1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Homework Helper

    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

    User Avatar
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Identity for |sinx-siny|
  1. Derivative of |sinx| (Replies: 6)

Loading...