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[tex]\cos(\frac{\pi}{7})-\cos(\frac{2\pi}{7})+\cos(\frac{3\pi}{7})=\frac12[/tex]

For å få hjernen på et annet spor i denne galskapens tid, forsøker jeg noe her;
setter

[tex]z=\cos(\pi/7)+i\sin(\pi/7)\\ z^7=(\cos(\pi/7)+i\sin(\pi/7))^7 \\ \text vha de Movries formel \\ z^7=\cos(\pi)+i\sin(\pi) = -1 \\ z^7 + 1 = 0 \\ (z+1)(z^6-z^5+z^4-z^3+z^2-z+1)=0 \\ z+1=0\,\,og \\1+z^2+z^4+z^6=z+z^3+z^5[/tex]

dvs

[tex]1+[\cos(2\pi/7)+i\sin(2\pi/7)+\cos(4\pi/7)+i\sin(4\pi/7)+\cos(6\pi/7)+i\sin(6\pi/7)]=\cos(\pi/7)+i\sin(\pi/7)+\cos(3\pi/7)+i\sin(3\pi/7)+\cos(5\pi/7)+i\sin(5\pi/7) \\ \text pga symmetri \\ 1= 2\cos(\pi/7)-2\cos(2\pi/7)+2\cos(3\pi/7) \\ 1/2= \cos(\pi/7)-\cos(2\pi/7)+\cos(3\pi/7)[/tex]