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A question about trigonometric identities

  1. Aug 29, 2012 #1
    Is this identity possible?

    [tex]cot 2x = \frac{cos3x + cosx}{sin 3x + sinx}[/tex]

    Thanks!
     
  2. jcsd
  3. Aug 29, 2012 #2

    Mark44

    Staff: Mentor

    Is this homework?
     
  4. Aug 30, 2012 #3
    Not homework. I'm just asking if it's possible for the expressions on both sides to be identical or the equation is an identity or not. Because I've tried to work on it but I couldn't make them identical.
     
  5. Aug 30, 2012 #4

    HallsofIvy

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    One obvious point is that if x= 0, the left side is 1 but the right side does not exist. And if you don't like "does not exist", try evaluating both sides at x= 0.01.
     
  6. Aug 30, 2012 #5

    uart

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    Try cross multiplying to get:

    [tex]\cos 2x (\sin 3x + \sin x) = \sin 2x (\cos 3x + \cos x)[/tex]

    and then apply the half sum/difference identities to all of the sin cos products on each side of the equation. It comes out fairly easy with this step.

    Use:
    [tex]\sin a \cos b = \frac{1}{2} \left[ \sin(a+b) + \sin(a-b) \right][/tex]
     
  7. Aug 30, 2012 #6

    mathman

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    An alternate approach is to convert to a polynomial equation in u = eix.
    For example: cot(2x) = i(u2+u-2)/(u2-u-2) = i(u4+1)/(u4-1).
    Do the similar steps for the right hand side, clear the denominators and end up with an identity.
     
  8. Aug 31, 2012 #7
    [tex]\cos 2x (\sin 3x + \sin x) = \sin 2x (\cos 3x + \cos x)[/tex]
    [tex]\cos 2x \sin3x + \cos2x sinx = \sin2x \cos3x + \sin2x cosx[/tex]

    Should I end here? I think both doesn't equal up. Or should I go further by extracting cos2x, sin2x, cos3x, and sin3x ?
     
  9. Aug 31, 2012 #8

    mathman

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    You need to end up with sinx and cosx only, using identities for cos2x, etc. I think my method (using representation in eix, etc.) might be easier.
     
  10. Aug 31, 2012 #9

    uart

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    No, apply the half sum-difference formula (that I gave above) to each of the sin-cos products.
     
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