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Homework Help: Very quick trig question

  1. Oct 23, 2007 #1


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    Just a quick trig. question:

    What's the best way to expand sin(A)sin^3(B) into a suitable form for integration? (A and B are both functions of x here. As it happens, A = j*pi*x/a and B = pi*x/a)

    I have written an expression in terms of elements such as cos(A-B), cos(A+B) etc., but it's a little long, and I'm inclined to think there might be a shorter way of doing it -- something obvious that I'm not thinking of (?).

    My final result: 3/8*cos(A-B) - 3/8*cos(A+B) + 1/8*cos(A+3B) - 1/8*cos(A-3B)

    (I used identities for sinAsinB, and for cosAcosB several times, and for cos 2B).

    Last edited: Oct 23, 2007
  2. jcsd
  3. Oct 23, 2007 #2


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