Trig factor formula proof help.

  1. 1. The problem statement, all variables and given/known data

    I dont understand the example in my book,

    it says; use the formula for sin(A+B) and sin(A-B) to derive the result that;

    [itex] sinP + sinQ = 2sin\frac{P+Q}{2} cos\frac{P-Q}{2} [/itex]

    [itex] sin(A+B) = sinAcosB + cosAcosB [/itex]

    [itex] sin(A-B) = sinAcosB-cosAsinB [/itex]

    Add the two intenties to get;

    [itex] sin(A+B) + sin(A-B) 2sinAcosB [/itex]

    let A+B = P and A-B=Q

    then [itex] A = \frac{p+q}{2} [/itex] and [itex] B = \frac{P-Q}{2} [/itex]

    This is the bit I dont get, How did they get this bit ^^. I understand that the LHS becomes sinP+sinQ but don't understand how they got the fraction?

    [itex] sinP + sinQ = 2sin\frac{P+Q}{2} cos\frac{P-Q}{2} [/itex]
     
  2. jcsd
  3. tiny-tim

    tiny-tim 26,054
    Science Advisor
    Homework Helper

    Hi tweety1234! :smile:

    I'm not sure what you're not getting …

    you have sin(A+B) + sin(A-B) = 2sinAcosB,

    and sinAcosB = sin((P+Q)/2)cos((P-Q)/2)
     
  4. LCKurtz

    LCKurtz 8,283
    Homework Helper
    Gold Member

    Add the equations A+B = P and A-B=Q, giving 2A = P+Q

    A = (P+Q)/2.

    Now subtract those two equations instead of adding them to get B.
     
  5. I dont get how they got P+Q and P-Q ?
     
  6. Oh I get it now.

    Thanks.

    so it would be, A+B=P -A-B =Q

    2B=p-q
     
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