# Find the Angle Between (A+B) and (A-B) is (A+B)=73(A-B)

1. Sep 21, 2013

### AltruistKnight

1. The problem statement, all variables and given/known data
Vector A and B have the same magnitude. Given that the magnitude of A+B is 73 times greater than the magnitude of A-B, find the angle between them.

2. Relevant equations
$sqrt[(A+Acosθ)^2+(ASinθ)^2)]=73sqrt[(A-Acosθ)^2+(-Asinθ)^2]$

3. The attempt at a solution
I managed to start the question using the equation above, which was derived from the question details. I numbered my steps as follows in an effort to keep track of the process:

1)Square both sides-this eliminates the square root over the majority of the terms, and also squares the "73" coefficient.

2)Expand all brackets. This is where the mess comes in, and results in an equation as follows:
$$(A^2+2A^2cosθ+Acos^2θ)+(A^2sin^2θ)=5329[(A^2-2A^2cos+Acos^2θ)+(A^2sin^2θ)]$$

3)Here's where I began to have a little difficulty. Firstly, I worked on the left half of the equation without the coefficient. Specifically, I dropped the brackets and factored out "A", giving me $$A+2Acosθ+cos^2θ+A+2Asinθ+sin^2θ$$ on the left side of the equation. I noticed that there's a trigonometric identity for "$cos^2θ-sin^2θ=2cosθ"$, and since those two terms appear in a positive form on either side of the equation, I'm guessing I can switch the "$sin^2θ$" terms on either side and get "$cos2θ$" on both sides, then move one over and cancel them both.

But from there I'm unsure of how to make the equation simplify any further, barring distributing the 5329 amongst all the left-side terms, then dividing the entire equation by 5329 to remove them (which probably has to be done before switching the $sin^2θ$ on either side). I get the feeling cancelling further will be the key here, as well as distribution and factor addition, but I'm running into difficulty proceeding with calculations due to the mass number of terms and variables, so I would greatly appreciate some assistance in how to proceed without making an even bigger mess.

Last edited: Sep 21, 2013
2. Sep 21, 2013

### Jolb

I believe your "relevant equation" is incorrect. If A and B have the same magnitude, then we could choose the x-axis along A and the y-axis perpendicular to A, so A, in component form, would be A=(A,0), where A is the magnitude of A, and B = (A cosθ, A sinθ).

Vector addition would say A+B = (A + A cosθ, A sinθ), while A-B=(A - A cosθ, -A sinθ).

So your relevant equation has the wrong y-components plugged into the pythagorean theorem.

3. Sep 21, 2013

### AltruistKnight

Ah, just noticed that, thanks a lot. That said, would it affect my process significantly? (It evidently cuts down a fair number of terms, thankfully)

That said, I'm re-working it now using the new equation, so I may update the question in a few minutes.

Last edited: Sep 21, 2013
4. Sep 21, 2013

### Jolb

Hmm did you edit your OP? I'm a little confused now since now your relevant equation is correct. But yes, once you make that correction, it should work out with some algebra.

Edit: Okay, now I see you edited your second post. This is quite a whirlwind of edits! Next time I'd recommend making an all-new post with further progress.

5. Sep 21, 2013

### AltruistKnight

Yeah, I updated it to fix the error. That said, given that the right side has that bothersome coefficient, would it be feasible to simply divide both sides by the same number prior to expanding the brackets?

Also, when I -do- expand the left side and attempt to add terms, I can't seem to make $A^2+2A^2cosθ+Acos^2θ+Asin^2θ$ produce a usable trig identity. I do recall at some point I had to factor the A's out, but that still leaves me with $$A(1+2Acosθ+cos^2θ+Asin^θ)$$

Just to be sure, $(Acosθ)^2$ IS $(A^2cos^2θ)$, right? If so, I can't seem to get rid of the "A" in that and the corresponding sine term, since there's one term on the right side that ends up with only one "A" coefficient, keeping me from factoring out "A^2" without making it even messier...

6. Sep 21, 2013

### Jolb

Well, I was able to solve this problem. You can factor A out pretty quickly... You can easily write:
A+B = (A + A cosθ, A sinθ) = A(1+cosθ, sinθ) while A-B=(A - A cosθ, -A sinθ) = A (1-cosθ, -sinθ). Then you have:

|A+B|=73|A-B|
|A(1+cosθ, sinθ)| = 73 |A(1-cosθ, -sinθ)|
Where the bars denote the length of the vector given by your pythagorean expression. So,

A |(1+cosθ, sinθ)|=73A|(1-cosθ, -sinθ)|
and divide out A yielding
|(1+cosθ, sinθ)| = 73 |(1-cosθ, -sinθ)|

This is basically equivalent to saying that these vectors would have the same angle even if we first rescaled A and B at the beginning of the calculation such that they have unit length. (i.e., set A equal to 1.)

The one trig identity that you seem to be missing is cos2θ+sin2θ=1

Last edited: Sep 21, 2013
7. Sep 28, 2013

### AltruistKnight

Thanks,the explanation and trig function really helped!