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Question Number 88503 by jagoll last updated on 11/Apr/20

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

Commented by john santu last updated on 11/Apr/20

\cos \frac{7 π}{8} = \cos (π - \frac{π}{8}) = - \cos \frac{π}{8}

\cos \frac{5 π}{8} = - \cos \frac{3 π}{8}

\Rightarrow (1 + \cos \frac{π}{8}) (1 - \cos \frac{π}{8}) = 1 - \cos^{2} \frac{π}{8} = \sin^{2} (\frac{π}{8})

\Rightarrow (1 + \cos \frac{3 π}{8}) (1 - \cos \frac{3 π}{8}) = 1 - \cos^{2} (\frac{3 π}{8}) = \sin^{2} (\frac{3 π}{8})

∴ {[\sin \frac{3 π}{8} . \sin \frac{π}{8}]}^{2} =

\frac{1}{4} {[\cos \frac{4 π}{8} - \cos \frac{π}{4}]}^{2} = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8}

Commented by peter frank last updated on 11/Apr/20

t h a n k y o u

Commented by peter frank last updated on 15/Apr/20

h e l p Q n 88937