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Homework Help: Identity for |sinx-siny|

  1. Jan 24, 2006 #1
    does anybody know the identity for |sinx-siny| and |cosx-cosy|?
     
  2. jcsd
  3. Jan 24, 2006 #2

    StatusX

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    What exactly do you want these identities to contain? sin(x+y)'s and cos(x+y)'s? I don't see how you could make these expressions much simpler.
     
  4. Jan 25, 2006 #3

    VietDao29

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    Are you looking for some Sum-to-product identities?
    If yes, then here are the four identities:
    [tex]\cos \alpha + \cos \beta = 2 \cos \left( \frac{\alpha + \beta}{2}
    \right) \cos \left( \frac{\alpha - \beta}{2} \right)[/tex].
    [tex]\cos \alpha - \cos \beta = -2 \sin \left( \frac{\alpha + \beta}{2} \right) \sin \left( \frac{\alpha - \beta}{2} \right)[/tex].
    [tex]\sin \alpha + \sin \beta = 2 \sin \left( \frac{\alpha + \beta}{2} \right) \cos \left( \frac{\alpha - \beta}{2} \right)[/tex].
    [tex]\sin \alpha - \sin \beta = 2 \cos \left( \frac{\alpha + \beta}{2} \right) \sin \left( \frac{\alpha - \beta}{2} \right)[/tex].
    --------------
    From the 4 identities above, one can easily show that:
    [tex]| \sin \alpha - \sin \beta | = 2 \left| \cos \left( \frac{\alpha + \beta}{2} \right) \sin \left( \frac{\alpha - \beta}{2} \right) \right|[/tex].
    and:
    [tex]| \cos \alpha - \cos \beta | = \left| -2 \sin \left( \frac{\alpha + \beta}{2} \right) \sin \left( \frac{\alpha - \beta}{2} \right) \right| = 2 \left| \sin \left( \frac{\alpha + \beta}{2} \right) \sin \left( \frac{\alpha - \beta}{2} \right) \right|[/tex].
    Is that what you are looking for?
    And that's not any simpler than your original expressions.
     
  5. Jan 25, 2006 #4

    matt grime

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    It is simpler, because you can drop with abs value signs using the odd or evenness of sin and cos respectively.
     
  6. Jan 25, 2006 #5

    arildno

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    As has been stated, it is crucial that you specify what sort of identity you'
    re after.

    For example, the following identity holds (for all x,y):
    |sin(x)-sin(y)|=|sin(x)-sin(y)|+0
     
  7. Jan 25, 2006 #6
    Hi,

    This is an inequality ..


    [tex] | \sin x - \sin y | \leq | x - y | [/tex]
     
  8. Jan 25, 2006 #7
    Maybe this is what you're looking for...

    |sinx - siny| = 2 * |{sin(x-y)/2} * {cos(x+y)/2}|

    |cosx-cosy| = 2 * |{sin(x+y)/2} * {sin(x-y)/2}|


    Spacetime
    Physics
     
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