How Can I Derive This Wave 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.