Question

What is 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. $\frac{1}{2}$
• B. $\frac{1}{2}+\frac{1}{2\sqrt{2}}$
• C. $\frac{1}{2}-\frac{1}{2\sqrt{2}}$
• D. $\frac{1}{8}$

Answer 1

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

$\cos (\pi -\theta )=-\cos \theta$

$\therefore \cos \left(\frac{5\pi }{8}\right)=\cos (\pi -\frac{3\pi }{8})=-\cos \frac{3\pi }{8}$...(1)

$\therefore \cos \left(\frac{7\pi }{8}\right)=\cos (\pi -\frac{\pi }{8})=-\cos \frac{\pi }{8}$...(2)

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

$=(1+\cos \frac{\pi }{8})(1+\cos \frac{3\pi }{8})(1-\cos \frac{3\pi }{8})(1-\cos \frac{\pi }{8})$ ...(from 1 and 2)

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

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

$={\sin }^2\frac{\pi }{8}{\sin }^2\frac{3\pi }{8}$

$=\frac{(2\sin \frac{\pi }{8}\sin \frac{3\pi }{8}{)}^2}{4}$ ...(multiplying and dividing by 4)

$=\frac{\left(\cos \right(\frac{\pi }{8}-\frac{3\pi }{8})-\cos (\frac{\pi }{8}+\frac{3\pi }{8}){)}^2}{4}$ ...($2\sin A\sin B=\cos (A-B)-\cos (A+B)$

$=\frac{(\cos \frac{-\pi }{4}-\cos \frac{\pi }{2}{)}^2}{4}$

$=\frac{(\frac{1}{\sqrt{2}}-0{)}^2}{4}$

$=\frac{1}{8}$

Answer is option (D)