Results 1 to 4 of 4

Math Help - Exam question

  1. #1
    Newbie
    Joined
    Jun 2008
    Posts
    2

    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Aug 2007
    From
    USA
    Posts
    3,110
    Thanks
    2
    Try a = -3 and b = 1. I really don't know how humans are supposed to find that.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member flyingsquirrel's Avatar
    Joined
    Apr 2008
    Posts
    802
    Hello

    \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 \frac{1}{\sin(2x)} :

    \frac{\sin x}{\sin (2x)}<br />
\left( \frac{1}{\sin x}+\frac{1}{\sin (3x)}\right)=\frac{\cos(ax) \sin(bx)+\sin(ax)\cos(bx)}{\sin(ax)\sin(bx)}

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

    \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
    \begin{aligned}\sin (3x)+\sin x<br />
&=\sin (2x+x)+\sin(2x-x)\\<br />
&=\sin(2x)\cos x+\sin x \cos(2x)+\sin(2x)\cos x-\sin x\cos(2x)\\<br />
&=2\sin(2x)\cos x<br />
\end{aligned}
    thus the equality becomes

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

    which can be written
    \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 a and b without knowing them seems to be a bit more difficult.
    Last edited by flyingsquirrel; June 18th 2008 at 08:59 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Jun 2008
    Posts
    2
    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; June 21st 2008 at 02:11 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: February 10th 2013, 02:11 PM
  2. Exam Question
    Posted in the Business Math Forum
    Replies: 3
    Last Post: July 30th 2012, 05:03 AM
  3. Need help with exam question...
    Posted in the Calculus Forum
    Replies: 1
    Last Post: January 21st 2010, 01:00 PM
  4. Exam Question
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: January 8th 2009, 03:48 PM
  5. exam question
    Posted in the Calculus Forum
    Replies: 4
    Last Post: July 1st 2007, 03:19 AM

Search Tags


/mathhelpforum @mathhelpforum