Help derivation of this equation.

[yungman]

Given a wave:
$$e(t)=\hat x E_{x0}\cos \omega t+\hat y E_{y0}\cos( \omega t+ \delta)$$(1)
The book claimed:
$$\sin^2\delta\;=\; \left[\frac {e_x(t)}{E_{x0}}\right]^2-2\left[\frac {e_x(t)}{E_{x0}}\right]\left[\frac {e_y(t)}{E_{y0}}\right]\cos\delta+ \left[\frac {e_y(t)}{E_{y0}}\right]^2$$
I can not get this. Below is my work.

From (1)
$$e_x(t)=E_{x0}\cos\omega t\;\Rightarrow \;\cos \omega t=\frac{e_x(t)}{E_{x0}}$$
$$\Rightarrow\; \sin\omega t\;=\;\sqrt{1-\left(\frac{e_x(t)}{E_{x0}}\right)^2}$$
$$e_y(t)=E_{y0}\cos(\omega t+\delta) \;\Rightarrow \;\frac{e_y(t)}{E_{y0}}\;=\;\cos\omega t \cos\delta-\sin\omega t sin\delta\;=\; \frac{e_x(t)}{E_{x0}}\cos\delta-\sqrt{1-\left(\frac{e_x(t)}{E_{x0}}\right)^2}sin\delta$$
$$\Rightarrow\;\sqrt{1-\left(\frac{e_x(t)}{E_{x0}}\right)^2}sin\delta \;=\; \frac{e_x(t)}{E_{x0}}\cos\delta-\frac{e_y(t)}{E_{y0}}$$
$$\Rightarrow\;\left(1-(\frac{e_x(t)}{E_{x0}})^2\right) sin^2\delta \;=\;\left[\frac{e_x(t)}{E_{x0}}\right]^2 \cos^2\delta-2\frac{e_x(t)}{E_{x0}}\frac{e_y(t)}{E_{y0}}\cos \delta + \left[\frac{e_y(t)}{E_{y0}}\right]^2$$
It is obvious that I am not going to get what the book gave:
$$\sin^2\delta\;=\; \left[\frac {e_x(t)}{E_{x0}}\right]^2-2\left[\frac {e_x(t)}{E_{x0}}\right]\left[\frac {e_y(t)}{E_{y0}}\right]\cos\delta+ \left[\frac {e_y(t)}{E_{y0}}\right]^2$$

Please help my derivation.

Thanks

 

[tiny-tim]

hi yungman!

(i can't see where you've gone wrong, but try it this way …)

you are essentially asked to prove that cos²A + cos²B - 2cosAcosBcos(A-B) = sin²(A-B)

just expand that! ​

when you've done that, try drawing the diameter PQ of a unit circle, with two points A and B such that angleAPD = A, angle BPD = B, and using simple geometry to prove that AB = sin(A-B)

 

[yungman]

Thanks Tiny Tim, I'll try it and see.

You think you can look at the other questions I posted in page 3 of the "Classical Physics" on Balanis a few days ago, I have no luck at all.

Many thanks

 

[yungman]

Thanks Tiny Tim. I worked out using $A=\omega t +\delta\;\hbox{ and }\; B=\omega t$. This is a little tricky and unobvious.

 

[paranormal]

for sharing your work and asking for help with your derivation. It's always good to check your work and make sure you understand the steps involved in solving a problem.

I have looked at your work and I see that you are on the right track, but there are a few mistakes or missing steps that are causing you to not get the same result as the book.

First, when you substitute in for e_x(t) and e_y(t), you need to use the expressions you have derived for them, not just their values. So for e_y(t), you should have e_y(t) = E_{y0} \cos(\omega t + \delta) = E_{y0} \cos(\omega t)\cos(\delta) - E_{y0} \sin(\omega t)\sin(\delta).

Next, when you substitute in for \cos(\omega t) and \sin(\omega t), you need to use the expressions you have derived for them as well. So for \cos(\omega t), you should have \cos(\omega t) = \frac{e_x(t)}{E_{x0}} and for \sin(\omega t), you should have \sin(\omega t) = \sqrt{1 - \left(\frac{e_x(t)}{E_{x0}}\right)^2}.

Finally, when multiplying out the terms and rearranging, it looks like you have made a mistake in the last step. You should have:

\left(1 - \left(\frac{e_x(t)}{E_{x0}}\right)^2\right)\sin^2\delta = \left(\frac{e_x(t)}{E_{x0}}\right)^2 \cos^2\delta - 2\frac{e_x(t)}{E_{x0}}\frac{e_y(t)}{E_{y0}} \cos \delta + \left(\frac{e_y(t)}{E_{y0}}\right)^2

Then, you can use the identities \sin^2\delta + \cos^2\delta = 1 and \cos\delta = \frac{e_x(t)}{E_{x0}} to simplify this expression to the one given in the book:

\sin^2\delta = \left(\frac{e_x(t)}{E_{x0}}\right)^2 - 2\frac