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Mathematics Question on Trigonometric Functions

(1+cosπ8)(1+cos3π8)(1+cos5π8)(1+cos7π8)\Bigg(1+cos\frac{\pi}{8}\Bigg)\Bigg(1+cos\frac{3\pi}{8}\Bigg)\Bigg(1+cos\frac{5\pi}{8}\Bigg)\Bigg(1+cos\frac{7\pi}{8}\Bigg) is equal to

A

12\frac{1}{2}

B

cosπ8cos\frac{\pi}{8}

C

18\frac{1}{8}

D

1+222\frac{1+\sqrt{2}}{2\sqrt{2}}

Answer

18\frac{1}{8}

Explanation

Solution

=(1+cosπ8)(1+cos3π8)(1+cos5π8)(1+cos7π8)=\Bigg(1+cos\frac{\pi}{8}\Bigg)\Bigg(1+cos\frac{3\pi}{8}\Bigg)\Bigg(1+cos\frac{5\pi}{8}\Bigg)\Bigg(1+cos\frac{7\pi}{8}\Bigg)
=(1+cosπ8)(1+cos3π8)(1−cos3π8)(1−cosπ8)=\Bigg(1+cos\frac{\pi}{8}\Bigg)\Bigg(1+cos\frac{3\pi}{8}\Bigg)\Bigg(1-cos\frac{3\pi}{8}\Bigg)\Bigg(1-cos\frac{\pi}{8}\Bigg)
=(1−cos2π8)(1−cos23π8)=\Bigg(1-cos^2\frac{\pi}{8}\Bigg)\Bigg(1-cos^2\frac{3\pi}{8}\Bigg)
=14(2−1−cosπ4)(2−1−cos33π4)=\frac{1}{4}\Bigg(2-1-cos\frac{\pi}{4}\Bigg)\Bigg(2-1-cos3\frac{3\pi}{4}\Bigg)
=14(1−cosπ4)(1−cos33π4)=\frac{1}{4}\Bigg(1-cos\frac{\pi}{4}\Bigg)\Bigg(1-cos3\frac{3\pi}{4}\Bigg)
=14(1−12)(1+12)=14(1−12)=18=\frac{1}{4}\Bigg(1-\frac{1}{\sqrt{2}}\Bigg)\Bigg(1+\frac{1}{\sqrt{2}}\Bigg)=\frac{1}{4}\Bigg(1-\frac{1}{2}\Bigg)=\frac{1}{8}