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 :$1,\left(\cos \left(\frac{2 \pi}{5}\right)+i \sin \left( \frac{2\pi}{5}\right)\right),\left(\cos \left(\frac{4 \pi}{5}\right)+i \sin \left(\frac{4 \pi}{5}\right)\right),\left(\cos \left(\frac{6 \pi}{5}\right)+i \sin \left(\frac{6 \pi}{5}\right)\right),  \left(\cos \left(\frac{8 \pi}{3}\right)+i \sin \left(\frac{8 \pi}{5}\right)\right)$

$$z^5=1$$
$$z^5-1=0$$
$$(z-1)(z^4+z^3+z^2+z+1)=0$$
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 $e^{\frac{2\pi i }{5}}=z$
$$(e^{\frac{2\pi i }{5}}-1)(e^{\frac{8\pi i }{5}}+ e^{\frac{6\pi i }{5}}+ e^{\frac{4\pi i }{5}}+ e^{\frac{2\pi i }{5}}+1)=0$$
