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    Use the formula for sin(A+B) and sin(A-B) to derive the result that...

    sinP + sinQ = 2sin (\frac{P+Q}{2})cos(\frac{P-Q}{2})

    It's example 20 from the C3 edexcel textbook.

    I get the part up to Let A + B = P, A - B = Q


    But not this next step:


    then A = \frac{P+Q}{2} and B = \frac{P-Q}{2}

    Surely A is P - B? I can't see why they divided by 2...thanks!
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    (a + B = P) + (a - B = Q)

    A + B + A - B = P + Q

    2a = P + Q

    A = (p + Q)/2
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    (Original post by gottastudy)
    Use the formula for sin(A+B) and sin(A-B) to derive the result that...

    sinP + sinQ = 2sin (\frac{P+Q}{2})cos(\frac{P-Q}{2})

    It's example 20 from the C3 edexcel textbook.

    I get the part up to Let A + B = P, A - B = Q


    But not this next step:


    then A = \frac{P+Q}{2} and B = \frac{P-Q}{2}

    Surely A is P - B? I can't see why they divided by 2...thanks!
    If P= A+B and Q=A-B

    Make A the subject in P so A= P-B
    Then sub it into Q so now:
    Q= (P-B)-B
    now make B the subject so -B = \frac{Q-P}{2} therefore B= \frac{P-Q}{2}

    And do the same for P
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    I understand now! Thanks for your help!!
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    (Original post by gottastudy)
    I understand now! Thanks for your help!!
    You could always leave rep...
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    (Original post by JTeighty)
    You could always leave rep...
    I've already used mine today. You'll get it soon
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    (Original post by gottastudy)
    I've already used mine today. You'll get it soon
    Thanks but don't worry about it.
 
 
 
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Updated: January 18, 2010
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