If a + 1/a = 2 Cos 6 then a^1000 + 1/a^1000 + 1 = ??

Question

If in a triangle ABC:
Cos A + Cos B + Cos C = 3/2
then what is so special about this triangle?

Answer 1

a + 1/a = 2 cos (6) => 6 is in radians
a^2 + 1/a^2 = (a + 1/a)^2 - 2 = 4 cos^2(6) - 2 = 2 cos(12)
a^3 +1/a^3 = (a+ 1/a) * (a^2 + 1/a^2) - (a + 1/a)
= 2 cos(6) * [4 cos^2(6) - 2] - 2 cos(6) = 2 cos(18)
a^4 + 1/a^4 = (a + 1/a) * (a^3 + 1/a^3) - (a^2 + 1/a^2)
= 2 cos(6) * 2 cos (18) - 2 cos(12) = 2 cos(24)
in general, for x > 1
a^x + 1/a^x = 2 cos(6x)
hence:
a^1000 + 1/a^1000 + 1 = 2 cos(6000) +1