# Euler's Relationship! Need Help!

## Main Question or Discussion Point

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 Last edited:

chroot
Staff Emeritus
Gold Member
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... Tide
Homework Helper
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 $e^{-2i\theta}[/tex] is also [itex]\left(e^{-i\theta}\right)^2$ which gives $\cos^2\theta-\sin^2\theta-2i\sin\theta\cos\theta$. HallsofIvy
Homework Helper
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 Hannah