Trig factor formula proof help.

In summary, to derive the result sinP + sinQ = 2sin\frac{P+Q}{2} cos\frac{P-Q}{2}, we use the formulas for sin(A+B) and sin(A-B) and set A+B = P and A-B = Q. By adding the two equations and subtracting them, we get the values for A and B, which allows us to rewrite the equation in terms of sin((P+Q)/2) and cos((P-Q)/2). This results in the desired formula.
  • #1
tweety1234
112
0

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;

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

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

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

Add the two intenties to get;

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

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

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

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?

[itex] sinP + sinQ = 2sin\frac{P+Q}{2} cos\frac{P-Q}{2} [/itex]
 
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  • #2
Hi tweety1234! :smile:

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)
 
  • #3
tweety1234 said:
let A+B = P and A-B=Q

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

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.
 
  • #4
tiny-tim said:
Hi tweety1234! :smile:

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 ?
 
  • #5
LCKurtz said:
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
 

What is the trig factor formula?

The trig factor formula, also known as the trigonometric identity, is a mathematical equation that relates different trigonometric functions such as sine, cosine, and tangent.

Why is it important to know the trig factor formula?

The trig factor formula is important because it allows us to simplify and solve trigonometric equations, which are commonly used in many fields of science, such as physics, engineering, and astronomy.

How can I prove the trig factor formula?

There are several ways to prove the trig factor formula, but the most common method is using the unit circle and trigonometric identities to show that both sides of the equation are equal.

Can you provide an example of using the trig factor formula?

Sure, for example, if we have the equation sin(x)cos(x) = 1/2, we can use the trig factor formula sin(2x) = 2sin(x)cos(x) to simplify it to sin(2x) = 1, and then solve for x to find that x = π/6, π/2, or 5π/6.

Are there any common mistakes when using the trig factor formula?

Yes, some common mistakes include not using the correct trigonometric identity or not simplifying the equation correctly. It is important to carefully follow the steps and double check your work when using the trig factor formula.

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