Question

A general expression for an electromagnetic plane wave can be written as E Asin( k-r-cot) + B cos( k . r-ot) or E Dsin( k . r-dot + α). Please find D and α in terms of A and B.

Answer 1

Answer:$D = \sqrt{A^{2}+B^{2 }}$ and $\aplha = arctan(\frac{B}{A} )$

Explanation:

Hi! Since the notation is a little bit messed up, I am going to suppose that

$E = A sin(kr-\omega t)+B cos(kr-\omega t)$   --- (1)

and :

$E = D sin(kr-\omega t +\alpha)$  --- (2)

Here we are going to use a trigonometric identity of the sine of the sum of two angles, namely:

$sin(a+b)=cos(b)sin(a)+sin(b)cos(a)$ --- (3)

Lets set:

$a = kr - \omega t\\b = \alpha$

So now (2) becomes:

$E = Dcos( \alpha ) sin(kr - \omega t) + Dsin(\alpha)cos(kr - \omega t))$  --- (4)

Now (1) and (4) must be equal, and in particular we must have the following identities:

$Dcos(\alpha) = A\\Dsin(\alpha) = B$   --- (5)

If we square these two identities and sum them we got:

$D^{2} (cos(\alpha)^{2} +sin(\alpha)^{2}) = A^{2} +B^{2}$

And since:

$cos(a)^{2} +sin(a)^{2} =1$

We got the first solution:

$D = \sqrt{A^{2}+B^{2 }}$

For the second part we must divide the identies (5)

We got:

$\frac{sin(\alpha)}{cos(\alpha)}=\frac{B}{A}$

And since:

$\frac{sin(\alpha)}{cos(\alpha)}=tan(\alpha)$

We use the inverse of the tan function:

$\alpha =arctan(\frac{B}{A} )$

Greetings!