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Prove that: \(2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13}+\cos \frac{3 \pi}{13}+\cos \frac{5 \pi}{13}=0.\)
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Prove that: \(2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13}+\cos \frac{3 \pi}{13}+\cos \frac{5 \pi}{13}=0.\)

Answer

\(\text{LHS} =2 \cos \frac{\pi}{13} \cos \frac{9 \cdot \pi}{13}+\cos \frac{3 \pi}{13}+\cos \frac{5 \pi}{13} \)
\(=\cos \left(\frac{9 \pi}{13}+\frac{\pi}{13}\right)+\cos \left(\frac{9 \pi}{13}-\frac{\pi}{13}\right)+\cos \frac{3 \pi}{13}+\cos \frac{5 \pi}{13}\)
\(=\cos \frac{10 \pi}{13}+\cos \frac{8 \pi}{13}+\cos \frac{3 \pi}{13}+\cos \frac{5 \pi}{13}\)
\(=\cos \left(\pi-\frac{3 \pi}{13}\right)+\cos \left(\pi-\frac{5 \pi}{13}\right)+\cos \frac{3 \pi}{13}+\cos \frac{5 \pi}{13}\)
\(=-\cos \frac{3 \pi}{13}-\cos \frac{5 \pi}{13}+\cos \frac{3 \pi}{13}+\cos \frac{5 \pi}{13}=0=\mathrm{RHS}\) \([\because \cos (\pi-x)=-\cos x]\)
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