Assertion - sin 3x-sin 3 x-2-3 -sin 3 x-4-3 -3sin xsin x-2-3 -sin x-4-3 Reason- If a-b-c-0- then a3-b3-c3-3abc

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Definite Integral as Limit of Sum

Assertion : ...

Question

Assertion : ${sin}^{3} x + {sin}^{3} (x + \frac{2 π}{3}) + {sin}^{3} (x + \frac{4 π}{3}) = 3 sin x sin (x + \frac{2 π}{3}) \cdot sin (x + \frac{4 π}{3})$ Reason: If $a + b + c = 0$ , then $a^{3} + b^{3} + c^{3} = 3 a b c$

Both Assertion & Reason are individually true & Reason is correct explanation of Assertion,

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Both Assertion & Reason are individually true but Reason is not the ,correct (proper) explanation of Assertion,

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Assertion is true but Reason is false,

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Assertion is false but Reason is true.

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Solution

The correct option is A Both Assertion & Reason are individually true & Reason is correct explanation of Assertion,
Assertion is true as
${sin}^{3} x + {sin}^{3} (x + \frac{2 π}{3}) + {sin}^{3} (x + \frac{4 π}{3}) {sin}^{3} A = \frac{3 sin A - sin 3 A}{4} = \frac{1}{4} (3 sin x - sin 3 x + 3 sin (x + \frac{2 π}{3}) - sin (3 x + 2 π) + 3 sin (x + \frac{4 π}{3}) - sin (3 x + 4 π)) = \frac{1}{4} (3 sin x - sin 3 x + 3 sin (x + 120) - sin 3 x + 3 sin (x + 240) - sin 3 x) = \frac{1}{4} (3 sin x - 3 sin 3 x + 3 (sin (x + 120) + sin (x + 240))) = \frac{1}{4} (3 sin x - 3 sin 3 x + 3 (sin (180 - (60 - x)) + sin (180 + (60 + x)))) = \frac{1}{4} (3 sin x - 3 sin 3 x + 3 (sin (60 - x) - sin (60 + x))) = \frac{1}{4} (3 sin x - 3 sin 3 x + 3 (- 2 cos 60 sin x)) = \frac{1}{4} (3 sin x - 3 sin 3 x + 3 (- 2 \frac{1}{2} sin x)) = - \frac{3 sin 3 x}{4} 3 sin x sin (x + \frac{2 π}{3}) sin (x + \frac{4 π}{3}) = 3 sin x sin (x + 120) sin (x + 240) = - 3 sin x sin (60 - x) sin (60 + x) = - 3 sin x ({sin}^{2} 60 {cos}^{2} x - {cos}^{2} 60 {sin}^{2} x) = - 3 sin x (\frac{3}{4} {cos}^{2} x - \frac{1}{4} {sin}^{2} x) = - \frac{3 sin 3 x}{4}$
Reason is true as for
$sin x + sin (x + \frac{2 π}{3}) + sin (x + \frac{4 π}{3}) = sin x + 2 sin (x + π) cos (- \frac{π}{3}) = sin x - 2 sin x cos 60 = 0$
we get
${(sin x)}^{3} + {(sin (x + \frac{2 π}{3}))}^{3} + {sin (x + \frac{4 π}{3})}^{3} = 3 sin x . sin (x + \frac{2 π}{3}) . sin (x + \frac{4 π}{3})$

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Similar questions

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Assertion :sin3x+sin3(x+2π3)+sin3(x+4π3)=3sinxsin(x+2π3)⋅sin(x+4π3) Reason: If a+b+c=0, then a3+b3+c3=3abc

Assertion : ${sin}^{3} x + {sin}^{3} (x + \frac{2 π}{3}) + {sin}^{3} (x + \frac{4 π}{3}) = 3 sin x sin (x + \frac{2 π}{3}) \cdot sin (x + \frac{4 π}{3})$ Reason: If $a + b + c = 0$ , then $a^{3} + b^{3} + c^{3} = 3 a b c$