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Trig factor formula proof help. 
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#1
Dec2009, 04:39 PM

P: 112

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(AB) to derive the result that; [itex] sinP + sinQ = 2sin\frac{P+Q}{2} cos\frac{PQ}{2} [/itex] [itex] sin(A+B) = sinAcosB + cosAcosB [/itex] [itex] sin(AB) = sinAcosBcosAsinB [/itex] Add the two intenties to get; [itex] sin(A+B) + sin(AB) 2sinAcosB [/itex] let A+B = P and AB=Q then [itex] A = \frac{p+q}{2} [/itex] and [itex] B = \frac{PQ}{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{PQ}{2} [/itex] 


#2
Dec2009, 04:44 PM

Sci Advisor
HW Helper
Thanks
P: 26,148

Hi tweety1234!
I'm not sure what you're not getting … you have sin(A+B) + sin(AB) = 2sinAcosB, and sinAcosB = sin((P+Q)/2)cos((PQ)/2) 


#3
Dec2009, 04:46 PM

HW Helper
Thanks
PF Gold
P: 7,719

A = (P+Q)/2. Now subtract those two equations instead of adding them to get B. 


#4
Dec2009, 04:55 PM

P: 112

Trig factor formula proof help.



#5
Dec2009, 05:12 PM

P: 112

Thanks. so it would be, A+B=P AB =Q 2B=pq 


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