Is it possible to show that the fifth roots of 1 add up to 0 simply by using trigonometric identities? You can't use geometric sums, minimal polynomials, pentagon, and exact values with radicals.
All the five,  fifth-roots of unity are :$1,\left(\cos \left(\frac{2 \pi}{5}\right)+i \sin \left( \frac{2\pi}{5}\right)\right),\left(\cos \left(\frac{4 \pi}{5}\right)+i \sin \left(\frac{4 \pi}{5}\right)\right),\left(\cos \left(\frac{6 \pi}{5}\right)+i \sin \left(\frac{6 \pi}{5}\right)\right),  \left(\cos \left(\frac{8 \pi}{3}\right)+i \sin \left(\frac{8 \pi}{5}\right)\right)$
 A: If you want a simple trigonometric proof, recall
$$\sin a-\sin b=2\sin\frac{a-b}2\cos\frac{a+b}2.$$
Therefore
\begin{align}
0&=\sin(a+2\pi)-\sin a\\
&=(\sin(a+2\pi)-\sin(a+8/\pi/5))+(\sin(a+8/\pi/5)-\sin(a+6/\pi/5))\\
&+(\sin(a+6/\pi/5)-\sin(a+4/\pi/5))+(\sin(a+4/\pi/5)-\sin(a+2/\pi/5))\\
&+(\sin(a+2/\pi/5)-\sin a)\\
&=2(\sin\pi/5)\left[\cos(a+9\pi/5)+\cos(a+7\pi/5)+\cos(a+\pi)+\cos(a+3\pi/5)+\cos(a+\pi/5)\right]
\end{align}
and
$$\cos(a+9\pi/5)+\cos(a+7\pi/5)+\cos(a+\pi)+\cos(a+3\pi/5)+\cos(a+\pi/5)=0.$$
Taking $a=-\pi/5$ gives
$$\cos(8\pi/5)+\cos(6\pi/5)+\cos(4\pi/5)+\cos(2\pi/5)+1=0.$$
and taking $a=-7\pi/10$ gives
$$\sin(8\pi/5)+\sin(6\pi/5)+\sin(4\pi/5)+\sin(2\pi/5)+0=0.$$
These are the real and imaginary parts of the sum of the fifth roots
of zero.
Of course, this method works for other values of $5$.
A: Let $S$ be the set of roots of $z^5-1$, then $$e^{\frac{2\pi i}{5}}S=\{e^{\frac{2\pi i}{5}}, e^{\frac{4\pi i}{5}},e^{\frac{6\pi i}{5}},e^{\frac{8\pi i}{5}},e^{\frac{10\pi i}{5}}\}=\{e^{\frac{2\pi i}{5}},e^{\frac{4\pi i}{5}},e^{\frac{6\pi i}{5}},e^{\frac{8\pi i}{5}},1\}=S$$
The sets are equal, so $$e^{\frac{2\pi i}{5}}\sum_{z\in S}z=\sum_{z\in S}z\\\Rightarrow \sum_{z\in S}z=0$$

More generally

Let $G$ be a multiplicative group of order $n$. Then $$\sum_{z\in G}z=0$$

Proof: Let $g\in G$ such that $g\neq e$. Let $\phi_g(x)=gx$, then $\phi_g\in\text{Aut}(G)$ since:

*

*injective: If $\phi_g(x)=\phi_g(y)$ then $gx=gy\Rightarrow x=y$.

*surjectivity: If $a\in G$, then $\phi_g(g^{-1}a)=a$.

This means that the sets $gG,G$ are equal. And so,
$$\sum_{z\in G}gz=\sum_{z\in G}g\iff (g-e)\sum_{z\in G}z=0\\\Rightarrow \sum_{z\in G}z=0$$
A: You don't need "geometric sum" to prove
$$
\sum_{j=0}^{n-1} \cos(\theta+j\phi)=\frac{\sum_{j=0}^{n-1}\cos(\theta+j\phi)\sin(\phi/2)}{\sin(\phi/2)}=\frac{\sum_{j=0}^{n-1}\left[\sin(\theta+(j+\tfrac12)\phi)-\sin(\theta+(j-\tfrac12)\phi)\right]}{\sin(\phi/2)}=\dots
$$
and similarly $\sum_{j=0}^{n-1}\sin(\theta+j\phi)$.
A: Good question. Shall we try?
So we have the five roots: $$1,\left(\cos \left(\frac{2 \pi}{5}\right)+i \sin \left( \frac{2\pi}{5}\right)\right),\left(\cos \left(\frac{4 \pi}{5}\right)+i \sin \left(\frac{4 \pi}{5}\right)\right),\left(\cos \left(\frac{6 \pi}{5}\right)+i \sin \left(\frac{6 \pi}{5}\right)\right),  \left(\cos \left(\frac{8 \pi}{3}\right)+i \sin \left(\frac{8 \pi}{5}\right)\right)$$
I am just going to write $\frac{2\pi}{5}$ as $\alpha$. It's going to save time and make this easier to read. In this notation, the roots are :
$1, \ \cos (\alpha) + i\cdot\sin(\alpha), \ \cos (2\alpha) + i\cdot\sin(2\alpha), \ \cos (3\alpha) + i\cdot\sin(3\alpha)$ and $\cos (4\alpha) + i\cdot\sin(4\alpha)$
OK let's add them all up, we'll start with the imaginary part because that is going to be easy. We have $$i \times \left[\sin \alpha+\sin 2\alpha+\sin 3\alpha+\sin 4\alpha\right]$$ Well we know that $5\alpha = 2\pi$ and so $\sin \alpha = \sin(2\pi-\alpha) = \sin(-4 \alpha) =-\sin (4\alpha$). Also $\sin(2 \alpha)=-\sin(3 \alpha)$ for similar reasons and so all of these terms cancel to zero.
In the real part, we have $$1+\cos(\alpha)+\cos(2\alpha)+\cos(3\alpha)+\cos(4\alpha)$$ $\cos(\alpha)=\cos(2\pi-\alpha)=\cos(4\alpha)$ Also $\cos(2 \alpha)=\cos(3 \alpha)$ for similar reasons. So we just need to show that $$1+2\cos(\alpha)+2\cos(2\alpha)=0$$
How hard can this be? We know, from the trig identities that $\cos(2\alpha)=2 \cos^{2} (\alpha)-1$. So now we need $1+2 \cos(\alpha)+4 \cos^{2}(\alpha)-2=0$ or $$4 \cos^{2}(\alpha)+2 \cos(\alpha)-1=0$$
Trigonometry tells us that $\cos \alpha = \frac{\sqrt 5-1}{4}$. If we pop this in then we get.
$$\frac{6-2 \sqrt 5}{4}+\frac{\sqrt 5-1}{2}-1=0$$
Hooray and Phew.
A: $$z^5=1$$
$$z^5-1=0$$
$$(z-1)(z^4+z^3+z^2+z+1)=0$$
Note that the parenthesis  with 5 terms are the roots but as we know the complex is a filed no zero divisors thus the sum of the roots is zero!
Plug in $e^{\frac{2\pi i }{5}}=z$
$$(e^{\frac{2\pi i }{5}}-1)(e^{\frac{8\pi i }{5}}+ e^{\frac{6\pi i }{5}}+ e^{\frac{4\pi i }{5}}+ e^{\frac{2\pi i }{5}}+1)=0$$
