Trig factor formula proof help.


by tweety1234
Tags: factor, formula, proof, trig
tweety1234
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#1
Dec20-09, 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(A-B) to derive the result that;

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

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

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

Add the two intenties to get;

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

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

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

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?

[latex] sinP + sinQ = 2sin\frac{P+Q}{2} cos\frac{P-Q}{2} [/latex]
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tiny-tim
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Dec20-09, 04:44 PM
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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)
LCKurtz
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Dec20-09, 04:46 PM
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Quote Quote by tweety1234 View Post
let A+B = P and A-B=Q

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

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.

tweety1234
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#4
Dec20-09, 04:55 PM
P: 112

Trig factor formula proof help.


Quote Quote by tiny-tim View Post
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 ?
tweety1234
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#5
Dec20-09, 05:12 PM
P: 112
Quote Quote by LCKurtz View Post
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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