If cos3⁡xsin⁡2x=∑r=0n arsin(rx),∀x∈R then

cos3⁡xsin⁡2x=cos2⁡x⋅cos⁡xsin⁡2x=1−cos⁡2x22sin⁡2xcos⁡x2=14(1−cos⁡2x)(sin⁡3x+sin⁡x)=14sin⁡3x+sin⁡x−12(2sin⁡3xcos⁡2x)−12(2cos⁡2xsin⁡x)=14sin⁡3x+sin⁡x−12(sin⁡5x+sin⁡x)−12(sin⁡3x−sin⁡x)=14sin⁡x+12sin⁡3x−12sin⁡5xa1=14;a3=18;n=5

If cos3⁡xsin⁡2x=∑r=0n arsin(rx),∀x∈R then

cos3⁡xsin⁡2x=cos2⁡x⋅cos⁡xsin⁡2x=1−cos⁡2x22sin⁡2xcos⁡x2=14(1−cos⁡2x)(sin⁡3x+sin⁡x)=14sin⁡3x+sin⁡x−12(2sin⁡3xcos⁡2x)−12(2cos⁡2xsin⁡x)=14sin⁡3x+sin⁡x−12(sin⁡5x+sin⁡x)−12(sin⁡3x−sin⁡x)=14sin⁡x+12sin⁡3x−12sin⁡5xa1=14;a3=18;n=5

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If $\cos^{3} xsin 2 x = \sum_{r = 0}^{n} a_{r} \sin (rx), \forall x \in R$ then

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

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

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

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

answer is B.

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Detailed Solution

$\cos^{3} xsin 2 x = \cos^{2} x \cdot \cos xsin 2 x$
$= (\frac{1 - \cos 2 x}{2}) (\frac{2 \sin 2 xcos x}{2}) = \frac{1}{4} (1 - \cos 2 x) (\sin 3 x + \sin x)$
$= \frac{1}{4} [\sin 3 x + \sin x - \frac{1}{2} (2 \sin 3 xcos 2 x) - \frac{1}{2} (2 \cos 2 xsin x)]$
$= \frac{1}{4} [\sin 3 x + \sin x - \frac{1}{2} (\sin 5 x + \sin x) - \frac{1}{2} (\sin 3 x - \sin x)] = \frac{1}{4} [\sin x + \frac{1}{2} \sin 3 x - \frac{1}{2} \sin 5 x]$
$a_{1} = \frac{1}{4}; a_{3} = \frac{1}{8}; n = 5$