Question

Consider the Polynomial $P\left( x \right) = \left( {x - \cos 36^\circ } \right)\left( {x - \cos 84^\circ } \right)\left( {x - \cos 156^\circ } \right)$
A) The coefficient of ${x^2}$ is zero.
B) The coefficient of $x$ is $\dfrac{{ - 3}}{4}$ .
C) Constant term is $\dfrac{{\sqrt 5 - 1}}{{16}}$
D) $P\left( {\dfrac{1}{2}} \right) < 0$

Answer 1

Hint:
Here, we will find which of the following is true for the given Polynomial. We will expand the given polynomial by using the algebraic identity. Then we will simplify the required coefficient by using the trigonometric identity to find which of the following is true for the given polynomial. A polynomial is defined as an expression with constants and variables.

Formula Used:
We will use the following formula:
1) The Product of the three Binomials is given by the Algebraic Identity $\left( {x + a} \right)\left( {x + b} \right)\left( {x + c} \right) = {x^3} + \left( {a + b + c} \right){x^2} + \left( {ab + bc + ca} \right)x + abc$
2) The sum of cosines of two angles is given by the formula $\cos A + \cos B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)$

Complete step by step solution:
We are given with a Polynomial $P\left( x \right) = \left( {x - \cos 36^\circ } \right)\left( {x - \cos 84^\circ } \right)\left( {x - \cos 156^\circ } \right)$
Now, by using the formula $\left( {x + a} \right)\left( {x + b} \right)\left( {x + c} \right) = {x^3} + \left( {a + b + c} \right){x^2} + \left( {ab + bc + ca} \right)x + abc$, we get
$ P\left( x \right) = {x^3} - \left( {\cos 36^\circ + \cos 84^\circ + \cos 156^\circ } \right){x^2} + \left( {\cos 36^\circ \cos 84^\circ + \cos 84^\circ \cos 156^\circ + \cos 156^\circ \cos 36^\circ } \right)x \\
   - \cos 36^\circ \cos 84^\circ \cos 156^\circ \\ $
Now, we will find the coefficient of ${x^2}$ in the Polynomial.
We know that the coefficient of ${x^2}$ in the given Polynomial is $ - \left( {\cos 36^\circ + \cos 84^\circ + \cos 156^\circ } \right)$.
Now, by using the sum of cosines of two angles formula $\cos A + \cos B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)$ for the last two cosines in the coefficient of ${x^2}$, we get
$\cos 84^\circ + \cos 156^\circ = 2\cos \left( {\dfrac{{84 + 156}}{2}} \right)\cos \left( {\dfrac{{156 - 84}}{2}} \right)$
By simplifying, we get
$ \Rightarrow \cos 84^\circ + \cos 156^\circ = 2\cos \left( {\dfrac{{240}}{2}} \right)\cos \left( {\dfrac{{72}}{2}} \right)$
Dividing the terms, we get
$ \Rightarrow \cos 84^\circ + \cos 156^\circ = 2\cos \left( {120^\circ } \right)\cos \left( {36^\circ } \right)$
By substituting $\cos \left( {120^\circ } \right) = - \dfrac{1}{2}$ in the above equation, we get
$ \Rightarrow \cos 84^\circ + \cos 156^\circ = 2 \times \dfrac{{ - 1}}{2} \times \cos 36^\circ $
Multiplying the terms, we get
$ \Rightarrow \cos 84^\circ + \cos 156^\circ = - \cos 36^\circ $
Now, by substituting the value in the coefficient of ${x^2}$, we get
$ - \left( {\cos 36^\circ + \cos 84^\circ + \cos 156^\circ } \right) = - \left( {\cos 36^\circ - \cos 36^\circ } \right)$
$ \Rightarrow - \left( {\cos 36^\circ + \cos 84^\circ + \cos 156^\circ } \right) = 0$
Thus, the coefficient of ${x^2}$ in the given Polynomial is 0.
Therefore, the coefficient of ${x^2}$ in the Polynomial $P\left( x \right) = \left( {x - \cos 36^\circ } \right)\left( {x - \cos 84^\circ } \right)\left( {x - \cos 156^\circ } \right)$ is 0.

Thus, Option (A) is the correct answer.

Note:
Here, we should note that sine and tangent are odd functions since both the functions are symmetric about the origin. Cosine is an even function since the functions are symmetric about the $y$ axis. So, we take the arguments in the negative sign for odd functions and positive signs for even functions. We should also remember that Polynomials can be expanded only by using algebraic identities.