Question

Consider the polynomial $P\left(x\right)=\left(x-\cos {36}^{\circ }\right)\left(x-\cos {84}^{\circ }\right)\left(x-\cos {156}^{\circ }\right)$ , then coefficient of $x^2$ is

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

Answer 1

The correct option is A 0

$P\left(x\right)=(x-\cos {36}^{\circ })(x-\cos {84}^{\circ })(x-\cos {156}^{\circ })$

Coefficient of $x^2=-(\cos {36}^{\circ }+\cos {84}^{\circ }+\cos {156}^{\circ })$

$=-(\cos {36}^{\circ }+\cos ({90}^{\circ }-6^{\circ })+\cos ({180}^{\circ }-{24}^{\circ }))$

$=-(\cos {36}^{\circ }+\sin 6^{\circ }-\cos {24}^{\circ })$

$=-(\cos {36}^{\circ }-\cos {24}^{\circ }+\sin 6^{\circ })$

$=-(-2\sin {30}^{\circ }\sin 6^{\circ }+\sin 6^{\circ })$

$=-(-2\times \frac{1}{2}\sin 6^{\circ }+\sin 6^{\circ })$

$=-(-\sin 6^{\circ }+\sin 6^{\circ })=0$