Question

2cos(π/13) cos (9π/13) +3cos(π/13)cos(5π/13)=0​

Answer 1

Answer:

Step-by-step explanation:

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Answer 2

AnswEr:

$\\ \rm LHS = 2 \cos \frac{\pi}{13} \: \cos\frac{9\pi}{13} + \cos \frac{3\pi}{13} + \cos \frac{5\pi}{13} \\ \\ \\ \implies \rm \: cos \: ( \frac{9\pi}{13} + \frac{\pi}{13} ) + cos \: ( \frac{9\pi}{13} - \frac{\pi}{13} ) + \: cos \frac{3\pi}{13} \\ \\ \rm \: + cos \: \frac{5\pi}{13} \\ \\ \\ \implies \rm \: cos \: \frac{10\pi}{13} + \cos \: \frac{8\pi}{13} + \cos \frac{3\pi}{13} + \cos \frac{5\pi}{13} \\ \\ \\ \implies\rm \: \cos(\pi - \frac{3}{13}\pi ) + \cos(\pi - \frac{5\pi}{13} ) + \cos( \frac{3\pi}{13} ) \\ \\ \rm \: + \cos( \frac{5\pi}{13} ) \\ \\ \\ \rm \implies - \cos \frac{3\pi}{13} - \cos \frac{5\pi}{13} + \cos \frac{3\pi}{13} \\ \\ \rm \: + \cos \frac{5\pi}{13} = 0 = RHS \\ \\ \qquad \sf ( \because \: cos(\pi - x) = - cos \: x)$