# Prove that $\operatorname{Re}\left(\frac{1-z^{n+1}}{1-z}\right)=\frac{1}{2}+\frac{\sin((n+\frac{1}{2})\theta)}{2\sin(\frac{\theta}{2})}$.

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.

• What's the connection between $z$ and $\theta$? Apr 29 at 22:04
• @JoséCarlosSantos $z=\cos\theta +i\sin\theta$, with $|z|=1$. So, by the Moivre's Theorem, $z^k=\cos(k\theta)+i\sin(k\theta)$. Apr 29 at 22:08
• I think it's important to say to you is that the term inside the "Re", i took it multiplied by the conjulgated of $1-z$ up and down. Apr 29 at 22:11
• Does this answer your question? Proving complex series $1 + \cos\theta + \cos2\theta +... + \cos n\theta$ Apr 29 at 22:37
• @rtybase unfortunelly, no! This prove that you sent me uses exponential relations. Apr 29 at 22:44

If you are unable to continue, you can look at the following steps:

$$\dfrac{1}{2}-\dfrac{\cos\big((n+1)\theta\big)}{2}+\dfrac{\sin\theta\sin\big((n+1)\theta\big)}{2-2\cos\theta}=$$

$$=\dfrac{1}{2}+\dfrac{-\cos\big((n+1)\theta\big)(1-\cos\theta)+\sin\theta\sin\big((n+1)\theta\big)}{2(1-\cos\theta)}=$$

$$=\!\dfrac{1}{2}\!+\!\dfrac{-\!\cos((n\!+\!1)\theta)\!+\!\cos\theta\cos((n\!+\!1)\theta)\!+\!\sin\theta\sin((n\!+\!1)\theta)}{2(1-\cos\theta)}=$$

$$=\dfrac{1}{2}+\dfrac{-\cos\big((n+1)\theta\big)+\cos\big((n+1)\theta-\theta\big)}{2(1-\cos\theta)}=$$

$$=\dfrac{1}{2}+\dfrac{-\cos\big((n+1)\theta\big)+\cos(n\theta)}{2(1-\cos\theta)}=$$

$$=\dfrac{1}{2}+\dfrac{-2\sin\big((n+\frac12)\theta\big)\sin\left(-\frac12\theta\right)}{2(1-\cos\theta)}=$$

$$=\dfrac{1}{2}+\dfrac{\sin\big((n+\frac12)\theta\big)\sin\left(\frac12\theta\right)}{1-\cos\theta}=$$

$$=\dfrac{1}{2}+\dfrac{\sin\big((n+\frac12)\theta\big)\sin\left(\frac12\theta\right)}{2\left(\sqrt{\frac{1-\cos\theta}2}\right)^2}=$$

$$=\dfrac{1}{2}+\dfrac{\sin\big((n+\frac12)\theta\big)\sin\left(\frac{\theta}2\right)}{2\sin^2\left(\frac{\theta}2\right)}=$$

$$=\dfrac{1}{2}+\dfrac{\sin\big((n+\frac12)\theta\big)}{2\sin\left(\frac{\theta}2\right)}\;.$$

A bit long for a comment, but may be of help: $$\frac{1-z^{n+1}}{1-z}=1+z+z^2+\cdots+z^n.$$ Assuming $$z=re^{i\theta}$$ then you have $$1+r\cos\theta+r^2\cos 2\theta+\cdots+r^n\cos n\theta,$$ so presumably your $$z$$ are of unit size and $$r=1$$ since this sum is related to Lagrange's trigonometric identity.

So in your case, $$\sum_{k=0}^n\cos(k\theta)=\frac{1}{\sin(\theta/2)}\sum_{k=0}^n\sin(\theta/2)\cos(k\theta)\\=\frac{1}{\sin(\theta/2)}\sum_{k=0}^n\sin(\theta(k+1/2))-\sin(\theta(k-1/2)).$$ This sum telescopes, e.g. $$3+1/2=4-1/2$$, so the right hand side is equal to $$\frac{\sin \left(\frac{1}{2} (2 n+1) \theta\right)+\sin \left(\frac{\theta}{2}\right)}{2\sin(\theta/2)}=\frac{1}{2}+\frac{\sin\left(\frac{1}{2}(2n+1)\theta\right)}{2\sin(\theta/2)}.$$

• I started from this first equality, but my goal is not to use exponential, just trigonometric relations. Apr 29 at 22:25
• @Manatee there's more than one way to peel an orange 🍊. You can use trigonometric telescoping instead of exponentials, see e.g. en.m.wikipedia.org/wiki/… Apr 30 at 6:15
• As I said, I was looking for some prove that doesn't contain exponential relation. Thanks! Apr 30 at 14:30
• @Manatee what do you mean by "exponential relation" ? Do you refer to DeMoivre's theorem, Euler's identity, or simply $z^n$ ? Apr 30 at 18:58
• See the Angelo's proof above. That is what I needed. =) May 4 at 23:32