proving a complex number equation

Relz

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I need to prove that: e-i(theta) = cos (theta) - isin (theta)

I'm not really sure where to start.
 
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 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.

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?
 
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!
 
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!

Oh! Oh my goodness, it was so simple. Thank you!

One more question, if z= cos (theta) + isin(theta) show that 2 cos (theta) = z + 1/z

Using the equations: cos (theta) = e-i (theta) + i sin (theta) and sin (theta) = ei (theta) - cos (theta)
 
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One more question, if z= cos (theta) + isin(theta) show that 2 cos (theta) = z + 1/z
Never add a question. Always start a new thread.

If \(\displaystyle z=e^{i\theta} \) then \(\displaystyle \dfrac{1}{z}=e^{-i\theta}.\)
 
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