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Math Help - Deduce to inequality

  1. #1
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    Deduce to inequality

    By using the formulae expressing sin\theta and cos\theta in terms of t, where (t\equiv tan\frac{\theta}{2}) or otherwise, show that \frac{1+sin\theta}{5+4cos\theta}\equiv\frac{(1+t)^  2}{9+t^2}.
    Deduce that 0\leq\frac{1+sin\theta}{5+4cos\theta}\leq\frac{10}  {9}
    I have done the first part using the identities sin\theta\equiv\frac{2t}{1+t^2} and cos\theta\equiv\frac{1-t^2}{1+t^2} where t\equiv tan\frac{\theta}{2}, but I don't know how to do the second part.
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  2. #2
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    Quote Originally Posted by arze View Post
    By using the formulae expressing sin\theta and cos\theta in terms of t, where (t\equiv tan\frac{\theta}{2}) or otherwise, show that \frac{1+sin\theta}{5+4cos\theta}\equiv\frac{(1+t)^  2}{9+t^2}.
    Deduce that 0\leq\frac{1+sin\theta}{5+4cos\theta}\leq\frac{10}  {9}
    I have done the first part using the identities sin\theta\equiv\frac{2t}{1+t^2} and cos\theta\equiv\frac{1-t^2}{1+t^2} where t\equiv tan\frac{\theta}{2}, but I don't know how to do the second part.
    Thanks
    the left side of the inequality is clear. to prove the right side we have: \frac{(1+t)^2}{9+t^2} -\frac{10}{9}=\frac{9(1+t)^2 - 10(9+t^2)}{9(9+t^2)}=\frac{-t^2+18t-81}{9(9+t^2)}=\frac{-(t-9)^2}{9(9+t^2)} \leq 0.

    but what if the question was this: find the maximum value of f(t)=\frac{(1+t)^2}{9+t^2}, \ t \in \mathbb{R}? (no calculus allowed!)
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  3. #3
    MHF Contributor red_dog's Avatar
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    We have to prove that 0\leq\frac{(1+t)^2}{9+t^2}\leq\frac{10}{9}

    The first inequality is obviously true.

    For the second:

    \frac{(1+t)^2}{9+t^2}\leq\frac{10}{9}\Leftrightarr  ow 9+18t+9t^2\leq 90+10t^2\Leftrightarrow t^2-18t+81\geq 0\Leftrightarrow(t-9)^2\geq 0
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  4. #4
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     <br />
\frac{1 + \sin\theta}{5 + 4\cos{\theta}}<br />

     = \frac{ (\sin{\theta/2} + \cos{\theta/2})^2}{1 + 4(2)(\cos^2{\theta/2})}

     = \cos^2{\theta/2} \frac{ (1 + \tan{\theta/2})^2}{ \sin^2{\theta/2} + 9\cos^2{\theta/2}}

     = \frac{ (1+\tan{\theta/2})^2}{ 9 + \tan^2{\theta/2}}


    for part 2 , besides using  b^2 - 4ac \geq 0
    here is another method :

     \frac{(1+t)^2 }{t^2 + 9}

     = \frac{ t^2 + 9 + 2t - 8}{(t-4)^2 + 8(t-4) + 25}

     = 1 + \frac{2}{ (t-4) + 8 + \frac{25}{t-4}}

     \leq 1 + \frac{2}{2\sqrt{25} + 8}

     = \frac{ 10}{9}
    Last edited by simplependulum; July 15th 2009 at 12:57 AM.
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