Let f(x) = cos 5x + A cos 4x + B cos 3x + C c | vedclass

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(C) Given $f(x) = \cos 5x + A \cos 4x + B \cos 3x + C \cos 2x + D \cos x + E$. Since $f(x) = f(2\pi - x)$,we have $f\left(\frac{\pi}{5}\right) = f\lef

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depends on $B, D$,but independent of $A, C, E$

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(C) Given $f(x) = \cos 5x + A \cos 4x + B \cos 3x + C \cos 2x + D \cos x + E$. Since $f(x) = f(2\pi - x)$,we have $f\left(\frac{\pi}{5}\right) = f\left(\frac{9\pi}{5}\right)$,$f\left(\frac{2\pi}{5}\right) = f\left(\frac{8\pi}{5}\right)$,$f\left(\frac{3\pi}{5}\right) = f\left(\frac{7\pi}{5}\right)$,and $f\left(\frac{4\pi}{5}\right) = f\left(\frac{6\pi}{5}\right)$. The expression $T$ can be written as $T = f(0) - 2f\left(\frac{\pi}{5}\right) + 2f\left(\frac{2\pi}{5}\right) - 2f\left(\frac{3\pi}{5}\right) + 2f\left(\frac{4\pi}{5}\right) - f(\pi)$. Substituting the values of $f(x)$ at these points,the terms involving $A, C, E$ cancel out due to the symmetry of the cosine function around $\pi$. Specifically,$f(0) - f(\pi) = 2(1 + B + D)$. The remaining terms also involve only $B$ and $D$ coefficients. Thus,$T$ depends on $B$ and $D$ but is independent of $A, C, E$.

Let $f(x) = \cos 5x + A \cos 4x + B \cos 3x + C \cos 2x + D \cos x + E$,and $T = f(0) - f\left(\frac{\pi}{5}\right) + f\left(\frac{2\pi}{5}\right) - f\left(\frac{3\pi}{5}\right) + \dots + f\left(\frac{8\pi}{5}\right) - f\left(\frac{9\pi}{5}\right)$. Then,$T$

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