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Trig identity

  1. Oct 27, 2009 #1
    1. The problem statement, all variables and given/known data

    cos(α − β)cos(α + β) = cos2α - sin2 β
    2. Relevant equations

    cos(α + β) = cos α cos β − sin α sin β

    cos(α − β) = cos α cos β + sin α sin β

    3. The attempt at a solution
    I worked out the LHS which makes it
    cos2α cos2β - sin2α sin2β=RHS

    Then, I'm stuck, however, i tried to do what I think can work...But I don't think it's right.
    From here I think it's wrong.
    (cos2α - sin2 β) (cos2β + sin2α)=RHS
    (cos2α - sin2 β) (1) =RHSS

    My problem is can we use identity for items like cos2 a + sin2 B = 1?
    the variable is not the same :/
     
  2. jcsd
  3. Oct 27, 2009 #2

    lanedance

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    Homework Helper

    can you use complex exponentials?
     
  4. Oct 27, 2009 #3

    sylas

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    Science Advisor

    Good work.

    The identity only works with one variable, just as you suggest. So sin2α+cos2β cannot be reduced to 1. Also, your initial step in reducing the RHS has too many powers of 2 in it.

    You will do better not to use a difference of squares at this point. Basically, you have equations that involve four terms:
    [tex]\sin\alpha, \sin\beta, \cos\alpha, \cos\beta[/tex]​
    You can easily get that down to two terms using the identity
    [tex]\sin^2\alpha + \cos^2\alpha = 1[/tex]​
    and the same again for β.

    You will find this identity easier to use while you have the squares present in the equation. Since the RHS uses cos α and sin β, you can try eliminating the cos β and the sin α from the LHS.

    Cheers -- sylas
     
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