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:
- we know that $\cos b = \frac{{e}^{ib}+{e}^{-ib}}{2}$
- after converting to complex numbers I understand that we got here 2 different summations: $\frac{1}{2}+\frac{1}{2}+\sum_{t=1}^n\frac{(e)^{tib}}{2}+\sum_{t=1}^n\frac{(e)^{-tib}}{2}$
- each of those are a geometric series, so: $=1+\frac{\frac{{e}^{ib}}{2}({e}^{ibn}-1)}{{e}^{ib}-1}+\frac{\frac{{e}^{-ib}}{2}({e}^{-ibn}-1)}{{e}^{-ib}-1}$
- after opening the equation: $\frac{\frac{3}{2}-{e}^{ib}-{e}^{-ib}+\frac{{e}^{ibn}}{2}-\frac{{e}^{ib(n+1)}}{2}+\frac{{e}^{ib}}{2}+\frac{{e}^{-ibn}}{2}-\frac{{e}^{ibn-ib}}{2}-\frac{1}{2}+\frac{{e}^{-ib}}{2}}{1-{e}^{ib}-{e}^{-ib}}$
- 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??
Best Answer
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