Question

cos 2π/7+cos 4π/7+cos 6π/7=-1/2
Докажите равенство

Answer 1

cos 2+cos 4π/7+cos 6π/7=-1/2
-0,5=-1/2
-0,5=-0,5
Чтд
Пошаговое решение:
х=cos(π/7)+Isin(π/7)
x^7=cos(π)+I
sin(π)=-0,5
<br/>x^7+0,5=(x+0,5)
(x^6-x^5+x^4-x^3+x^2-x+0,5)=0
Т.к х неравен 0,5, то
x^6+x^4+x^2=x+x^3+x^5
<br/> cos(6π/7)+cos (4π/7)+cos (2π/7)=cos (π/7)+
cos (3π/7)+cos (5π/7)-0,5
<br/>
cos (π/7)=-cos (6π/7)
cos (3π/7)=-cos (4π/7)
cos (5π/7)=-cos (2π/7)
(cos (6π/7)+cos (4π/7)+cos (2π/7))=-0,5