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If $\cos ^3 \mathrm{x} \sin 4 \mathrm{x}=\sum_{\mathrm{r}=0}^{\mathrm{n}} \mathrm{a}_{\mathrm{r}} \sin \mathrm{rx} \forall \mathrm{x} \in \mathbb{R}$, then $a_3+a_5: a_1+a_7=$
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Verified Answer
The correct answer is:
$3: 1$
$\cos ^3 x \cdot \sin 4 x=\frac{1}{2}(2 \sin 4 x \cdot \cos x) \cos ^2 x$
$\begin{aligned}
& =\frac{1}{2}[\sin 5 x+\sin 3 x] \cdot \cos ^2 x \\
& =\frac{1}{4}[\sin 6 x+2 \sin 4 x+\sin 2 x] \cos x \\
& =\frac{1}{8}[\sin 7 x+3 \sin 5 x+3 \sin 3 x+\sin x] \\
& =a_1 \sin x+a_2 \sin 2 x+a_3 \sin 3 x+\ldots . \\
& \Rightarrow a_1=\frac{1}{8}, a_3=\frac{3}{8}, a_5=\frac{3}{8}, a_7=\frac{1}{8}
\end{aligned}$
Now, $a_3+a_5: a_1+a_7=3: 1$
$\begin{aligned}
& =\frac{1}{2}[\sin 5 x+\sin 3 x] \cdot \cos ^2 x \\
& =\frac{1}{4}[\sin 6 x+2 \sin 4 x+\sin 2 x] \cos x \\
& =\frac{1}{8}[\sin 7 x+3 \sin 5 x+3 \sin 3 x+\sin x] \\
& =a_1 \sin x+a_2 \sin 2 x+a_3 \sin 3 x+\ldots . \\
& \Rightarrow a_1=\frac{1}{8}, a_3=\frac{3}{8}, a_5=\frac{3}{8}, a_7=\frac{1}{8}
\end{aligned}$
Now, $a_3+a_5: a_1+a_7=3: 1$
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