Deriving sin(a-b) trig identity using Cross Product of Unit Vectors

[deedsy]

Homework Statement

A and B are two unit vectors in the x-y plane.
A = <cos(a), sin(a)>
B = <cos(b), sin(b)>

I need to derive the trig identity:
sin(a-b) = sin(a) cos(b) - sin(b) cos (a)

I'm told to do it using the properties of the cross product A x B

Homework Equations

A x B = |A||B| sinθ , where θ is the angle between the two vectors

The Attempt at a Solution

Well, |A|=|B|=1 *unit vectors
sinθ = sin(a-b) *for a > b
A x B = cos(a)sin(b) - cos(b)sin(a)

Putting this together, I get:
sin(a-b) = cos(a)sin(b) - cos(b)sin(a)

I can't figure out what I did wrong?

 

[Shinaolord]

Try using the determinant definition of the cross product.

Actually nevermind. I don't thin that'd work. It may though.

I'm going to try.

$\left( \begin{array}{cc} a & b \\ d & e \end{array} \right)$

Where a and b are the 1st and 2nd entries of a, and d and e being the same for b.
$det(A)= a*e-b*d$

Which is what you got I think. Hmm.

 

[deedsy]

well I did do the determinate, that's how I got:
A x B = cos(a)sin(b) - cos(b)sin(a)

But that by itself doesn't tell you the trig identity?

 

[Shinaolord]

Hmm. Are you sure it's $cos(\alpha), sin(\alpha)$and not the other way around?

Then it'd work.
Oh in the x y plane. I think that switching them around to (y,x) will give the negative of the first cross product. It would just "point" in the opposite direction. Possibly?
I'm not quite sure. Trying to help, though.

 

[deedsy]

yeah, the question has it how I wrote it

 

[Shinaolord]

I updated the above reply with a possibility. Since -(cos(a)sin(b)-sin(a)cos(b)) would give the identity

 

[deedsy]

yeah, you're right. That works if they are switched; it also works to derive the sin(a+b) trig identity if they are switched like that. I'm just thinking now if that is ok to switch the x and y components for both unit vectors..

 

[Shinaolord]

Well by the definition of the cross product, it would have the same magnitude as (x,y) just pointing in the opposite direction. So it's just taking the negative of the (x,y) determinant, you don't actually have to switch them. It just shows how I set up the second determinant

 

[Tanya Sharma]

Homework Statement

A and B are two unit vectors in the x-y plane.
A = <cos(a), sin(a)>
B = <cos(b), sin(b)>

I need to derive the trig identity:
sin(a-b) = sin(a) cos(b) - sin(b) cos (a)

I'm told to do it using the properties of the cross product A x B

Homework Equations

A x B = |A||B| sinθ , where θ is the angle between the two vectors

The Attempt at a Solution

Well, |A|=|B|=1 *unit vectors
sinθ = sin(a-b) *for a > b
A x B = cos(a)sin(b) - cos(b)sin(a)

Putting this together, I get:
sin(a-b) = cos(a)sin(b) - cos(b)sin(a)

I can't figure out what I did wrong?

The problem is that you have neglected the vectors in the last expression .

$A \times B = [cos(a)sin(b) - cos(b)sin(a)]\hat{k}$

$A \times B = -sin(a-b)\hat{k}$

Equate the above two and you would get the desired result .

 

[Shinaolord]

The problem is that you have neglected the vectors in the last expression .
$A \times B = [cos(a)sin(b) - cos(b)sin(a)]\hat{k}$
$A \times B = -sin(a-b)\hat{k}$
Equate the above two and you would get the desired result .

That works too, and is mathematically "valid"

Thanks for clearing that up.

 

[FactChecker]

I think that your "Relevant equation" needs theta to be the angle oriented positive starting at A and ending at B. You have it going the other way when a > b. Change the signs and you've got it.

 

[deedsy]

So I want to make sure I understand why there should be a negative sign in front of sin(a-b).

A x B gives me a vector with a magnitude equal to |A||B|sinθ. The direction of the vector, by the right hand rule, is perpendicular to these vectors (into the page). However, the determinate of A x B gives a direction k-hat (out of the page), so I need to make one of the terms negative to be consistent. Is this correct?

 

[Shinaolord]

Basically yes, but consistent because of the presence of k hat, no. . We're multiplying the entire expression of A x B by -1. See the post 2 above this one for an explanation by another user.

 

[deedsy]

How do you know which vector the angle should start from?

 

[Shinaolord]

That's why use use the determinant definition, avoiding the angle $\phi$ between A and B

 

[Shinaolord]

Actually for sin(a-b) b should be smaller than a so we get a positive angle $\phi$. Is that what you meant. ?

 

[deedsy]

wait, what? If you want the angle between the vectors θ to be positive, you'd want a > b since it is sin(a-b)

 

[deedsy]

The problem is that you have neglected the vectors in the last expression .

$A \times B = [cos(a)sin(b) - cos(b)sin(a)]\hat{k}$

$A \times B = -sin(a-b)\hat{k}$

Equate the above two and you would get the desired result .

thanks Tanya. but can you explain why there needs to be a negative sign in front of sin(a-b) ? Or was my explanation correct?

 

[Shinaolord]

wait, what? If you want the angle between the vectors θ to be positive, you'd want a > b since it is sin(a-b)

That's what I meant. Basically there's a negative sign so we can flip around the terms. There's not anything ore to it, as far as I know.

 

[FactChecker]

How do you know which vector the angle should start from?

In the cross product notation, AxB, the angle is always measured starting from the first vector, A, and ending at the second vector, B. So in your notation, theta = b-a always. That agrees with the right hand rule.

 

[Tanya Sharma]

So I want to make sure I understand why there should be a negative sign in front of sin(a-b).

A x B gives me a vector with a magnitude equal to |A||B|sinθ. The direction of the vector, by the right hand rule, is perpendicular to these vectors (into the page). However, the determinate of A x B gives a direction k-hat (out of the page), so I need to make one of the terms negative to be consistent. Is this correct?

thanks Tanya. but can you explain why there needs to be a negative sign in front of sin(a-b) ? Or was my explanation correct?

Right hand rule gives the cross product in $-\hat{k}$ direction.