# Euler's Formula

## Homework Statement

Derive...

x(t)=(exp^(-$$\zeta$$$$\omega$$t))*(a1(exp^(i$$\omega$$sqrt(1-$$\zeta^2$$)*t)))+a2(exp^(-i$$\omega$$sqrt(1-$$\zeta^2$$)*t))))

into

x(t)=(exp^(-$$\zeta$$$$\omega$$t))*(A sin ($$\omega$$*t + $$\varphi$$))

n/a

## The Attempt at a Solution

I've managed to get x(t) = (exp^(-$$\zeta$$$$\omega$$t))*((a1(cos$$\omega$$*t) + i sin ($$\omega$$*t))+(a2(cos$$\omega$$*t) - i sin ($$\omega$$*t))) then when i simplify things...sin terms cancel out and i end up geting...

exponential term * (a1+a2) * (2cos $$\omega$$*t)

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Tom Mattson
Staff Emeritus
Gold Member
Hi,

Is the $\zeta$ a constant here?

I've managed to get x(t) = (exp^(-$$\zeta$$$$\omega$$t))*((a1(cos$$\omega$$*t) + i sin ($$\omega$$*t))+(a2(cos$$\omega$$*t) - i sin ($$\omega$$*t))) then when i simplify things...sin terms cancel out and i end up geting...

exponential term * (a1+a2) * (2cos $$\omega$$*t)
No, the sine terms do not cancel out. You have the following.

$$x(t)=e^{-\zeta\omega t}\left[(a_1+a_2)\cos\left(\omega\sqrt{1-\zeta^2}t\right)+(a_1-a_2)\sin\left(\omega\sqrt{1-\zeta^2}t\right)\right]$$.

The only way the sine term will vanish is if $a_1=a_2$, and you did not say that that is the case in your problem statement.

Hi,

Is the $\zeta$ a constant here?

No, the sine terms do not cancel out. You have the following.

$$x(t)=e^{-\zeta\omega t}\left[(a_1+a_2)\cos\left(\omega\sqrt{1-\zeta^2}t\right)+(a_1-a_2)\sin\left(\omega\sqrt{1-\zeta^2}t\right)\right]$$.

The only way the sine term will vanish is if $a_1=a_2$, and you did not say that that is the case in your problem statement.
Hello. It is a constant.
Yeah, I figured out that I made a mistake.

that's as far as i can go but i need to combine that into sin like i've posted above.