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I really need an answer-Derive trig identity

  1. Aug 4, 2005 #1
    undefinedundefinedundefinedneed help=====)
  2. jcsd
  3. Aug 4, 2005 #2


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    Homework Helper

    You can use these to prove that trig identity:
    [tex]\cos \alpha + \cos \beta = 2 \cos \frac{\alpha + \beta}{2} \cos \frac{\alpha - \beta}{2}[/tex]
    [tex]\cos \alpha - \cos \beta = - 2 \sin \frac{\alpha + \beta}{2} \sin \frac{\alpha - \beta}{2}[/tex]
    [tex]\sin \alpha + \sin \beta = 2 \sin \frac{\alpha + \beta}{2} \cos \frac{\alpha - \beta}{2}[/tex]
    [tex]\sin \alpha - \sin \beta = 2 \cos \frac{\alpha + \beta}{2} \sin \frac{\alpha - \beta}{2}[/tex]
    Choose 2 of the above equations, and use them.
    Can you go from there?
    Viet Dao,
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