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Evaluate:$2\cos\large\frac{\pi}{13}$$\cos \large\frac{9\pi}{13}+$$\cos \large\frac{3\pi}{13}$$+\cos \large\frac{5\pi}{13}$

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  • $2\cos A\cos B=\cos(A+B)+\cos(A-B)$
$2\cos\large\frac{\pi}{13}$$\cos \large\frac{9\pi}{13}+$$\cos \large\frac{3\pi}{13}$$+\cos \large\frac{5\pi}{13}$
$\cos\large\frac{10\pi}{13}$$+\cos \large\frac{8\pi}{13}$$+\cos\large\frac{3\pi}{13}+$$\cos\large\frac{5\pi}{13}$
$\big(\cos\large\frac{10\pi}{13}$$+\cos \large\frac{3\pi}{13}\big)$$+\big(\cos\large\frac{8\pi}{13}+$$\cos\large\frac{5\pi}{13}\big)$
$2\cos\large\frac{\pi}{2}$$\cos \large\frac{7\pi}{13}$$+2\cos \large\frac{\pi}{2}$$\cos \large\frac{3\pi}{13}$
$2\cos \large\frac{\pi}{2}$$\big(\cos \large\frac{7\pi}{13}+$$\cos \large\frac{3\pi}{13}\big)$
$\cos \large\frac{\pi}{2}$$=0$
$0\times \big(\cos \large\frac{7\pi}{13}+$$\cos \large\frac{3\pi}{13}\big)$
$0$
Hence proved.
answered Apr 18, 2014 by sreemathi.v
 

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