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Using the concept of finite difference

  1. Aug 2, 2005 #1
    I am trying to find the ▲cos x. By using its definition.

    It simply turned out to be cos(x+1) - cos(x). How do I express this in
    terms of sine? (and as only one term)?
     
  2. jcsd
  3. Aug 2, 2005 #2

    HallsofIvy

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    cos(x+1)= cos(x)cos(1)- sin(x) sin(1) so cos(x+1)- cos(x)= cos(x)(cos(1)-1)- sin(x)sin(1). What makes you think you can write it in terms of sine only?
     
  4. Aug 2, 2005 #3

    lurflurf

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    That is a completely correct equation for the difference. It is also not a particularly useful form. The standard form for such things in finite calculus is
    [tex]{\Delta}^n\cos(x+a)=(2\sin(\frac{h}{2}))^n\cos(x+a+\frac{h}{2}n+\frac{\pi}{2}n)[/tex]
    in this specific case where a=0 n=1 h=1
    [tex]\Delta\cos(x)=-2\sin(\frac{1}{2})\sin(x+\frac{1}{2})[/tex]
    This is one term additive, but two terms multiplicative.
     
    Last edited: Aug 2, 2005
  5. Aug 2, 2005 #4
    [tex] sin(x+1 - \frac{\pi}{4}) - sin(x-\frac{\pi}{4}) [/tex]
     
  6. Aug 2, 2005 #5

    lurflurf

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    little off
    you may have been attempting to use the cofunctional identity
    [tex]\cos(x)=\sin(\frac{\pi}{2}-x)[/tex]
    getting
    [tex]\sin(\frac{\pi}{2}-x-1) - \sin(\frac{\pi}{2}-x)[/tex]
    or equivalently
    [tex]\sin(x-\frac{pi}{2})-\sin(x+1-\frac{\pi}{2})[/tex]
    but this form is not perfered for most purposes that would be
    [tex]-2\sin(\frac{1}{2})\sin(x+\frac{1}{2})[/tex]
     
    Last edited: Aug 2, 2005
  7. Aug 3, 2005 #6
    Excuse me, how was [tex]-2\sin(\frac{1}{2})\sin(x+\frac{1}{2})[/tex]
    derived? My derivation just ended with that of HallsofIvy.
     
  8. Aug 3, 2005 #7

    lurflurf

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    A well known identity in trigonometry is
    [tex]\cos(a)-\cos(b)=-2\sin(\frac{a-b}{2})\sin(\frac{a+b}{2})[/tex]
    The result is mode clear by taking a=x+1; b=x
    This identity can be derived by writing
    a=(a+b)/2+(a-b)/2
    b=(a+b)/2-(a-b)/2
    Then using the identity
    cos(a+b)=cos(a)cos(b)-sin(a)sin(b)
    and then adding the like terms
     
  9. Aug 4, 2005 #8
    Thank you very much.
     
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