Question

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

• A. $\frac{7}{2}$
• B. $\cos \frac{\pi }{8}$
• C. $\frac{1}{8}$
• D. $\frac{1+\sqrt{2}}{2\sqrt{2}}$

Answer 1

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

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

$=(1+\cos \frac{\pi }{8})(1+\cos \frac{3\pi }{8})(1-\cos \frac{3\pi }{8})(1-\cos \frac{\pi }{8})$

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

$=\frac{1}{4}(1-\cos \frac{\pi }{4})(1-\cos \frac{3\pi }{4})=\frac{1}{4}(1-\frac{1}{\sqrt{2}})(1+\frac{1}{\sqrt{2}})$

$=\frac{1}{8}$