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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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Correct option is B

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

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