The cubic equation whose roots are 2 cos-7- 2 cos3 -7- 2 cos5 -7- is

Let α = 2 cosπ7, β = 2 cos3π7, γ = 2 cos5π7 α +β +γ =2cosπ7+2cos3π7+2cos5π7 nbsp; = 1sinπ7 (2 sinπ7cosπ7 +2 sinπ7cos3π7+2 sinπ7cos5π7) = 1sinπ7 ( sin2π7 + si ...

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Standard XII

Mathematics

Conditions on the Parameters of Logarithm Function

The cubic equ...

Question

The cubic equation whose roots are $2 cos \frac{π}{7}, 2 cos \frac{3 π}{7}, 2 cos \frac{5 π}{7}$ , is

x^{3} - x^{2} - x + 1 = 0

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x^{3} - x^{2} + 2 x - 1 = 0

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x^{3} - x^{2} - 2 x + 1 = 0

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x^{3} - 2 x^{2} - 2 x + 1 = 0

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Solution

The correct option is C $x^{3} - x^{2} - 2 x + 1 = 0$
Let $α = 2 cos \frac{π}{7}, β = 2 cos \frac{3 π}{7}, γ = 2 cos \frac{5 π}{7}$

$α + β + γ$
$= 2 cos \frac{π}{7} + 2 cos \frac{3 π}{7} + 2 cos \frac{5 π}{7}$
$= \frac{1}{sin \frac{π}{7}} (2 sin \frac{π}{7} cos \frac{π}{7} + 2 sin \frac{π}{7} cos \frac{3 π}{7} + 2 sin \frac{π}{7} cos \frac{5 π}{7})$
$= \frac{1}{sin \frac{π}{7}} (sin \frac{2 π}{7} + sin \frac{4 π}{7} - sin \frac{2 π}{7} + sin \frac{6 π}{7} - sin \frac{4 π}{7})$
$= \frac{sin \frac{6 π}{7}}{sin \frac{π}{7}} = 1$

$α β + β γ + γ α$
$= 2 (2 cos \frac{π}{7} cos \frac{3 π}{7} + 2 cos \frac{3 π}{7} cos \frac{5 π}{7} + 2 cos \frac{5 π}{7} cos \frac{π}{7})$
$= 2 (cos \frac{4 π}{7} + cos \frac{2 π}{7} + cos \frac{8 π}{7} + cos \frac{2 π}{7} + cos \frac{6 π}{7} + cos \frac{4 π}{7})$
$= 2 (- cos \frac{4 π}{7} - cos \frac{5 π}{7} - cos \frac{π}{7} - cos \frac{5 π}{7} - cos \frac{π}{7} - cos \frac{3 π}{7})$
$= - 4 (cos \frac{π}{7} + cos \frac{3 π}{7} + cos \frac{5 π}{7})$
$= - 4 (\frac{1}{2}) = - 2$

$α β γ$
$= 8 cos \frac{π}{7} cos \frac{3 π}{7} cos \frac{5 π}{7}$
$= 8 cos \frac{π}{7} cos \frac{2 π}{7} cos \frac{4 π}{7}$
$= 8 \times \frac{sin \frac{8 π}{7}}{8 sin \frac{π}{7}} = - 1$

Required equation is
$x^{3} - (α + β + γ) x^{2} + (α β + β γ + γ α) x - α β γ = 0$
$\Rightarrow x^{3} - x^{2} - 2 x + 1 = 0$

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Similar questions

Q. The cubic equation whose roots are thrice the roots of

x^{3} + 2 x^{2} - 4 x + 1 = 0

, is:

Q. If the equation whose roots are the squares of the roots of the cubic

x^{3} - a x^{2} + b x - 1 = 0

is identical with the given cubic equation, then

Q. The equation whose roots are

{tan}^{2} \frac{2 π}{7}, {tan}^{2} \frac{4 π}{7}, {tan}^{2} \frac{6 π}{7}

Q. The cubic equation whose roots are the AM, GM and HM of the roots of

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Q. Formation of equations whose roots are given :
Find the equation whose roots are

c o s \frac{2 π}{7}, c o s \frac{4 π}{7} a n d c o s \frac{8 π}{7}

c o s \frac{2 π}{7}, c o s \frac{4 π}{7} a n d c o s \frac{6 π}{7}

Deduction :

c o s \frac{2 π}{4} c o s \frac{4 π}{7} c o s \frac{8 π}{7} = S_{3} = \frac{1}{8}

c o s \frac{2 π}{4} + c o s \frac{4 π}{7} + c o s \frac{8 π}{7} = S_{1} = - \frac{1}{2}

The equation whose roots are

s e c \frac{2 π}{7}, s e c \frac{4 π}{7}, s e c \frac{8 π}{7} (= s e c \frac{6 π}{7})

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The cubic equation whose roots are 2cosπ7,2cos3π7,2cos5π7, is

The cubic equation whose roots are $2 cos \frac{π}{7}, 2 cos \frac{3 π}{7}, 2 cos \frac{5 π}{7}$ , is