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I got the following problem:

I need to prove - using complex numbers:

$\sum_{t=0}^n \cos(tb) = \frac{\cos\frac{nb}{2}\sin\frac{nb+b}{2}}{\sin\frac{b}{2}}$

Ok so what I came up with so far:

  1. we know that $\cos b = \frac{{e}^{ib}+{e}^{-ib}}{2}$
  2. after converting to complex numbers I understand that we got here 2 different summations: $\frac{1}{2}+\frac{1}{2}+\sum_{t=1}^n\frac{(e)^{tib}}{2}+\sum_{t=1}^n\frac{(e)^{-tib}}{2}$
  3. each of those are a geometric series, so: $=1+\frac{\frac{{e}^{ib}}{2}({e}^{ibn}-1)}{{e}^{ib}-1}+\frac{\frac{{e}^{-ib}}{2}({e}^{-ibn}-1)}{{e}^{-ib}-1}$
  4. after opening the equation: $\frac{\frac{3}{2}-{e}^{ib}-{e}^{-ib}+\frac{{e}^{ibn}}{2}-\frac{{e}^{ib(n+1)}}{2}+\frac{{e}^{ib}}{2}+\frac{{e}^{-ibn}}{2}-\frac{{e}^{ibn-ib}}{2}-\frac{1}{2}+\frac{{e}^{-ib}}{2}}{1-{e}^{ib}-{e}^{-ib}}$
  5. and now I am stuck nothing I do (including switching back to cosine and sine) brings me to anything.

please help maybe I am approaching it wrong??

Andrei
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misha312
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1 Answers1

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There is a mistake from step 3 to 4. In the denominator $(e^{ib}-1)(e^{-ib}-1)$ should be $2-e^{ib}-e^{-ib}$. But I would not do that. $$(e^{ib}-1)(e^{-ib}-1)=e^{ib/2}(e^{ib/2}-e^{-ib/2})e^{-ib/2}(e^{-ib/2}-e^{ib/2})=-4\sin^2\frac{b}{2}$$ You need to apply the same procedure to the denominator

Andrei
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