<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] 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.
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