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Let $f(x)=\cos 5 x+A \cos 4 x+B \cos 3 x+C \cos 2 x+D \cos x+E$, and $T=f(0)-f\left(\frac{\pi}{5}\right)+f\left(\frac{2 \pi}{5}\right)-f\left(\frac{3 \pi}{5}\right)+\ldots+f\left(\frac{8 \pi}{5}\right)-f\left(\frac{9 \pi}{5}\right) .$ Then $T$
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depends on $B, D$ but independent of $A, C, E$
Clearly $\mathrm{f}(\pi+\mathrm{x})+\mathrm{f}(\pi-\mathrm{x}) \quad($ every term contain cosine $)$
$f\left(\frac{\pi}{5}\right)=f\left(\frac{9 \pi}{5}\right), f\left(\frac{2 \pi}{5}\right)=f\left(\frac{8 \pi}{5}\right), f\left(\frac{3 \pi}{5}\right)=f\left(\frac{7 \pi}{5}\right)$
$f\left(\frac{4 \pi}{5}\right)=f\left(\frac{6 \pi}{5}\right)$
$\mathrm{T}=\mathrm{f}(0)-2\left[f\left(\frac{\pi}{5}\right)+f\left(\frac{3 \pi}{5}\right)\right]+2\left[f\left(\frac{2 \pi}{5}\right)+f\left(\frac{4 \pi}{5}\right)\right]-f(\pi)$
$\mathrm{f}(0)-\mathrm{f}(\pi)=2(1+\mathrm{B}+\mathrm{D})$
$f\left(\frac{\pi}{5}\right)+f\left(\frac{3 \pi}{5}\right)=f\left(\frac{\pi}{5}\right)-f\left(\frac{4 \pi}{5}\right)=2\left(1+B \cos \frac{3 \pi}{5}+D \cos \frac{\pi}{5}\right)$
$f\left(\frac{2 \pi}{5}\right)+f\left(\frac{4 \pi}{5}\right)=f\left(\frac{2 \pi}{5}\right)-f\left(\frac{3 \pi}{5}\right)=2\left(1+B \cos \frac{6 \pi}{5}+D \cos \frac{2 \pi}{5}\right)$
$\mathrm{T} \Rightarrow$ contains only $\mathrm{B}, \mathrm{D}$ terms
$f\left(\frac{\pi}{5}\right)=f\left(\frac{9 \pi}{5}\right), f\left(\frac{2 \pi}{5}\right)=f\left(\frac{8 \pi}{5}\right), f\left(\frac{3 \pi}{5}\right)=f\left(\frac{7 \pi}{5}\right)$
$f\left(\frac{4 \pi}{5}\right)=f\left(\frac{6 \pi}{5}\right)$
$\mathrm{T}=\mathrm{f}(0)-2\left[f\left(\frac{\pi}{5}\right)+f\left(\frac{3 \pi}{5}\right)\right]+2\left[f\left(\frac{2 \pi}{5}\right)+f\left(\frac{4 \pi}{5}\right)\right]-f(\pi)$
$\mathrm{f}(0)-\mathrm{f}(\pi)=2(1+\mathrm{B}+\mathrm{D})$
$f\left(\frac{\pi}{5}\right)+f\left(\frac{3 \pi}{5}\right)=f\left(\frac{\pi}{5}\right)-f\left(\frac{4 \pi}{5}\right)=2\left(1+B \cos \frac{3 \pi}{5}+D \cos \frac{\pi}{5}\right)$
$f\left(\frac{2 \pi}{5}\right)+f\left(\frac{4 \pi}{5}\right)=f\left(\frac{2 \pi}{5}\right)-f\left(\frac{3 \pi}{5}\right)=2\left(1+B \cos \frac{6 \pi}{5}+D \cos \frac{2 \pi}{5}\right)$
$\mathrm{T} \Rightarrow$ contains only $\mathrm{B}, \mathrm{D}$ terms
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