    If \tan a and \tan b are the solutions of the equation x^2+px+q=0 then \tan a + \tan b = -p and \tan a\tan b = q. Thus \tan(a+b) = \frac{\tan a + \tan b}{1-\tan a\tan b} = \frac{-p}{1-q}. Also, \sec^2(a+b) = 1+\tan^2(a+b) = \frac{p^2+(1-q)^2}{(1-q)^2}, and so \cos^2(a+b) = \bigl(\sec^2(a+b)\bigr)^{-1} = \frac{(1-q)^2}{p^2+(1-q)^2}.

    Then \sin^2(a+b)+p\sin(a+b)\cos(a+b)+q\cos^2(a+b) = \cos^2(a+b)\bigl(\tan^2(a+b)+p\tan(a+b)+q\bigr), and you can check by using the results of the previous paragraph that this simplifies to q (which in this case is \sqrt2).
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