<prompt> Rahisishacos frac2 pi13 cos frac6 pi13 cos frac8 pi13.  <response>: Acha x cos frac2 pi13 cos frac6 pi13 cos frac8 pi13, na uruhusu omega e2 pi i13. Kisha omega13 e2 pi i 1. Tunaona kwamba x ndiyo sehemu halisi yaomega omega3 omega4.Tangu omega 1, overlineomega frac1omega. Hivyo, x pia ni sehemu halisi yaanzaalignumuhimu zaidiomega omega3 omega4 ujumlaomega ujumlaomega3 ujumlaomega4 frac1omega frac1omega3 frac1omega4 omega12 omega10 omega9.mwishoalignKwa hiyo,x fracomega omega3 omega4 omega9 omega10 omega122.Kutoka kwenye mlinganyo omega13 1, omega13 - 1 0, ambayo huchangia kamaomega - 1omega12 omega11 omega10 nukta 1 0.Tangu omega neq 1,1 omega omega2 dots omega12 0.Hebuanzaalignalfa omega omega3 omega4 omega9 omega10 omega12, beta omega2 omega5 omega6 omega7 omega8 omega11.mwishoalignKisha alpha beta omega omega2 omega3 dots omega12 -1.Pia, kwa kutumia ukweli kwamba omega13 1, bidhaa alpha beta hurahisishaalpha beta 3 omega 3 omega2 dots 3 omega12 -3.Kwa hivyo, alpha na beta ndio mizizi ya z 2 z - 3 0. Kwa fomula ya quadratic,z frac-1 pm sqrt132.Hivyo, x frac-1 sqrt134 au x frac -1 - sqrt134.Kumbuka hilocos frac8 pi13 -cos left pi - frac8 pi13 kulia -cos frac5 pi13 ,hivyox cos frac2 pi13 cos frac6 pi13 cos frac8 pi13 left cos frac 2 pi13 - cos frac5 pi13 kulia cos frac6 pi13 0.Kwa hivyo,x fracsqrt13 - 14. 