Prove the following trigonometric identity without a calculator involved [duplicate]

Let $O = (0,0)$ and $A_i = (\cos 2\pi i/5,\sin 2\pi i/5)$. Then $A_0A_1A_2A_3A_4$ is a regular pentagon with vertices on the unit circle. The sum you've written is the $x$-coordinate of the vector $u = \overrightarrow{OA_0} + \dots \overrightarrow{OA_4}$. If you apply a rotation centred at $O$ with angle $2\pi/5$, the pentagon remains invariant. Therefore $u$ doesn't change when rotated by this angle. That shows that $u = 0$.

Using complex exponential: $$ 1+e^{\frac{2\pi i}{5}}+e^{\frac{4\pi i}{5}}+e^{\frac{6\pi i}{5}}+e^{\frac{8\pi i}{5}}=\frac{(e^{\frac{2\pi i}{5}})^5 -1}{e^{\frac{2\pi i}{5}}-1}=0 $$ and its real part is $0$.

Note that $e^{2i\pi/5}=\omega_5$ is the fifth root of unity. Now $$\omega_5+\omega_5^2+\omega_5^3+\omega_5^4+\omega_5^5=x$$ so \begin{align} \omega_5x&=\omega_5(\omega_5+\omega_5^2+\omega_5^3+\omega_5^4+\omega_5^5)\\ &=\omega_5^2+\omega_5^3+\omega_5^4+\omega_5^5+\omega_5^6\\ &=\omega_5^2+\omega_5^3+\omega_5^4+\omega_5^5+\omega_5\\ &=x \end{align} so $\omega_5x=x$, and so $x\neq 0$ would imply $\omega_5=1$ which is false. So $x=0$. Now use $$e^{i\theta}=\cos(\theta)+i\sin(\theta)$$ Now take $\theta=\frac{2k\pi}5$ for $k=1,2,3,4,5$ to get $\omega_5^k$. So we get: \begin{align} 0&=\Re(0)\\ &=\Re(\omega_5+\omega_5^2+\omega_5^3+\omega_5^4+\omega_5^5)\\ &=\Re(e^{2i\pi/5}+e^{4i\pi/5}+e^{6i\pi/5}+e^{8i\pi/5}+e^{10i\pi/5})\\ &=\cos(\frac{2\pi}5)+\cos(\frac{4\pi}5)+\cos(\frac{6\pi}5)+\cos(\frac{8\pi}5)+\cos(\frac{10\pi}5)\\ &=\cos(\frac{2\pi}5)+\cos(\frac{4\pi}5)+\cos(\frac{6\pi}5)+\cos(\frac{8\pi}5)+1 \end{align}

Let's try this -

Using $$2\cos a \sin b = \sin(a+b) - \sin(a- b)$$ we get:

$$2\cos(2π/5)\sin(2π/5)=\sin(4π/5 )- \sin(0)$$

$$2\cos(4π/5)\sin(2π/5)=\sin(6π/5)-\sin(2π/5)$$

$$2\cos(6π/5)\sin(2π/5)=\sin(8π/5) - \sin(4π/5)$$

$$2\cos(8π/5)\sin(2π/5)=\sin(10π/5) -\sin(6π/5)$$

$$2\cos(10π/5)\sin(2π/5)=\sin(12π/5) - \sin(8π/5) $$

Adding: \begin{align} &2\sin(2π/5)\{\cos(10π/5)+\cos(2π/5)+\cos(4π/5)+\cos(6π/5)+\cos(8π/5)\}\\ &=2\sin(2π/5)\{1+\cos(2π/5)+\cos(4π/5)+\cos(6π/5)+\cos(8π/5)\} \\ &=\sin(12π/5) - \sin(2π/5)\\ &=2\sin(10π/5)\cos(14π/5)\\&=0\qquad [\;\because \sin a - \sin b = 2\sin(a-b)\cos(a+b)] \end{align}

$$1+\cos(2π/5)+\cos(4π/5)+ \cos(6π/5)+\cos(8π/5)=0$$