How Can I Simplify Trigonometric Expressions Using Euler's Formula?

[mewmew]

Express the following in the form z=Re[Ae^{i(\omega t+\alpha)}]

z=cos(\omega t - \frac{\pi}{3}) - cos (\omega t)
and
z=sin(\omega t) - 2cos(\omega t - \frac{\pi}{4}) + cos(\omega t)

I got a few of the problems correct by using trig. identities but it was pretty tough and two I can't get. Our teacher said you can use a tric to solve them easier but didn't have time to finish, I just know it has something to do with the polar form of e^{i \theta} I really have no clue on how to do these without using the really long method of trig. identities. Any help would be great. Thanks

 

[CarlB]

z = \textrm{Re}(e^{i(\omega t - \pi/3} - e^{i\omega t})
=\textrm{Re}( e^{i\omega t}( e^{-i\pi/3}-1))
= \textrm{Re}(( e^{-i\pi/3}-1) \;\;e^{i\omega t})

Does that help?

Don't be ashamed, it's far better to be conversant in trig than to know a few tricks.

Carl

 

[robphy]

Along the lines of CarlB, but without jumping straight into using Re(),
recall e^{i\theta}=\cos\theta+i\sin\theta, from which you can derive
\cos\theta=\frac{1}{2}(e^{i\theta}+e^{-i\theta}) and a similar expression for \sin\theta (which I left for you to do).

So, now:
<br /> z&amp;=\frac{1}{2}(e^{i[\omega t-\pi/3]}+e^{-i[\omega t-\pi/3]})-\frac{1}{2}(e^{i[\omega t]}+e^{-i[\omega t]})
then do some algebra.

Recall that Re(z)=\frac{1}{2}(z+z^*). Thus Re(e^{i\theta})=\cos\theta.

 

[mewmew]

Thanks a lot, that makes it much more simple.