What is the value of the expression $[1 + \cos(\frac{\pi}{8})][1 + \cos(\frac{3\pi}{8})][1 + \cos(\frac{5\pi}{8})][1 + \cos(\frac{7\pi}{8})]$? This is rather a simple problem that I'm posting ; looking forward not for the solution of it but the different ways it could be solved.
What is the value of the following expression? 

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

Edit: Do I manually put the values of $\cos\frac{\pi}{8}$ and all other cosine terms? Is there a better and shorter way? 

The answer is $\dfrac{1}{8}$
As $\cos(\pi-x)=-\cos x$ we can write
$$\left(1+\cos\frac{\pi}{8}
\right)\left(1+\cos\frac{3\pi}{8}
\right)\left(1+\cos\frac{5\pi}{8}
\right)\left(1+\cos\frac{7\pi}{8}
\right)=\left(1+\cos\frac{\pi}{8}
\right)\left(1+\cos\frac{3\pi}{8}
\right)\left(1-\cos\frac{3\pi}{8}
\right)\left(1-\cos\frac{\pi}{8}
\right)=
\left(1-\cos^2\frac{\pi}{8}
\right)\left(1-\cos^2\frac{3\pi}{8}
\right)=\sin^2\frac{\pi}{8}\,\sin^2\frac{3\pi}{8}=\left(\sin\frac{\pi}{8}\,\sin\frac{3\pi}{8}\right)^2$$
For the Werner's formula 
$$\sin x\sin y=\frac{1}{2} (\cos (x-y)-\cos (x+y))$$
so 
$$\sin\frac{\pi}{8}\,\sin\frac{3\pi}{8}=\frac{1}{2} \left(\cos \left(\frac{\pi}{8}-\frac{3\pi}{8}\right)-\cos \left(\frac{\pi}{8}+\frac{3\pi}{8}\right)\right)=\frac{1}{2}\left(\cos\frac{-\pi}{4}-\cos\frac{\pi}{2}\right)=\frac{\sqrt{2}}{4}$$
Remember that
$\cos\dfrac{-\pi}{4}=\cos\dfrac{\pi}{4}=\dfrac{\sqrt 2}{2}$
So the result is $\left(\dfrac{\sqrt{2}}{4}\right)^2=\dfrac{1}{2}$
