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If $\sin ^4 \theta \cos ^2 \theta=\sum_{n=0}^{\infty} a_{2 n} \cos 2 n \theta$, then the least $n$ for which $a_{2 n}=0$ is
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The correct answer is:
$1$
Given,
$\sin ^4 \theta \cos ^2 \theta=\sum_{n=0}^{\infty} a_{2 n} \cos 2 n \theta$
$\begin{aligned} & \Rightarrow\left(\sin ^2 \theta\right)^2 \cos ^2 \theta=\sum_{n=0}^{\infty} a_{2 n} \cos 2 n \theta \\ & \Rightarrow\left(1-\cos ^2 \theta\right)^2 \cos ^2 \theta=\sum_{n=0}^{\infty} a_{2 n} \cos 2 n \theta \\ & \Rightarrow\left(1+\cos ^4 \theta-2 \cos ^2 \theta\right) \cos ^2 \theta=\sum_{n=0}^{\infty} a_{2 n} \cos 2 n \theta\end{aligned}$
$\therefore \quad \cos ^2 \theta+\cos ^4 \theta \cos ^2 \theta-2 \cos ^4 \theta$
$=\sum_{n=0}^{\infty} a_{2 n} \cos 2 n \theta$
On comparing both sides, we get minimum value of $n$ is 1 .
$\sin ^4 \theta \cos ^2 \theta=\sum_{n=0}^{\infty} a_{2 n} \cos 2 n \theta$
$\begin{aligned} & \Rightarrow\left(\sin ^2 \theta\right)^2 \cos ^2 \theta=\sum_{n=0}^{\infty} a_{2 n} \cos 2 n \theta \\ & \Rightarrow\left(1-\cos ^2 \theta\right)^2 \cos ^2 \theta=\sum_{n=0}^{\infty} a_{2 n} \cos 2 n \theta \\ & \Rightarrow\left(1+\cos ^4 \theta-2 \cos ^2 \theta\right) \cos ^2 \theta=\sum_{n=0}^{\infty} a_{2 n} \cos 2 n \theta\end{aligned}$
$\therefore \quad \cos ^2 \theta+\cos ^4 \theta \cos ^2 \theta-2 \cos ^4 \theta$
$=\sum_{n=0}^{\infty} a_{2 n} \cos 2 n \theta$
On comparing both sides, we get minimum value of $n$ is 1 .
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