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

Answer 1

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

We have
$(1+cos\frac{\pi }{8})(1+cos\frac{3\pi }{8})(1+cos\frac{5\pi }{8})(1+cos\frac{7\pi }{8})=(1+cos\frac{\pi }{8})(1+cos\frac{3\pi }{8})(1+cos\left(\frac{\pi -3\pi }{8}\right))(1+cos\left(\frac{\pi -\pi }{8}\right))=(1+cos\frac{\pi }{8})(1+cos\frac{3\pi }{8})(1-cos\frac{3\pi }{8})(1-cos\frac{\pi }{8})=(1-co{s}^2\frac{\pi }{8})(1+cos\frac{3\pi }{8})=si{n}^2\frac{\pi }{8}si{n}^2\frac{3\pi }{8}=\frac{1}{4}{\left[2sin\frac{\pi }{8}sin(\frac{\pi }{2}-\frac{\pi }{8})\right]}^2=\frac{1}{4}{\left[2sin\frac{\pi }{8}cos\frac{\pi }{8}\right]}^2=\frac{1}{4}.si{n}^2\frac{\pi }{4}=\frac{1}{4}\times \frac{1}{2}=\frac{1}{8}$
$\therefore$ (c) is the correct answer.