InterviewSolution
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Evaluate: (\(1+cos\frac{\pi}{8}\)) (\(1+cos\frac{3\pi}{8}\)) (\(1+cos\frac{5\pi}{8}\)) (\(1+cos\frac{7\pi}{8}\)) |
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Answer» We have, (\(1+cos \frac{\pi}{8}\) ) (\(1+cos\frac{3\pi}{8}\) ) (\(1+cos\frac{5\pi}{8}\) ) (\(1+cos\frac{7\pi}{8}\) ) = (\(1+cos\frac{\pi}{8}\)) (\(1+sin(\frac{\pi}{2}-\frac{3\pi}{8})\)) (\(1+sin(\frac{\pi}{2}-\frac{5\pi}{8})\)) (\(1+cos(\pi -\frac{\pi}{8})\)) = (\(1+cos\frac{\pi}{8}\)) (\(1+sin\frac{\pi}{8}\)) . (\(1-sin\frac{\pi}{8}\)) (\(1-cos\frac{\pi}{8}\) ) = (\(1-cos^2\frac{\pi}{8}\) ) (\(1-sin^2\frac{\pi}{8}\) ) = \(sin^2\frac{\pi}{8}cos^2\frac{\pi}{8}\) = \(\frac{1}{4}(2sin\frac{\pi}{8}cos\frac{\pi}{8})^2\) =\(\frac{1}{4}sin^2(\frac{2\pi}{8})\) = \(\frac{1}{4}sin^2(\frac{2\pi}{8})\) =\(\frac{1}{4}(\frac{1}{\sqrt{2}})^2\) = \(\frac{1}{4}\times \frac{1}{2}\)= \(\frac{1}{8}\) |
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