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ChildishHambino
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\displaystyle\frac{1}{1-e^{j\theta}}=\frac{1}{2}(1+jcot(  \frac{\theta}{2})
Need help on this question, i've tried to rationalize after converting to mod arg form and I got;
\displaystyle\frac{1-cos(\theta)+jsin(\theta)}{2(1-cos(\theta)}
But I'm pretty sure I'm not going the right way about it because of the half theta on the RHS. Which leads me to thinking it's something to do with the identity \alpha + 1=2cos(\frac{\theta}{2})e^{j\fra  c{\theta}{2}}
and the other one. but how do I get it in that form?

Thanks in advance.
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Felix Felicis
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(Original post by ChildishHambino)
\displaystyle\frac{1}{1-e^{j\theta}}=\frac{1}{2}\left(1+  j\cot \tfrac{\theta}{2}\right)
Need help on this question, i've tried to rationalize after converting to mod arg form and I got;
\displaystyle\frac{1-cos(\theta)+jsin(\theta)}{2(1-cos(\theta)}
But I'm pretty sure I'm not going the right way about it because of the half theta on the RHS. Which leads me to thinking it's something to do with the identity \alpha + 1 = 2\cos \left(\frac{\theta}{2}\right)e^{  j\frac{\theta}{2}}
and the other one. but how do I get it in that form?

Thanks in advance.
lol j u wot m8

\displaystyle \frac{1}{1-e^{ix}} = \frac{1-\cos x + i \sin x}{(1-\cos x)^2 + \sin^2 x} = \frac{1-\cos x + i \sin x}{2(1-\cos x)}

Divide top by 1 - \cos x and recall what \dfrac{\sin x}{1-\cos x} is.
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ChildishHambino
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#3
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(Original post by Felix Felicis)
lol j u wot m8

\displaystyle \frac{1}{1-e^{ix}} = \frac{1-\cos x + i \sin x}{(1-\cos x)^2 + \sin^2 x} = \frac{1-\cos x + i \sin x}{2(1-\cos x)}

Divide top and bottom by 1 - \cos x and recall what \dfrac{\sin x}{1-\cos x} is.
MEI has got me in its grasp :/. And sinx/1-cosx? I'm guessing its something like tan(x/2) and thanks
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Felix Felicis
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(Original post by ChildishHambino)
MEI has got me in its grasp :/. And sinx/1-cosx? I'm guessing its something like tan(x/2) and thanks
Not quite, write \displaystyle \frac{\sin x}{1 - \cos x} = \frac{2\sin{\frac{x}{2}} \cos{\frac{x}{2}}}{1 - \left(\cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}\right)}.
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ChildishHambino
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(Original post by Felix Felicis)
Not quite, write \displaystyle \frac{\sin x}{1 - \cos x} = \frac{2\sin{\frac{x}{2}} \cos{\frac{x}{2}}}{1 - \left(\cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}\right)}.
Yeah haha just worked it out thanks
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