<s>[INST] 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? 
 [/INST] Recall that
$$
1+\cos\alpha=2\cos^2\frac{\alpha}{2}
$$
If $\beta=\pi/16$ (so $8\beta=\pi/2$), you get
$$
16\cos^2\beta\cos^23\beta\cos^25\beta\cos^27\beta=
(2\cos\beta\cos7\beta)^2(2\cos3\beta\cos5\beta)^2
$$
Now, with the product-to-sum formula,
$$
2\cos\beta\cos7\beta=\cos8\beta+\cos6\beta=\cos6\beta
$$
and
$$
2\cos3\beta\cos5\beta=\cos8\beta+\cos2\beta=\cos2\beta
$$
Hence your expression is
$$
(\cos6\beta\cos2\beta)^2=
\frac{1}{4}(\cos8\beta+\cos4\beta)^2=
\frac{1}{4}\cos^2\frac{\pi}{4}=\frac{1}{8}
$$
</s>