2011/050) If a and b are roots of the equation x^2 – 2x cos t +1 = 0 form the equation whose roots are a^n and b^n

As a and b are roots so

a + b = 2 cos t … 1
ab = 1 …2

we have (a-b)^2 = (a+b)^2 – 4ab = 4 cos^2 t – 4 = - 4 sin ^2 t

or a-b = 2 i sin t … 3

from 1 and 3 adding 2a = 2 (cos t + i sin t) = 2 e^it or a= e^it
subtracting 2b = 2 (cos t - i sin t) = 2 e^it or b = e^-it

now a^n + b^n = (e^int + e^-int) = 2 cos nt

and a^nb^n = (ab)^n = 1

so the equation = (x^2- 2x cos nt + 1) = 0