Parametric representation of a vector electric field

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jjand
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Homework Statement



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Homework Equations



None I can think of

The Attempt at a Solution



I'm fairly certain that phi_yx is zero
Also I tried factoring out the cos and splitting up the equation into it's respective components, but to no avail. Am I even going about this correctly?
 
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  • Typeset homework problem: vector electric field E(r,t) = x-hat A_x cos(kz - wt + phi_x) + y-hat A_y cos(kz - wt + phi_y); for phi_y = 0, prove (E_x(tau)/A_x)^2 + (E_y(tau)/A_y)^2 - 2 cos phi_yx (E_x(tau)/A_x)(E_y(tau)/A_y) = sin^2 phi_yx, where tau = wt - kz and phi_yx = phi_y - phi_x.
    Screen shot 2011-10-23 at 3.35.20 AM.png
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The last bit is a typo. It should be [itex]\phi_{xy} = \phi_x[/itex].

I'm not sure what you mean by "splitting up the equation into its respective components" since the equation you're supposed to show is true is a scalar equation.
 
Thanks for the reply Vela! Actually I think I figured it out. Maybe you or someone else can check my work? I assumed that it the phi's were not typos, but they could very well be. I went ahead and attached my work.
 
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Your expressions for Ex and Ey are slightly wrong. For example,
[tex]E_x = A_x\cos(kz-\omega t+\phi_x) = A_x\cos(-r+\phi_x) = A_x\cos(r-\phi_x)[/tex]since cosine is an even function. You have an extra negative signs floating around.

You must have made a mistake somewhere in the middle because the cross term has the wrong sign. The sign doesn't flip the way you did it because, again, [itex]\cos (-\theta)=\cos\theta[/itex].

You actually made it more complicated by keeping the E's and A's around. You know that
\begin{align*}
\frac{E_x}{A_x} &= \cos(\phi_x-r) \\
\frac{E_y}{A_y} &= \cos(r) \\
\end{align*}so you could have just written the lefthand side in terms of cosines and then shown it simplifies down to the righthand side.
 
Ah that I forgot about that cosine property and I found a convenient sign mistake half way through the problem. Thanks Vela!