Question

1.Show that (1+2cos$) is a factor of f($)= 8cos$^3-4cos$-1
2.If $= 36° is a solution of f($)=0, deduce that cos36° = 1\4(√5+1)

Answer 1

Let c = cos θ and f(θ) = 8c^3 − 4c − 1.

1) Check that c = −1/2 is a root:
f(−1/2) = 8(−1/8) − 4(−1/2) − 1 = −1 + 2 − 1 = 0,
so (2c + 1) (equivalently 1 + 2 cos θ) is a factor. Dividing gives
f(θ) = (2c + 1)(4c^2 − 2c − 1).

2) If θ = 36° is a solution, then c = cos 36° is a root of f, and since cos 36° ≠ −1/2 we must have
4c^2 − 2c − 1 = 0.
Solve:
c = [2 ± √(4 + 16)] / 8 = [2 ± 2√5] / 8 = (1 ± √5)/4.
Because cos 36° > 0, choose the plus sign:
cos 36° = (1 + √5) / 4.