Computing an infinite product $\prod_{n=1}^{\infty} \frac{1}{2}(1+\cos\frac{x}{2^n})$ I would like to compute the infinite product
$\displaystyle f(x)=\prod_{n=1}^{N\rightarrow\infty} \frac{1}{2}\left({1+\cos\frac{x}{2^n}}\right)$ for a given real $x$.
Since the terms in the product are positive and smaller than $1$, the product is bounded by $0$ and decreases. Therefore the limit $N\rightarrow\infty$ is well defined.
I know that 
$\displaystyle \prod_{n=1}^{\infty} \cos\frac{x}{2^n}=\frac{\sin(x)}{x}$ and I wonder if $f$ also has a simple analytical expression.
Thank you.
EDIT : Using the expression $\cos^2(x)=(1+\cos(2x)/2)$ I can reducethe first sum to the second and get
$f(x)=\left(\frac{\sin(x)}{x \cos(x/2)}\right)^2$

Take the well known product and square:
$$\prod_{n=1}^{\infty} \cos\frac{x}{2^n}=\frac{\sin(x)}{x}$$
$$\prod_{n=1}^{\infty} \cos^2\frac{x}{2^n}=\frac{\sin^2(x)}{x^2}$$
Now use the formula:
$$\cos^2 \frac{t}{2}=\frac{1}{2} (1+\cos t)$$
To get:
$$\prod_{n=0}^\infty \frac{1}{2} \left( 1+\cos \frac{x}{2^n} \right)=\frac{\sin^2 x}{x^2}$$
