If cos theta + i sin theta, theta in R, is a | vedclass

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(C) Given the equation $a_0 x^n + a_1 x^{n-1} + \ldots + a_{n-1} x + a_n = 0$. Since $x = \cos 	heta + i \sin 	heta = e^{i 	heta}$ is a root, we ha

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$0$

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(C) Given the equation $a_0 x^n + a_1 x^{n-1} + \ldots + a_{n-1} x + a_n = 0$. Since $x = \cos 	heta + i \sin 	heta = e^{i 	heta}$ is a root, we have $a_0 (e^{i 	heta})^n + a_1 (e^{i 	heta})^{n-1} + \ldots + a_n = 0$. Using De Moivre's Theorem, $e^{i k 	heta} = \cos k 	heta + i \sin k 	heta$. Substituting this into the equation: $a_0(\cos n 	heta + i \sin n 	heta) + a_1(\cos(n-1) 	heta + i \sin(n-1) 	heta) + \ldots + a_n = 0$. Equating the imaginary part to zero: $a_0 \sin n 	heta + a_1 \sin(n-1) 	heta + \ldots + a_{n-1} \sin 	heta = 0$. However, the expression requested is $a_1 \sin 	heta + a_2 \sin 2 	heta + \ldots + a_n \sin n 	heta$. By considering the conjugate root $x = \cos 	heta - i \sin 	heta = e^{-i 	heta}$ (since coefficients are real), we have $a_0 e^{-i n 	heta} + a_1 e^{-i(n-1) 	heta} + \ldots + a_n = 0$. Subtracting the imaginary parts of the two equations leads to the result $0$.

If $\cos \theta + i \sin \theta, \theta \in R$, is a root of the equation $a_0 x^n + a_1 x^{n-1} + \ldots + a_{n-1} x + a_n = 0$, where $a_0, a_1, \ldots, a_n \in R$ and $a_0 \neq 0$, then the value of $a_1 \sin \theta + a_2 \sin 2 \theta + \ldots + a_n \sin n \theta$ is:

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