Question

Consider the polynomial P(x) = (x - cos 36°)(x - cos 84°)(x - cos156°)

• A. 0
• B. 1
• C. -$\frac{1}{2}$
• D. $\frac{\sqrt{5}-1}{2}$

Answer 1

Answer: (A)

0

Answer 2

The roots of the polynomial are $\cos 36^\circ, \cos 84^\circ, \cos 156^\circ$. These are of the form $\cos \theta, \cos(120^\circ - \theta), \cos(120^\circ + \theta)$ with $\theta = 36^\circ$. Such values are the roots of $4x^3 - 3x - \cos(3\theta) = 0$. Dividing by 4, we get $x^3 - \frac{3}{4}x - \frac{1}{4}\cos(3 \times 36^\circ) = 0$. In this equation, the coefficient of $x^2$ is 0.