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

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Euler's formula states that for any real number x, e^(ix) = cos(x) + i*sin(x). In the case of e^(-2i*theta), substituting -2*theta into Euler's formula yields e^(-2i*theta) = cos(-2*theta) + i*sin(-2*theta). This simplifies to e^(-2i*theta) = cos(2*theta) - 2i*sin(theta)*cos(theta), confirming the relationship between exponential functions and trigonometric identities. The discussion also highlights the use of TeX for formatting mathematical equations.

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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
:blushing:
 
Last edited:
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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
 
Thanks, that helps!
 
Just solved it, after 45 minutes... :frown:
 
chroot said:
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 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:" />
 
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.
 
THANK YOU SO MUCH GUYS... you've all been too helpful :blushing:

Hannah
 
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!
 

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