If sin^4 theta cos^2 theta = sum_{n=0}^{infty | vedclass

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(D) Given,$\sin^4 	heta \cos^2 	heta = \sum_{n=0}^{\infty} a_{2n} \cos 2n 	heta$. We know that $\sin^2 	heta = \frac{1 - \cos 2	heta}{2}$ and $\c

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$4$

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(D) Given,$\sin^4 	heta \cos^2 	heta = \sum_{n=0}^{\infty} a_{2n} \cos 2n 	heta$. We know that $\sin^2 	heta = \frac{1 - \cos 2	heta}{2}$ and $\cos^2 	heta = \frac{1 + \cos 2	heta}{2}$. Substituting these,we get: $\sin^4 	heta \cos^2 	heta = (\sin^2 	heta)^2 \cos^2 	heta = \left(\frac{1 - \cos 2	heta}{2}ight)^2 \left(\frac{1 + \cos 2	heta}{2}ight)$ $= \frac{1}{8} (1 - 2\cos 2	heta + \cos^2 2	heta)(1 + \cos 2	heta)$ $= \frac{1}{8} (1 + \cos 2	heta - 2\cos 2	heta - 2\cos^2 2	heta + \cos^2 2	heta + \cos^3 2	heta)$ $= \frac{1}{8} (1 - \cos 2	heta - \cos^2 2	heta + \cos^3 2	heta)$ Using $\cos^2 2	heta = \frac{1 + \cos 4	heta}{2}$ and $\cos^3 2	heta = \frac{3\cos 2	heta + \cos 6	heta}{4}$: $= \frac{1}{8} \left(1 - \cos 2	heta - \frac{1 + \cos 4	heta}{2} + \frac{3\cos 2	heta + \cos 6	heta}{4}ight)$ $= \frac{1}{8} \left(1 - \cos 2	heta - \frac{1}{2} - \frac{1}{2}\cos 4	heta + \frac{3}{4}\cos 2	heta + \frac{1}{4}\cos 6	hetaight)$ $= \frac{1}{8} \left(\frac{1}{2} - \frac{1}{4}\cos 2	heta - \frac{1}{2}\cos 4	heta + \frac{1}{4}\cos 6	hetaight)$ $= \frac{1}{16} - \frac{1}{32}\cos 2	heta - \frac{1}{16}\cos 4	heta + \frac{1}{32}\cos 6	heta$. Comparing this with $\sum_{n=0}^{\infty} a_{2n} \cos 2n 	heta$,we have $a_0 = \frac{1}{16}$,$a_2 = -\frac{1}{32}$,$a_4 = -\frac{1}{16}$,$a_6 = \frac{1}{32}$,and $a_{2n} = 0$ for $n \ge 4$. The least $n$ for which $a_{2n} = 0$ is $4$.

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

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