Let A-B-C- and - - sin3B-C-sin2C-A- - sin3A-C-sin2A-B- -sin3A-B-sin2B-Care roots of the cubic equation x3-ax2 -bx-c - 0- then the value of a is

Α = sin^3(B+C)·sin(2C+A) =sin^3(π A)·sin(2C+π B C) =sin^3A·sin(B C) =sin^2Asin Asin(B C) =sin^2Asin(B+C)sin(B C) =sin^2A(sin^2B sin^2C) =sin^2A(sin^ ...

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Byju's Answer

Standard XII

Mathematics

Conditional Identities

Let A+B+C=π a...

Question

Let $A + B + C = π$ and $α = {sin}^{3} (B + C) \cdot sin (2 C + A), β = {sin}^{3} (A + C) \cdot sin (2 A + B), γ = {sin}^{3} (A + B) \cdot sin (2 B + C)$
are roots of the cubic equation $x^{3} + a x^{2} + b x + c = 0$ , then the value of $a$ is

0

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sin A sin B sin C

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1

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cos A cos B cos C

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Solution

The correct option is A $0$
$α = {sin}^{3} (B + C) \cdot sin (2 C + A) = {sin}^{3} (π - A) \cdot sin (2 C + π - B - C) = {sin}^{3} A \cdot sin (B - C) = {sin}^{2} A sin A sin (B - C) = {sin}^{2} A sin (B + C) sin (B - C) = {sin}^{2} A ({sin}^{2} B - {sin}^{2} C) = {sin}^{2} A ({sin}^{2} B - {sin}^{2} C)$
Similarly,
$β = {sin}^{2} B ({sin}^{2} C - {sin}^{2} A) γ = {sin}^{2} C ({sin}^{2} A - {sin}^{2} B)$

Now,
$- a = α + β + γ = {sin}^{2} A ({sin}^{2} B - {sin}^{2} C) + {sin}^{2} B ({sin}^{2} C - {sin}^{2} A) + {sin}^{2} C ({sin}^{2} A - {sin}^{2} B) ∴ - a = 0 \Rightarrow a = 0$

Alternate solution:
If we assume $A = B = C = 60^{\circ}$ , we get
$α = {sin}^{3} 120^{\circ} sin 180^{\circ} = 0$
Similarly,
$β = 0, γ = 0$
Sum of roots
$α + β + γ = - a \Rightarrow 0 = - a ∴ a = 0$

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Similar questions

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A + B + C = π

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are roots of the cubic equation

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Let A+B+C=π and α=sin3(B+C)⋅sin(2C+A),β=sin3(A+C)⋅sin(2A+B),γ=sin3(A+B)⋅sin(2B+C) are roots of the cubic equation x3+ax2+bx+c=0, then the value of a is

Let $A + B + C = π$ and $α = {sin}^{3} (B + C) \cdot sin (2 C + A), β = {sin}^{3} (A + C) \cdot sin (2 A + B), γ = {sin}^{3} (A + B) \cdot sin (2 B + C)$
are roots of the cubic equation $x^{3} + a x^{2} + b x + c = 0$ , then the value of $a$ is