Question

The value of $(1+\cos \frac{\pi }{8})(1+\cos \frac{3\pi }{8})(1+\cos \frac{5\pi }{8})(1+\cos \frac{7\pi }{8})$ is

• A. $\frac{1}{4}$
• B. $\frac{1}{8}$
• C. $\frac{1}{16}$
• D. None of these

Answer 1

Answer: (B)

$\frac{1}{8}$

The given expression is
$=(1+\cos \frac{\pi }{8})(1+\sin \frac{\pi }{8})(1-\sin \frac{\pi }{8})(1-\cos \frac{\pi }{8})$
$[\because \cos \frac{3\pi }{8}=\cos (\frac{\pi }{2}-\frac{\pi }{8})=\sin \frac{\pi }{8},\cos \frac{5\pi }{8}=\cos (\frac{\pi }{2}+\frac{\pi }{8})=-\sin \frac{\pi }{8},$
$\cos \frac{7\pi }{8}=\cos (\pi -\frac{\pi }{8})=-\cos \frac{\pi }{8}]$
$=(1-{\cos }^2\frac{\pi }{8})(1-{\sin }^2\frac{\pi }{8})={\sin }^2\frac{\pi }{8}{\cos }^2\frac{\pi }{8}$
$=\frac{1}{4}{\left[2\sin \frac{\pi }{8}\cos \frac{\pi }{8}\right]}^2=\frac{1}{4}{\sin }^2\frac{\pi }{4}=\frac{1}{4}{\left(\frac{1}{\sqrt{2}}\right)}^2=\frac{1}{8}$