If cos ^3 x sin 4 x = sum_{r=0}^{n} a_{r} sin | vedclass

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(D) Given $\cos ^3 x \sin 4 x = \sum_{r=0}^{n} a_{r} \sin rx$. Using the identity $\cos ^3 x = \frac{1}{4}(3 \cos x + \cos 3 x)$,we have: $\cos ^3 x \

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$3 : 1$

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(D) Given $\cos ^3 x \sin 4 x = \sum_{r=0}^{n} a_{r} \sin rx$. Using the identity $\cos ^3 x = \frac{1}{4}(3 \cos x + \cos 3 x)$,we have: $\cos ^3 x \sin 4 x = \frac{1}{4}(3 \cos x + \cos 3 x) \sin 4 x$ $= \frac{3}{4} \cos x \sin 4 x + \frac{1}{4} \cos 3 x \sin 4 x$ Using $2 \sin A \cos B = \sin(A+B) + \sin(A-B)$: $= \frac{3}{8}(\sin 5 x + \sin 3 x) + \frac{1}{8}(\sin 7 x + \sin x)$ $= \frac{1}{8} \sin x + \frac{3}{8} \sin 3 x + \frac{3}{8} \sin 5 x + \frac{1}{8} \sin 7 x$ Comparing coefficients,we get $a_1 = \frac{1}{8}, a_3 = \frac{3}{8}, a_5 = \frac{3}{8}, a_7 = \frac{1}{8}$. Therefore,$a_3 + a_5 = \frac{3}{8} + \frac{3}{8} = \frac{6}{8}$ and $a_1 + a_7 = \frac{1}{8} + \frac{1}{8} = \frac{2}{8}$. Thus,$\frac{a_3 + a_5}{a_1 + a_7} = \frac{6/8}{2/8} = 3 : 1$.

If $\cos ^3 x \sin 4 x = \sum_{r=0}^{n} a_{r} \sin rx$ for all $x \in R$,then $a_3+a_5 : a_1+a_7 = $

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