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((B)/(A) )`

Step-by-step 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:

`(sin(\alpha))/(cos(\alpha))=(B)/(A)`

And since:

`(sin(\alpha))/(cos(\alpha))=tan(\alpha)`

We use the inverse of the tan function:

`\alpha =arctan((B)/(A) )`

Greetings!