Evaluating $(1+\cos\frac{\pi}{8})(1+\cos\frac{3\pi}{8}) (1+\cos\frac{5\pi}{8})(1+\cos\frac{7\pi}{8})$ 
How to find the value of $$\left(1+\cos\frac{\pi}{8}\right)\left(1+\cos\frac{3\pi}{8}\right) \left(1+\cos\frac{5\pi}{8}\right)\left(1+\cos\frac{7\pi}{8}\right)$$


I used: $ 1+ \cos \theta=2\sin^2(\theta/2)$ to get $$2^4 \cdot\prod_{n=0}^3 \sin\left(\frac{(2n+1)\pi}{16}\right)$$
After that I tried using the formula: 
$$\sin\left(1 \frac{\pi}n\right)\sin\left(2 \frac{\pi}n\right) \cdots \sin\left((n-1) \frac{\pi}n\right)= \frac n {2^{n-1}}$$
to find the value of the expression in the parenthesis in the first equation, but currently I'm stuck. I don't think this method will lead me to the correct answer.
Any help would be appreciated!

First notice that $\dfrac{7\pi }{8}=\pi-\dfrac{\pi}{8}$ and also  $\dfrac{5\pi }{8}=\pi-\dfrac{3\pi}{8}$.
Substituting these values in the original expression, you'll get:
$$\left(1+\cos {\pi \over 8}\right)\left(1+\cos {3\pi \over 8}\right)\left(1-\cos {3\pi \over 8}\right)\left(1-\cos {\pi \over 8}\right)$$
Using the obvious algebraic identity, you get
$$ \left(1-\cos^2 {\pi \over 8}\right)\left(1-\cos^2 {3\pi \over 8}\right)$$
Which is nothing but:
$$\sin^2{\pi \over 8}.sin^2{3\pi \over 8}$$
Now finally notice that $ \dfrac{3\pi}{8}=\dfrac{\pi}{2}-\dfrac{\pi}{8}$ and use the identity $\sin2\theta=2\sin\theta \cos\theta$ to get the simple form
$$\frac{1}{4}\sin^2{\pi \over 4}$$
