<s>[INST] 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 :$1,\left(\cos \left(\frac{2 \pi}{5}\right)+i \sin \left( \frac{2\pi}{5}\right)\right),\left(\cos \left(\frac{4 \pi}{5}\right)+i \sin \left(\frac{4 \pi}{5}\right)\right),\left(\cos \left(\frac{6 \pi}{5}\right)+i \sin \left(\frac{6 \pi}{5}\right)\right),  \left(\cos \left(\frac{8 \pi}{3}\right)+i \sin \left(\frac{8 \pi}{5}\right)\right)$
 [/INST] 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 $5$.
</s>