I don't know where to start either.I need to prove that: e-i(theta) = cos (theta) - isin (theta) I'm not really sure where to start.
I don't know where to start either.
Not because I do not know how to do this, I do.
But because I don't know what tools you have to use.
If you know that \(\displaystyle e^{\theta i }=\cos(\theta)+i\,\sin(\theta)\) then it is easy.
Just note that \(\displaystyle \cos(-\theta)=\cos(\theta)\), an even function.
And \(\displaystyle \sin(-\theta)=-\sin(\theta)\), an odd function.
I told you, the sine function is odd and the cosine is even.Okay, I do know that \(\displaystyle e^{\theta i }=\cos(\theta)+i\,\sin(\theta)\) what does even and odd functions have to do with it?
I told you, the sine function is odd and the cosine is even.
\(\displaystyle \sin(-t)=-\sin(t)~\&~\cos(-t)=\cos(t)\).
You are done!
Never add a question. Always start a new thread.One more question, if z= cos (theta) + isin(theta) show that 2 cos (theta) = z + 1/z