If cos3x.sin2x =∑r=0narsinrx ∀x∈R then

cos3x.sin2x=1−cos2x2.2sin2x.cosx2 =14[1−cos2x](sin3x+sinx) =14[sin3x+sinx−12(sin5x+sinx)−12(sin3x−sixx)] =14[sinx+12sin3x−12sin5x]

If cos3x.sin2x =∑r=0narsinrx ∀x∈R then

cos3x.sin2x=1−cos2x2.2sin2x.cosx2 =14[1−cos2x](sin3x+sinx) =14[sin3x+sinx−12(sin5x+sinx)−12(sin3x−sixx)] =14[sinx+12sin3x−12sin5x]

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If $\cos^{3} x . \sin 2 x = \sum_{r = 0}^{n} a_{r} \sin r x \forall x \in R$ then

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$n = 5, a_{2} = \frac{1}{8}$

$n = 5, a_{1} = \frac{1}{4}$

$n = 5, a_{2} = \frac{1}{4}$

$n = 5, a_{1} = \frac{1}{2}$

answer is B.

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Detailed Solution

$\cos^{3} x . \sin 2 x = \frac{1 - \cos 2 x}{2} . \frac{2 \sin 2 x . \cos x}{2} = \frac{1}{4} [1 - \cos 2 x] (\sin 3 x + \sin x) = \frac{1}{4} [\sin 3 x + \sin x - \frac{1}{2} (\sin 5 x + \sin x) - \frac{1}{2} (\sin 3 x - s i x x)] = \frac{1}{4} [\sin x + \frac{1}{2} \sin 3 x - \frac{1}{2} \sin 5 x]$