Question

find the value of (1+cos π/9)(1+cos3π/9)(1+cos5π/9)(1+cos7π/9)

Answer 1

$= > (1 + cos \frac{\pi}{9} )(1 + cos \frac{3\pi}{9} ) \\ (1 + cos \frac{5\pi}{9} )(1 + cos \frac{7\pi}{9} ) \\ \\ = > (2 {cos}^{2} \frac{\pi}{18} )(2 {cos}^{2} \frac{3\pi}{18} ) \\ (2 {cos}^{2} \frac{5\pi}{18} )(2 {cos}^{2} \frac{7\pi}{18} ) \\ \\ = > {(4sin \frac{4\pi}{9} \times sin \frac{3\pi}{9} \times sin \frac{2\pi}{9} \times sin \frac{\pi}{9} )}^{2}$
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Now if you see :-

$= > 4sin \frac{\pi}{9} \times sin( \frac{\pi}{9} + \frac{\pi}{3} ) \times sin( \frac{\pi}{3} - \frac{\pi}{9} ) \\ \\ = > sin(3 \times \frac{\pi}{9} )$
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Hence the given expression :-

$= > {(sin \frac{3\pi}{9} \times sin \frac{3\pi}{9} )}^{2} \\ \\ = > {( \frac{3}{4} ) }^{2} \\ \\ = > \frac{9}{16}$
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Answer 2

Answer:

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