Finding Coefficients for Vector Equivalence: A Trigonometric Approach

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Homework Statement


A screenshot of the problem:
http://img246.imageshack.us/img246/9194/homeworkkd6.jpg

Homework Equations


Not sure... possibly the dot product of two vectors?

v*w = a(1)a(2) + b(1)b(2)

The Attempt at a Solution


Part of the problem is that I'm not entirely sure what the question is asking for. I think it's talking about the coefficents for the two vectors which will allow those vectors to produce the same vector as X.

I tried plugging (2*pi/5) into the equations above and adding them together giving me (using decimal approximations because I don't know how to find the exact values):

U + V = -.642 I + 1.260 J

Then I try replacing X for:

8I + 3J = -.642 I + 1.260 J

Which doesn't make much sense to me... should I treat I and J as variables and solve for them? I think I'm on the completely wrong track.
 
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By dumb luck, it appears that you have found that u=1 and v=1 is the solution to your problem. Now you need to go back to find that in some rational manner, rather than simply guessing at the answer.

How about starting by writing
X = u U + v V and then taking a dot product with U? What will that get you?
 
heh, well the problem is that isn't the right answer (it's an online submission process, and apparently 1 is not the correct answer)

Going from another angle, I thought about trying to reduce it to two equations of two variables... something like:

ucos(a)-vsin(a)=8
usin(a)+vcos(a)=3

where a = theta.

If I do that, though, how do I find exact values for sin(2pi/5) and cos(2pi/5)?
 
If you do as I suggested, and write
X = u U + v V = 8I + 3J
then
X.U = u = (8I + 3J).(cos theta I + sin theta J) = 8 cos theta + 3 sin theta
which you can then evaluate. Does that do anything for you?
 
Ah, actually yeah.

solving all the way through

u ~ 5.325
v ~ -6.682

Thanks!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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