Trig factor formula proof help.

[tweety1234]

Homework Statement

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

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

 

[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)

 

[LCKurtz]

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 don't 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]

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 don't get how they got P+Q and P-Q ?

 

[tweety1234]

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