1+cosπ81+cos3π81+cos5π81+cos7π8 is equal to

1+cosπ81+cos3π81+cos5π81+cos7π8 Expand the value of 1+cos5π8 & 1+cos7π8We have, 1+cosπ81+cos3π81+cosπ-3π81+cosπ-π8 =1+cosπ81+cos3π81-cos3π81-cosπ8 =1.2-cos2 π81.2-cos2 3π8 =141-cosπ41-cos3π8 =14(1-12) =1412 =18

1+cosπ81+cos3π81+cos5π81+cos7π8 is equal to

1+cosπ81+cos3π81+cos5π81+cos7π8 Expand the value of 1+cos5π8 & 1+cos7π8We have, 1+cosπ81+cos3π81+cosπ-3π81+cosπ-π8 =1+cosπ81+cos3π81-cos3π81-cosπ8 =1.2-cos2 π81.2-cos2 3π8 =141-cosπ41-cos3π8 =14(1-12) =1412 =18

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$(1 + \cos \frac{π}{8}) (1 + \cos \frac{3 π}{8}) (1 + \cos \frac{5 π}{8}) (1 + \cos \frac{7 π}{8})$ is equal to

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$\frac{1}{8}$

$\frac{1 + \sqrt{2}}{2 \sqrt{2}}$

$\frac{1}{2}$

$\cos \frac{π}{8}$

answer is C.

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Detailed Solution

$(1 + \cos \frac{π}{8}) (1 + \cos \frac{3 π}{8}) (1 + \cos \frac{5 π}{8}) (1 + \cos \frac{7 π}{8})$

Expand the value of $(1 + \cos \frac{5 π}{8}) & (1 + \cos \frac{7 π}{8})$

We have,

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