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Question Number 186741 by cortano12 last updated on 09/Feb/23

\cos (\frac{π}{18}) . \cos (\frac{3 π}{18}) . \cos (\frac{5 π}{18}) . \cos (\frac{7 π}{18}) = ?

Answered by pablo1234523 last updated on 09/Feb/23

\frac{1}{2} [\cos \frac{8 π}{18} + \cos \frac{6 π}{18}] \cdot \frac{1}{2} [\cos \frac{8 π}{18} + \cos \frac{2 π}{18}]

= \frac{1}{4} (\cos^{2} \frac{4 π}{9} + \cos \frac{4 π}{9} \cos \frac{π}{9} + \cos \frac{3 π}{9} \cos \frac{4 π}{9} + \cos \frac{3 π}{9} \cos \frac{π}{9})

= \frac{1}{8} (1 + \cos \frac{8 π}{9} + \cos \frac{5 π}{9} + \cos \frac{3 π}{9} + \cos \frac{7 π}{9} + \cos \frac{π}{9} + \cos \frac{4 π}{9} + \cos \frac{2 π}{9})

= \frac{1}{8} (1 - \cos \frac{π}{9} - \cos \frac{4 π}{9} + \cos \frac{3 π}{9} - \cos \frac{2 π}{9} + \cos \frac{π}{9} + \cos \frac{4 π}{9} + \cos \frac{2 π}{9})

= \frac{1}{8} (1 + \cos \frac{3 π}{9})

= \frac{1}{8} (1 + \cos \frac{π}{3})

= \frac{1}{8} (1 + \frac{1}{2}) = \frac{1}{8} (\frac{3}{2})

= \frac{3}{16}