# Trig factor formula proof help.

by tweety1234
Tags: factor, formula, proof, trig
 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(A-B) to derive the result that; $sinP + sinQ = 2sin\frac{P+Q}{2} cos\frac{P-Q}{2}$ $sin(A+B) = sinAcosB + cosAcosB$ $sin(A-B) = sinAcosB-cosAsinB$ Add the two intenties to get; $sin(A+B) + sin(A-B) 2sinAcosB$ let A+B = P and A-B=Q then $A = \frac{p+q}{2}$ and $B = \frac{P-Q}{2}$ 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? $sinP + sinQ = 2sin\frac{P+Q}{2} cos\frac{P-Q}{2}$
 Sci Advisor HW Helper Thanks P: 26,167 Hi tweety1234! 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)
HW Helper
Thanks
PF Gold
P: 7,187
 Quote by tweety1234 let A+B = P and A-B=Q then $A = \frac{p+q}{2}$ and $B = \frac{P-Q}{2}$ 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?
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.

P: 112

## Trig factor formula proof help.

 Quote by tiny-tim Hi tweety1234! 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)
I dont get how they got P+Q and P-Q ?
P: 112
 Quote by LCKurtz 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.
Oh I get it now.

Thanks.

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

2B=p-q

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