If sin 3 x sin 3 x-C0-C1cos 2 x-C2cos 4 x-Cncos 2 n x- where C0- C1- Cn are constants and Cn -0- then which of thefollowing is-are correct-

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Question

If ${sin}^{3} x sin 3 x = C_{0} + C_{1} cos 2 x + C_{2} cos 4 x + \dots + C_{n} cos 2 n x$ , where $C_{0}, C_{1} . . . . . . C_{n}$ are constants and $(C_{n} \neq 0)$ , then which of the following is/are correct?

n = 3

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C_{0} = - \frac{1}{8}

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C_{3} = \frac{1}{8}

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Sum of all the coefficients is

0

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Solution

The correct options are
A $n = 3$
B $C_{0} = - \frac{1}{8}$
C $C_{3} = \frac{1}{8}$
D Sum of all the coefficients is $0$
Given :
${sin}^{3} x sin 3 x = C_{0} + C_{1} cos 2 x + C_{2} cos 4 x + \dots + C_{n} cos 2 n x$

Also,
${sin}^{3} x sin 3 x = \frac{3 sin x - sin 3 x}{4} \times sin 3 x = \frac{3 sin x sin 3 x - {sin}^{2} 3 x}{4} = \frac{1}{4} [3 sin x sin 3 x - {sin}^{2} 3 x] = \frac{1}{4} \times [\frac{3}{2} (cos 2 x - cos 4 x) - \frac{1}{2} (1 - cos 6 x)] = \frac{1}{8} \times [cos 6 x - 3 cos 4 x + 3 cos 2 x - 1] = - \frac{1}{8} + \frac{3}{8} cos 2 x - \frac{3}{8} cos 4 x + \frac{1}{8} cos 6 x$

Therefore,
$n = 3 C_{0} = - \frac{1}{8} C_{3} = \frac{1}{8}$
Sum of all coefficients
$= \frac{1}{8} - \frac{3}{8} + \frac{3}{8} - \frac{1}{8} = 0$

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If sin3xsin3x=C0+C1cos2x+C2cos4x+⋯+Cncos2nx, where C0,C1......Cn are constants and (Cn≠0), then which of the following is/are correct?

If ${sin}^{3} x sin 3 x = C_{0} + C_{1} cos 2 x + C_{2} cos 4 x + \dots + C_{n} cos 2 n x$ , where $C_{0}, C_{1} . . . . . . C_{n}$ are constants and $(C_{n} \neq 0)$ , then which of the following is/are correct?