If sin3x sin3x=∑r=0narcosrx is an identity then n+a1=

a0cos0x+a1cosx+..........+ancosnx=sin3xsin3x=143sinx−sin3xsin3x=382sin3xsinx-181−cos6x=38cos2x−cos4x-181−cos6x=-18+38cos2x-38cos4x+18cos6xComparing,we get a0=-18, a1=0, n=6 ∴a1+n=6

If sin3x sin3x=∑r=0narcosrx is an identity then n+a1=

a0cos0x+a1cosx+..........+ancosnx=sin3xsin3x=143sinx−sin3xsin3x=382sin3xsinx-181−cos6x=38cos2x−cos4x-181−cos6x=-18+38cos2x-38cos4x+18cos6xComparing,we get a0=-18, a1=0, n=6 ∴a1+n=6

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If $\sin^{3} x \sin 3 x = \sum_{r = 0}^{n} a_{r} \cos r x$ is an identity then $n + a_{1} =$

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$- 6$

answer is B.

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Detailed Solution

$\begin{array}{l} a_{0} \cos 0 x + a_{1} \cos x + .......... + a_{n} \cos n x \\ = \sin^{3} x \sin 3 x = \frac{1}{4} (3 \sin x - \sin 3 x) \sin 3 x \\ = \frac{3}{8} (2 \sin 3 x \sin x) - \frac{1}{8} (1 - \cos 6 x) \\ = \frac{3}{8} (\cos 2 x - \cos 4 x) - \frac{1}{8} (1 - \cos 6 x) \\ = - \frac{1}{8} + \frac{3}{8} \cos 2 x - \frac{3}{8} \cos 4 x + \frac{1}{8} \cos 6 x \\ C o m p a r i n g, w e g e t a_{0} = - \frac{1}{8}, a_{1} = 0, n = 6 \\ ∴ a_{1} + n = 6 \end{array}$