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Very hard trig solution that is way too long

  1. Sep 21, 2009 #1
    1. The problem statement, all variables and given/known data

    Prove that
    sin(5A)+sin(2A)-sin(A) = sin(2A)*(2*cos(3A)+1)


    2. Relevant equations

    sin(2A) = 2(sin(A)cos(A))
    cos(2B) = cos2(A)-sin2(A)
    cos(A+B) = cos(A)cos(B) - sin(A)sin(B)
    sin(A+B) = sin(A)cos(B) + sin(B)cos(A)

    3. The attempt at a solution

    I can't find any other way than to just decompose the whole thing into sinA or cosA, and that is really long and usually full of mistakes. Is there something that i'm missing here? Something that makes this proof a lot easier?
     
  2. jcsd
  3. Sep 21, 2009 #2

    Dick

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    Well, yeah, somewhat easier. Expand the right side and cancel the sin(2A) on both sides. Now expand sin(5A)=sin(2A+3A). Then maybe use sin(3A)cos(2A)-cos(3A)sin(2A)=sin(A)? It goes a bit easier, yes? Hope I don't have a typo in there.
     
    Last edited: Sep 21, 2009
  4. Sep 21, 2009 #3

    ehild

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    Do you know the identity

    sin(A)*cos(B) = sin(A+B)+sin(A-B) ?

    Use on the right side for 2*sin(2A)*cos(3A).

    ehild
     
  5. Sep 21, 2009 #4
    are you sure that's correct?

    d21208f87b9c55b68e4cb36e4ec1cc8f.png

    that's what's on wikipedia.

    if it is however, it could help.
     
  6. Sep 22, 2009 #5

    ehild

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    My mistake, I forgot the division by two. Use Wiki's formula, it really helps! You will be surprised, how easy the solution is. :)

    ehild
     
  7. Sep 22, 2009 #6
    aight, got it. all is well! Close thread please!
     
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