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Verifying a Hard Trigonometric Identitiy (please help)

  1. Dec 14, 2005 #1
    My Pre-Cal teacher gave us this problem today. I have worked on it for a very long time and have goten no where :confused:. I was wondering if anyone had any ideas on how to do it, or even where to start. I started it myself with using the pythagorean idenities for the left side, then the difference of square; however after that I didn't know where to go.

    cos²5x - cos²x = -sin4x x sin6x

    Thanks a lot for your help,

    Edit: (Damit... sorry I think I posted this in the wrong place)
    Last edited: Dec 14, 2005
  2. jcsd
  3. Dec 14, 2005 #2


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    Are you sure that's true?

    It's kind of impossible. Because as x goes to infinity the left side remains contained in a small range, where as the right side blows up because you are multiplying by x.
  4. Dec 14, 2005 #3


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    I think that is a multiplication sign and not the variable x.


    HINT: Write the left side as

    [tex](\cos 5x - \cos x)(\cos 5x + \cos x)[/tex]
  5. Dec 14, 2005 #4
    Good hint. I see the difference of squares. Should I factor out a cosx now?

    BTW that x is a multiplication sign.... sorry about that.
    Last edited: Dec 14, 2005
  6. Dec 14, 2005 #5


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    No, you don't want to factor out a cos x.

    Do you know any trig identities involving [itex]\cos a - b[/itex] and [itex]\cos a + b[/itex]?
  7. Dec 14, 2005 #6
    Yes I know both of thoes. Thoes can be used on [tex](\cos 5x - \cos x)(\cos 5x + \cos x)[/tex] ? Whouldn't a cosx have to be on the outside of the () in order to use it?
    Last edited: Dec 14, 2005
  8. Dec 14, 2005 #7


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    cos(5x)= cos(4x+ x)= cos(4x)cos(x)- sin(4x)sin(x)

    cos(4x)= cos(3x+ x)= cos(3x)cos(x)- sin(3x)sin(x) and
    sin(4x)= sin(3x+ x)= sin(3x)cos(x)+ cos(3x)sin(x) so
    cos(5x)= (cos(3x)cos(x)- sin(3x)sin(x))cos(x)- (sin(3x)cos(x)+ cos(3x)sin(x))sin(x)
    = cos(3x)cos2(x)- sin(3x)sin(x)cos(x)- sin(3x)sin(x)cos(x)- cos(3x)sin2(x)= cos(3x)(cos2x- sin2(x))

  9. Dec 14, 2005 #8
    You lost me.:eek: :confused:
  10. Dec 14, 2005 #9
    Whould this bring me anyware:

    cos²5x - cos²x = (-sin4x)(sin6x)
    (1-sin^2(5x)) - (1-sin^2x) =

  11. Dec 14, 2005 #10


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    It's basically the answer minus two-three steps.

    I'd recommend to go through it all step-by-step. Write them all if you have to.
  12. Dec 15, 2005 #11


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    Hmm,... have you learnt the following 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]?
    Back to your problem, we have:
    [tex]\cos ^ 2 (5x) - \cos ^ 2 x = (\cos (5x) - \cos x)(\cos (5x) + \cos x)[/tex]. Now with the 4 identities above, and the double-angle formulas, can you prove that [tex]\cos ^ 2 (5x) - \cos ^ 2 x = - \sin (4x) \sin (6x)[/tex]?
    Can you go from here?
    Last edited: Dec 15, 2005
  13. Aug 17, 2010 #12
    The key is to change the right hand side:

    cos²5x - cos²x = -sin6x.sin4x

    So cos²5x - cos²x = -sin(5x+x).sin(5x-x)

    then expand and simplify the rhs.

    The rest is:

    - sin4x.sin6x = - (sin5x - x).(sin(5x + x)

    = - [(sin5x.cosx - cos5x.sinx).(sin5x.cosx - cos5x.sinx)]

    Tip #3, Difference of two squares

    = - (sin²5x.cos²x - cos²5x.sin²x)

    = - [(1 - cos²5x).cos²x - cos²5x.(1 - cos²x)]

    = - [cos²x - cos²5x.cos²x - cos²5x + cos²5x.cos²x]

    = - (cos²x - cos²5x)

    = cos²5x - cos²x

    = LHS


    The tip numbers refer to trig proof tips in an article I have written,


    Good luck,
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