Question

Find the value of $(1+\cos \frac{\pi }{8})(1+\cos \frac{3\pi }{8})(1+\cos \frac{5\pi }{8})(1+\cos \frac{7\pi }{8})$.

• A. $\frac{-1}{8}$
• B. $\frac{1}{8}$
• C. $\frac{3}{2}$
• D. $\frac{3}{4}$

Answer 1

The correct option is B $\frac{1}{8}$

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

$=(1+\cos \frac{\pi }{8})(1+\cos \frac{3\pi }{8})(1+\cos (3-\frac{3\pi }{8}\left)\right)(1+\cos (\pi -\frac{\pi }{8}\left)\right)$
$=(1+\cos \frac{\pi }{8})(1+\cos \frac{3\pi }{8})(1-\cos \frac{3\pi }{8})(1-\cos \frac{\pi }{8})$
$=(1-{\cos }^2\frac{\pi }{8})(1-{\cos }^2\frac{3\pi }{8})$
$={\sin }^2\frac{\pi }{8}{\sin }^2\frac{3\pi }{8}$
$=\frac{1}{4}{\left(2\sin \frac{\pi }{8}\sin \frac{3\pi }{8}\right)}^2$
$=\frac{1}{4}{(\cos \frac{\pi }{4}-\cos \frac{\pi }{2})}^2$
$=\frac{1}{4}{(\frac{1}{\sqrt{2}}-0)}^2$
$=\frac{1}{8}$