Consider the polynomial P ( x )=( x - cos 36°)( x - cos 84°)( x - cos 156°) The absolute term in P ( x ) has the value equal to

Question

Consider the polynomial P ( x )=( x - cos 36°)( x - cos 84°)( x - cos 156°) The absolute term in P ( x ) has the value equal to (A) (√5-1/4) (B) (√5-1/16) (C) (√5+1/16) (D) (1/16)

Tardigrade Admin · Accepted Answer

text Using - cos 36° cos 84° cos 156° cos 24° cos 36° cos 84° cos θ cos (60°-θ) cos (60°-θ)=( cos 3 θ/4) =( cos (3 ⋅ 24)/4)=( cos 72°/4)=(√5-1/4 ⋅ 4)=(√5-1/16)

Q. Consider the polynomial $P (x) = (x - cos 3 6^{\circ}) (x - cos 8 4^{\circ}) (x - cos 15 6^{\circ})$
The absolute term in $P (x)$ has the value equal to

$\frac{5 - 1}{4}$

$\frac{5 - 1}{16}$

$\frac{5 + 1}{16}$

$\frac{1}{16}$

$Using - cos 3 6^{\circ} cos 8 4^{\circ} cos 15 6^{\circ}$
$cos 2 4^{\circ} cos 3 6^{\circ} cos 8 4^{\circ}$
$cos θ cos (6 0^{\circ} - θ) cos (6 0^{\circ} - θ) = \frac{c o s 3 θ}{4}$
$= \frac{c o s ( 3 \cdot 24 )}{4} = \frac{c o s 7 2 ^{\circ}}{4} = \frac{5 - 1}{4 \cdot 4} = \frac{5 - 1}{16}$

Q. Consider the polynomial P(x)=(x−cos36∘)(x−cos84∘)(x−cos156∘) The absolute term in P(x) has the value equal to

45​−1​

165​−1​

165​+1​

161​