If $\sin ^4 heta \cos ^2 heta=\sum_{n=0}^{\infty} a_{2 n} \cos 2 n heta$, then the least $n$ for which $a_{2 n}=0$ is

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If $\sin ^4 	heta \cos ^2 	heta=\sum_{n=0}^{\infty} a_{2 n} \cos 2 n 	heta$, then the least $n$ for which $a_{2 n}=0$ is

Aicharya · Accepted Answer

Given, $\sin ^4 	heta \cos ^2 	heta=\sum_{n=0}^{\infty} a_{2 n} \cos 2 n 	heta$ $\begin{aligned} & \Rightarrow\left(\sin ^2 	hetaight)^2 \cos ^2 	heta=\sum_{n=0}^{\infty} a_{2 n} \cos 2 n 	heta \ & \Rightarrow\left(1-\cos ^2 	hetaight)^2 \cos ^2 	heta=\sum_{n=0}^{\infty} a_{2 n} \cos 2 n 	heta \ & \Rightarrow\left(1+\cos ^4 	heta-2 \cos ^2 	hetaight) \cos ^2 	heta=\sum_{n=0}^{\infty} a_{2 n} \cos 2 n 	heta\end{aligned}$ $	herefore \quad \cos ^2 	heta+\cos ^4 	heta \cos ^2 	heta-2 \cos ^4 	heta$ $=\sum_{n=0}^{\infty} a_{2 n} \cos 2 n 	heta$ On comparing both sides, we get minimum value of $n$ is 1 .

If $\sin ^4 \theta \cos ^2 \theta=\sum_{n=0}^{\infty} a_{2 n} \cos 2 n \theta$, then the least $n$ for which…

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