If f(x) = sin x cdot sin 2x cdot sin 3x and f | vedclass

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(B) $f(x) = \sin x \sin 2x \sin 3x$ $= \frac{1}{2} (2 \sin x \sin 2x) \sin 3x$ $= \frac{1}{2} (\cos x - \cos 3x) \sin 3x$ $= \frac{1}{2} (\cos x \sin

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-$8$

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(B) $f(x) = \sin x \sin 2x \sin 3x$ $= \frac{1}{2} (2 \sin x \sin 2x) \sin 3x$ $= \frac{1}{2} (\cos x - \cos 3x) \sin 3x$ $= \frac{1}{2} (\cos x \sin 3x - \cos 3x \sin 3x)$ $= \frac{1}{4} (2 \sin 3x \cos x) - \frac{1}{4} (2 \sin 3x \cos 3x)$ $= \frac{1}{4} (\sin 4x + \sin 2x) - \frac{1}{4} \sin 6x$ $f'(x) = \frac{1}{4} (4 \cos 4x + 2 \cos 2x) - \frac{6}{4} \cos 6x = \cos 4x + \frac{1}{2} \cos 2x - \frac{3}{2} \cos 6x$ $f''(x) = -4 \sin 4x - \sin 2x + 9 \sin 6x$ Comparing with $f''(x) = a \sin bx + c \sin dx + e \sin kx$,we get: $a = -4, b = 4, c = -1, d = 2, e = 9, k = 6$ $(a+c+e) - (b+d+k) = (-4 - 1 + 9) - (4 + 2 + 6) = 4 - 12 = -8$

If $f(x) = \sin x \cdot \sin 2x \cdot \sin 3x$ and $f''(x) = a(\sin bx) + c(\sin dx) + e(\sin kx)$,then the value of $(a+c+e) - (b+d+k)$ equals

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