Tricky Quadratic formula / trig identities

[binbagsss]

Im to solve $(k+l)^{2}e^{-ila}-(k-l)^{2}e^{ila}=0$, for $k$,

The solution is $k=l(e^{ial}-1)/(e^{ial}+1)=il tan(al/2)$

FIRST QUESTION

So it's a quadratic in k, should be simple enough, my working so far using the quad. formula is $k= (4l^{2}(e^{-ila}+e^{ila})\pm \sqrt{4l^{2}(e^{-ila}+e^{ila})^{2}-4l^{2}(e^{-ila}-e^{ila})^{2}})/2(e^{-ila}-e^{ila})$
=$(4l^{2}(e^{-ila}+e^{ila})\pm \sqrt{16l^{2}})/2(e^{-ila}-e^{ila})$
... which doesn't give the correct answer.

SECOND QUESTION.

Even after attaining the correct solution, I'm struggling to show that $l(e^{ial}-1)/(e^{ial}+1)=il tan(al/2)$
LHS: $cos (al) + isin (al) -1 / cos (al) + isin (al) +1=(cos^{2}(al/2)-sin^{2}(al/2)+isin(al/2)cos(al/2)-1)/(cos^{2}(al/2)-sin^{2}(al/2)+isin(al/2)cos(al/2)+1)$

using$cos 2x= cos^{2}x-sin^{2}x$
and $sin 2x=2sinxcosx,$

but this doesn't seem to be going anywhere...

Any help on either question really appreciated,
Thanks in advance.

 

[SammyS]

Im to solve $(k^{\color{red}{2}}+l)^{2}e^{-ila}-(k-l)^{2}e^{ila}=0$, for $k$,

The solution is $k=l(e^{ial}-1)/(e^{ial}+1)=il tan(al/2)$

FIRST QUESTION

So it's a quadratic in k, should be simple enough, my working so far using the quad. formula is $k= (4l^{2}(e^{-ila}+e^{ila})\pm \sqrt{4l^{2}(e^{-ila}+e^{ila})^{2}-4l^{2}(e^{-ila}-e^{ila})^{2}})/2(e^{-ila}-e^{ila})$
=$(4l^{2}(e^{-ila}+e^{ila})\pm \sqrt{16l^{2}})/2(e^{-ila}-e^{ila})$
... which doesn't give the correct answer.

SECOND QUESTION.

Even after attaining the correct solution, I'm struggling to show that $l(e^{ial}-1)/(e^{ial}+1)=il \tan(al/2)$
LHS: $\cos (al) + i\sin (al) -1 / \cos (al) + i\sin (al) +1=(\cos^{2}(al/2)-\sin^{2}(al/2)+i\sin(al/2)\cos(al/2)-1)/(\cos^{2}(al/2)-\sin^{2}(al/2)+i\sin(al/2)\cos(al/2)+1)$

using$cos 2x= cos^{2}x-sin^{2}x$
and $sin 2x=2sinxcosx,$

but this doesn't seem to be going anywhere...

Any help on either question really appreciated,
Thanks in advance.

The given solution is consistent with there being a typo in your stated problem statement.

It seems there should be no exponent on the k inside the first parenthesis .

 

[binbagsss]

The given solution is consistent with there being a typo in your stated problem statement.

It seems there should be no exponent on the k inside the first parenthesis .

apologies, yes that was a typo,
edited ta.

 

[SammyS]

apologies, yes that was a typo,
edited ta.

Does that help you in solving the problem, or are you still stuck?

 

[binbagsss]

Does that help you in solving the problem, or are you still stuck?

The original question remains.
it was a typo.

 

[epenguin]

For the first part, less tedious and error-prone than using your quadratic solution maybe, you have got there a slightly disguised difference of two squares.

If you can't see that, you can also take a term from right to the left hand side and see how you could then best simplify.

 

[binbagsss]

For the first part,less tedious and error-prone than using your quadratic solution maybe, you have got there a slightly disguised difference of two squares.

If you can't see that, you can also take a term from right to the left hand side and see how you could then best simplify.

I think what you are saying is equivalent to the step I took in simpifying the part inside the square root of the quadratic formula, the last line where I get it simplifies to square rt of 16l^{2} ?

 

[SammyS]

The original question remains.

Second Question:
Multiply the numerator & denominator by $\ e^{-i\, l a/2}\ .$

 

[SammyS]

First Question:

Multiply the original equation by $\ e^{ila} \$ .

Then it's pretty quick to get difference of squares else, add $\displaystyle \ (k-l)^{2}e^{2ila} \$ to both sides & take square root.

The given solution only uses one of the two possible sign choices.

 

[theodoros.mihos]

how many variables have your equation?