How Does Euler's Formula Apply to e^(-2i*theta)?

[Emc2brain]

Help! Euler's Relationship!

if e^(i*theta) = cos(theta) + i*sin(theta)

then what is e^(-2i*theta) = ?

I attempted to derive this and got the following for the +2i:
e^(+2*theta) = cos(2*theta) + 2i*cos(theta)sin(theta)

Not even sure if this may be correct, but I believe the answer to my question with negative 2 (-2i) must be simple... Help please, thanx.


Because I am attempting to derive 2sin^2(theta) = 1-cos(2theta) from euler's relationship: e^(i*theta) = cos(theta) + i*sin(theta)


Hannah

 

[chroot]

Euler's relation is that

$$e^{ix} = \cos(x) + i \sin(x)$$

where x can be anything at all. In your example, x would be $-2 \theta$, so plug it in:

$$e^{-2 i \theta} = \cos(-2 \theta) + i \sin(-2 \theta)$$

- Warren

 

[Emc2brain]

Thanks, that helps!

 

[Emc2brain]

Just solved it, after 45 minutes...

 

[Tide]

Euler's relation is that

$$e^{ix} = \cos(x) + i \sin(x)$$

where x can be anything at all. In your example, x would be $-2 \theta$, so plug it in:

$$e^{-2 i \theta} = \cos(-2 \theta) + i \sin(-2 \theta)$$

- Warren

And [itex]e^{-2i\theta}[/tex] is also $\left(e^{-i\theta}\right)^2$ which gives $\cos^2\theta-\sin^2\theta-2i\sin\theta\cos\theta$. <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f642.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":smile:" title="Smile :smile:" data-smilie="1"data-shortname=":smile:" />[/itex]

 

[HallsofIvy]

Since you arrived at e^(+2*theta) = cos(2*theta) + 2i*cos(theta)sin(theta)
I'm surprised you could continue: using -θ instead of θ just replaces &theta; with -θ and cos(-θ)= cos(θ), sin(-&theta;)= -sin(&theta;).

Also, since you clearly replaced sin(2θ) with 2sin(θ)cos(&theta), why not also replace cos(2&theta) with cos²(θ)- sin²(θ)?

Putting those together, $$e^{-2\theta}= cos(-2\theta)+ i sin(-2\theta)$$
$$= cos^2(-\theta)- sin^2(-\theta)+ 2i sin(-\theta)cos(-\theta)$$
$$= cos^(\theta)+ sin^2(\theta)- 2i sin(\theta)cos(\theta)$$,
exactly what Tide got by squaring.

 

[Emc2brain]

THANK YOU SO MUCH GUYS... you've all been too helpful

Hannah

 

[Emc2brain]

Hello there helpful bunch! ;)

How are you guys able to write out the equations?? Because I tried to copy and past them into this email however it simply would not do that...Thanx for all the assistance!

 

[chroot]

The equations are written with in a language called TeX, which our forum software parses and turns into images.

See more here: https://www.physicsforums.com/showthread.php?t=8997

- Warren