My objective is to prove that:
$$\operatorname{Re}\left(\frac{1-z^{n+1}}{1-z}\right)=\frac{1}{2}+\frac{\sin((n+\frac{1}{2})\theta)}{2\sin(\frac{\theta}{2})}\text{ , where $$z is a complex number }.$$
I have developed a good reasoning, but I cannot conclude. Let's go:
$$\operatorname{Re}\left(\frac{1-z^{n+1}}{1-z}\right)=\frac{[1-\cos((n+1)\theta)](1-\cos\theta)+\sin((n+1)\theta)\sin\theta}{[1-2\cos\theta + cos^2\theta +sen^2\theta]}=$$
$$=\frac{1-\cos\theta-\cos((n+1)\theta)+cos\theta\cos((n+1)\theta)+\sin\theta\sin((n+1)\theta)}{2-2\cos\theta}=$$
$$=\frac{1-\cos\theta}{2-2\cos\theta}+\frac{\cos((n+1)\theta)(\cos\theta-1)+\sin\theta\sin((n+1)\theta)}{2-2\cos\theta}=$$
$$=\frac{1}{2}-\frac{\cos((n+1)\theta)}{2}+\frac{\sin\theta\sin((n+1)\theta)}{2-2\cos\theta}.$$
After that, i was unable to continue. I tried to go the other way, that is, try to develop the right side of equality. However, I was not successful. Does anyone have any idea how I can make progress?
Note: I need to do it using only trigonometric relations. I cannot use exponential rules.
Note: The previous steps I did not put in, because I am sure that it is right and it is not necessary for the continuation. I just need to know how to continue to develop to get to the right side of the requested equality.
Best Answer
If you are unable to continue, you can look at the following steps:
$\dfrac{1}{2}-\dfrac{\cos\big((n+1)\theta\big)}{2}+\dfrac{\sin\theta\sin\big((n+1)\theta\big)}{2-2\cos\theta}=$
$=\dfrac{1}{2}+\dfrac{-\cos\big((n+1)\theta\big)(1-\cos\theta)+\sin\theta\sin\big((n+1)\theta\big)}{2(1-\cos\theta)}=$
$=\!\dfrac{1}{2}\!+\!\dfrac{-\!\cos((n\!+\!1)\theta)\!+\!\cos\theta\cos((n\!+\!1)\theta)\!+\!\sin\theta\sin((n\!+\!1)\theta)}{2(1-\cos\theta)}=$
$=\dfrac{1}{2}+\dfrac{-\cos\big((n+1)\theta\big)+\cos\big((n+1)\theta-\theta\big)}{2(1-\cos\theta)}=$
$=\dfrac{1}{2}+\dfrac{-\cos\big((n+1)\theta\big)+\cos(n\theta)}{2(1-\cos\theta)}=$
$=\dfrac{1}{2}+\dfrac{-2\sin\big((n+\frac12)\theta\big)\sin\left(-\frac12\theta\right)}{2(1-\cos\theta)}=$
$=\dfrac{1}{2}+\dfrac{\sin\big((n+\frac12)\theta\big)\sin\left(\frac12\theta\right)}{1-\cos\theta}=$
$=\dfrac{1}{2}+\dfrac{\sin\big((n+\frac12)\theta\big)\sin\left(\frac12\theta\right)}{2\left(\sqrt{\frac{1-\cos\theta}2}\right)^2}=$
$=\dfrac{1}{2}+\dfrac{\sin\big((n+\frac12)\theta\big)\sin\left(\frac{\theta}2\right)}{2\sin^2\left(\frac{\theta}2\right)}=$
$=\dfrac{1}{2}+\dfrac{\sin\big((n+\frac12)\theta\big)}{2\sin\left(\frac{\theta}2\right)}\;.$