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Thread: Deduce to inequality

  1. #1
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    Deduce to inequality

    By using the formulae expressing $\displaystyle sin\theta$ and $\displaystyle cos\theta$ in terms of t, where $\displaystyle (t\equiv tan\frac{\theta}{2})$ or otherwise, show that $\displaystyle \frac{1+sin\theta}{5+4cos\theta}\equiv\frac{(1+t)^ 2}{9+t^2}$.
    Deduce that $\displaystyle 0\leq\frac{1+sin\theta}{5+4cos\theta}\leq\frac{10} {9}$
    I have done the first part using the identities $\displaystyle sin\theta\equiv\frac{2t}{1+t^2}$ and $\displaystyle cos\theta\equiv\frac{1-t^2}{1+t^2}$ where $\displaystyle t\equiv tan\frac{\theta}{2}$, 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 $\displaystyle sin\theta$ and $\displaystyle cos\theta$ in terms of t, where $\displaystyle (t\equiv tan\frac{\theta}{2})$ or otherwise, show that $\displaystyle \frac{1+sin\theta}{5+4cos\theta}\equiv\frac{(1+t)^ 2}{9+t^2}$.
    Deduce that $\displaystyle 0\leq\frac{1+sin\theta}{5+4cos\theta}\leq\frac{10} {9}$
    I have done the first part using the identities $\displaystyle sin\theta\equiv\frac{2t}{1+t^2}$ and $\displaystyle cos\theta\equiv\frac{1-t^2}{1+t^2}$ where $\displaystyle t\equiv tan\frac{\theta}{2}$, 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: $\displaystyle \frac{(1+t)^2}{9+t^2} -\frac{10}{9}=\frac{9(1+t)^2 - 10(9+t^2)}{9(9+t^2)}=\frac{-t^2+18t-81}{9(9+t^2)}=\frac{-(t-9)^2}{9(9+t^2)} \leq 0.$

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

    The first inequality is obviously true.

    For the second:

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

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

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

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


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

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

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

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

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

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