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:
$$z&=\frac{1}{2}(e^{i[\omega t-\pi/3]}+e^{-i[\omega t-\pi/3]})-\frac{1}{2}(e^{i[\omega t]}+e^{-i[\omega t]})$$
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.