If $sin$ A $sin (60^{o} - A) sin (60^{o} + A) = k sin 3 A$ , then what is k equal to?

\frac{1}{4}

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\frac{1}{2}

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1

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4

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Solution

The correct option is A $\frac{1}{4}$
Given $sin A sin (60 - A) sin (60 + A)$
$= sin A [sin (60 - A) sin (60 + A)]$

$= sin A [(sin 60 cos A - cos 60 sin A) (sin 60 cos A + cos 60 sin A)]$

$= sin A ({sin}^{2} 60 {cos}^{2} A - {cos}^{2} 60 {sin}^{2} A)$

$= sin A (\frac{3}{4} {cos}^{2} A - \frac{1}{4} {sin}^{2} A)$

$= sin A (\frac{3}{4} (1 - {sin}^{2} A) - \frac{1}{4} {sin}^{2} A)$

$= sin A (\frac{3}{4} - {sin}^{2} A)$
$= \frac{1}{4} (3 sin A - 4 {sin}^{3} A)$

$= \frac{1}{4} sin 3 A$

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Q. If

s i n^{- 1} x + s i n^{- 1} y + s i n^{- 1} z = π

, then

x^{4} + y^{4} + z^{4} + 4 x^{2} y^{2} z^{2} = k (x^{2} y^{2} + y^{2} z^{2} + z^{2} x^{2})

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If true then write 1 and if false then write 0

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If sin A sin(60o−A)sin(60o+A)=ksin3A, then what is k equal to?

The correct option is A 14Given sinAsin(60−A)sin(60+A) =sinA[sin(60−A)sin(60+A)]=sinA[(sin60cosA−cos60sinA)(sin60cosA+cos60sinA)]=sinA(sin260cos2A−cos260sin2A)=sinA(34cos2A−14sin2A)=sinA(34(1−sin2A)−14sin2A)=sinA(34−sin2A)=14(3sinA−4sin3A)=14sin3A

If $sin$ A $sin (60^{o} - A) sin (60^{o} + A) = k sin 3 A$ , then what is k equal to?