Question
What is the value of $$\left( {1 + \cos \frac{\pi }{8}} \right)\left( {1 + \cos \frac{{3\pi }}{8}} \right)\left( {1 + \cos \frac{{5\pi }}{8}} \right)\left( {1 + \cos \frac{{7\pi }}{8}} \right)?$$
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 :
$$\frac{1}{8}$$
Solution :
$$\left[ {1 + \cos \frac{\pi }{8}} \right]\left[ {1 + \cos \frac{{3\pi }}{8}} \right]\left[ {1 + \cos \frac{{5\pi }}{8}} \right]\left[ {1 + \frac{{\cos 7\pi }}{8}} \right]$$
We have,
$$\eqalign{
& \cos \frac{{7\pi }}{8} = \cos \left[ {\pi - \frac{\pi }{8}} \right] = - \cos \frac{\pi }{8}\,{\text{and}}\,\,{{\cos}}\frac{{5\pi }}{8} = \cos \left[ {\pi - \frac{{3\pi }}{8}} \right] = - \cos \frac{{3\pi }}{8} \cr
& \therefore \left[ {1 + \cos \frac{\pi }{8}} \right]\left[ {1 + \cos \frac{{3\pi }}{8}} \right]\left[ {1 - \cos \frac{\pi }{8}} \right]\left[ {1 - \cos \frac{{3\pi }}{8}} \right] \cr
& = \left[ {1 - {{\cos }^2}\frac{\pi }{8}} \right]\left[ {1 - {{\cos }^2}\frac{{3\pi }}{8}} \right] = {\sin ^2}\frac{\pi }{8} \cdot {\sin ^2}\frac{{3\pi }}{8} \cr
& = \frac{1}{4}\left[ {2\,{{\sin }^2}\frac{\pi }{8} \cdot 2\,{{\sin }^2}\frac{{3\pi }}{8}} \right] \cr
& = \frac{1}{4}\left[ {\left( {1 - \cos \frac{\pi }{4}} \right)\left( {1 - \cos \frac{{3\pi }}{4}} \right)} \right]\,\,\,\,\,\,\,\,\,\left( {\because 1 - \cos \theta = 2\,{{\sin }^2}\frac{\theta }{2}} \right) \cr
& = \frac{1}{4}\left[ {\left( {1 - \frac{1}{{\sqrt 2 }}} \right)\left( {1 + \frac{1}{{\sqrt 2 }}} \right)} \right] = \frac{1}{8} \cr} $$