Answer:[tex]D = \sqrt{A^{2}+B^{2 }}[/tex] and [tex] \aplha = arctan(\frac{B}{A} )[/tex]
Explanation:
Hi! Since the notation is a little bit messed up, I am going to suppose that
[tex]E = A sin(kr-\omega t)+B cos(kr-\omega t)[/tex] --- (1)
and :
[tex]E = D sin(kr-\omega t +\alpha)[/tex] --- (2)
Here we are going to use a trigonometric identity of the sine of the sum of two angles, namely:
[tex]sin(a+b)=cos(b)sin(a)+sin(b)cos(a)[/tex] --- (3)
Lets set:
[tex]a = kr - \omega t\\b = \alpha[/tex]
So now (2) becomes:
[tex]E = Dcos( \alpha ) sin(kr - \omega t) + Dsin(\alpha)cos(kr - \omega t))[/tex] --- (4)
Now (1) and (4) must be equal, and in particular we must have the following identities:
[tex]Dcos(\alpha) = A\\Dsin(\alpha) = B[/tex] --- (5)
If we square these two identities and sum them we got:
[tex]D^{2} (cos(\alpha)^{2} +sin(\alpha)^{2}) = A^{2} +B^{2}[/tex]
And since:
[tex]cos(a)^{2} +sin(a)^{2} =1[/tex]
We got the first solution:
[tex]D = \sqrt{A^{2}+B^{2 }}[/tex]
For the second part we must divide the identies (5)
We got:
[tex]\frac{sin(\alpha)}{cos(\alpha)}=\frac{B}{A}[/tex]
And since:
[tex]\frac{sin(\alpha)}{cos(\alpha)}=tan(\alpha)[/tex]
We use the inverse of the tan function:
[tex]\alpha =arctan(\frac{B}{A} )[/tex]
Greetings!
