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Homework Help: Trigonometric Identities

  1. Sep 11, 2010 #1
    1. The problem statement, all variables and given/known data

    I got this expression while solving a problem.

    (y+1)/(y-1)=sin2x/sin2a

    we need to find the range of y

    2. Relevant equations



    3. The attempt at a solution

    here y=(sin2x+sin2a)/(sin2x-sin2a)

    Numerator of RHS lies between sin2a-1 and 1+sin2a
    Denominator lies between -1-sin2a and 1-sin2a

    so RHS lies between (sin2a-1)/(-1-sin2a) and (1+sin2a)/(1-sin2a)

    on solving I got y lies between tan^2 (45-a) and tan^2 (45+a)

    but the answer is opposite (i.e. y does not lie between them).
    Please help.
     
  2. jcsd
  3. Sep 11, 2010 #2

    HallsofIvy

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    Did you remember that if a< x< b then (1/b)< 1/x< (1/a)? That, of course, is assuming that all numbers are positive. Here, since the denominator lies between -1- sin(2a) and 1- sin(2a) it can be both negative and positive and, for some x, will be 0. Of course, if the denominator goes to 0, the fraction goes to infinity so the domain can't be "between" two values.
     
  4. Sep 11, 2010 #3
    We have to find the range of y. Not its domain.
     
  5. Sep 12, 2010 #4

    hunt_mat

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    One thing that could possibly help you with this is the formulae:
    [tex]
    \sin A+\sin B=2\sin\left(\frac{a+b}{2}\right)\cos\left(\frac{a-b}{2}\right)
    [/tex]
    And
    [tex]
    \sin A-\sin B=2\cos\left(\frac{a+b}{2}\right)\sin\left(\frac{a-b}{2}\right)
    [/tex]
    The only question is now, is what is the domain of y? Once you can tell us this then we can determine the range of y
     
  6. Sep 12, 2010 #5

    vela

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    Let [itex]\beta=\sin 2x/\sin 2a[/itex]. Then

    [tex]y=\frac{\beta+1}{\beta-1}[/tex]

    Try plotting y vs. β and then consider what the range of β is. Then perhaps you'll understand HallsofIvy's point.
     
  7. Sep 12, 2010 #6
    Thanks, I understood this time.
     
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