<s>[INST] How do we prove $\cos(\pi/5) - \cos(2\pi/5) = 0.5$ ?. How do we prove $\cos(\pi/5) - \cos(2\pi/5) = 0.5$ without using a calculator.
Related question: how do we prove that $\cos(\pi/5)\cos(2\pi/5) = 0.25$, also without using a calculator
 [/INST] $$\cos(\pi/5) - \cos(2\pi/5)= {\rm Re}(e^{i\frac{\pi}{5}}-e^{i\frac{2\pi}{5}})=\frac{1}{2}( e^{i\frac{\pi}{5}}-e^{i\frac{2\pi}{5}}+ \overline{e^{i\frac{\pi}{5}}-e^{i\frac{2\pi}{5}}})=\frac{1}{2}( e^{i\frac{\pi}{5}}-e^{i\frac{2\pi}{5}}+ e^{-i\frac{\pi}{5}}-e^{-i\frac{2\pi}{5}})=\frac{e^{\frac{-2i\pi}{5}}}{2}( -e^{i\frac{4\pi}{5}}+e^{i\frac{3\pi}{5}}+ e^{i\frac{\pi}{5}}-1)$$
To simplify the computations, let $\omega=e^{\frac{i \pi}{5}}$. Note that $\omega^5=-1$.
Then
$$\cos(\pi/5) - \cos(2\pi/5)= \frac{-1}{2\omega^2}(\omega^4-\omega^3-\omega+1)= \frac{-1}{2\omega^2}(\omega^4-\omega^3+\omega^2-\omega+1-\omega^2)$$
$$\cos(\pi/5) - \cos(2\pi/5)=  \frac{-1}{2\omega^2}(\frac{\omega^5+1}{\omega+1} -\omega^2)=\frac{-1}{2\omega^2}(0 -\omega^2)=\frac{1}{2}$$
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