Solving Derivation Trouble with Euler's Relationships

[Emc2brain]

Derivation Trouble!

Please read the attachment with this posting: Here's the problem, I have been attempting to derive this for a couple of days now... However, it seems that whatever I do all that I end up deriving is itself again; meaning I get back to where I started. Can anyone give me a few pointers because I'm flat out of luck here. Here I use Euler's Relationships...but no help...hmm..?

e^(i*theta)-1=2i*sin(theta/2)*e^(i*(theta/2))

I have the pretty version in the attachment.

Crazed Hannah

 

[Emc2brain]

liz this is how one posts

 

[Tide]

Note that $1=e^{2\pi i}$ and $a^2-b^2 = (a-b)(a+b)$

 

[mathman]

I will use x for theta.

e^ix-1=cosx+isinx-1
=cos²x/2-sin²x/2+2icosx/2sinx/2-1
=-2sin²x/2+2icosx/2sinx/2
=2isinx/2(cosx/2+isinx/2)
=2isinx/2e^ix/2

 

[Emc2brain]

I'm still a little confused on how you got the half angle in there? I seemed to have missed a step

Hannah

 

[Emc2brain]

Because I know that e^(2ix) = cos2x/2-sin2x/2+2icosx/2sinx/2. Where'd you get e^(2ix), because all that I see is e^(ix)? Or is it emplied that 1=e^(2ix)? If that is true then why does your answer still contain a -1? Looking like this: cos2x/2-sin2x/2+2icosx/2sinx/2 -1?

Hannah

 

[Tide]

Note that $1=e^{2\pi i}$ and $a^2-b^2 = (a-b)(a+b)$

Even simpler:

$$e^{i \theta} - 1 = e^{i \frac{\theta}{2}} \left(e^{i \frac{\theta}{2}} - e^{-i \frac{\theta}{2}}\right)$$

It's just factoring.

 

[Emc2brain]

Thanx so much...