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Thread: Exam question

  1. #1
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    Cool Exam question

    Hi everyone,

    This was one of the questions on my exam. Nobody was able to solve it, so I'm starting to wonder if there is a solution to this problem. Please have a look at it:


    sinx .(1/sinx . 1/sin2x + 1/sin2x . 1/sin3x) = cot (ax) + cot (bx)

    What's the value of a and b?

    I hope for you people who put effort in this that there is a solution to this question. Would be kind of a stupid exam question if there isn't, wouldn't it?

    Thanks
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  2. #2
    MHF Contributor
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    Try a = -3 and b = 1. I really don't know how humans are supposed to find that.
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  3. #3
    Super Member flyingsquirrel's Avatar
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    Hello

    $\displaystyle \sin x\left(\frac{1}{\sin x}\cdot\frac{1}{\sin (2x)}+\frac{1}{\sin (2x)}\cdot\frac{1}{\sin (3x)}\right)=\frac{\cos(ax)}{\sin(ax)}+\frac{\cos( bx)}{\sin(bx)}$

    Let's factor $\displaystyle \frac{1}{\sin(2x)}$ :

    $\displaystyle \frac{\sin x}{\sin (2x)}
    \left( \frac{1}{\sin x}+\frac{1}{\sin (3x)}\right)=\frac{\cos(ax) \sin(bx)+\sin(ax)\cos(bx)}{\sin(ax)\sin(bx)}$

    Using $\displaystyle \sin[(a+b)x]=\cos(ax) \sin(bx)+\sin(ax)\cos(bx)$ :

    $\displaystyle \frac{\sin x}{\sin (2x)}\cdot\frac{\sin (3x)+\sin x}{\sin(3x)\sin x}=\frac{\sin[(a+b)x]}{\sin(ax)\sin(bx)}$

    Notice that
    $\displaystyle \begin{aligned}\sin (3x)+\sin x
    &=\sin (2x+x)+\sin(2x-x)\\
    &=\sin(2x)\cos x+\sin x \cos(2x)+\sin(2x)\cos x-\sin x\cos(2x)\\
    &=2\sin(2x)\cos x
    \end{aligned}$
    thus the equality becomes

    $\displaystyle \frac{2\sin (2x)\cos x}{\sin(3x)\sin(2x)}=\frac{\sin[(a+b)x]}{\sin(ax)\sin(bx)}$

    which can be written
    $\displaystyle \sin(ax)\times 2\sin(bx)\cos x=\sin(3x)\times\sin[(a+b)x]$

    From this last equality one can see more easily that the solution given by TKHunny is correct... but finding the values of $\displaystyle a$ and $\displaystyle b$ without knowing them seems to be a bit more difficult.
    Last edited by flyingsquirrel; Jun 18th 2008 at 08:59 AM.
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  4. #4
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    Thank you very much for the help. It's much appreciated.

    This forum rules

    By the way, there are two possible solutions to this problem:

    a=1 and b=-3
    or
    a=-3 and b=1
    Last edited by MathMatters; Jun 21st 2008 at 02:11 AM.
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