If $x_0 = \cos(s)$, $0 < s < \pi$, and $x_{n+1} = \sqrt{\dfrac{1+x_n}{2}}$, then $x_{n} = \cos(s/2^n)$. Now $$ \prod_{n=0}^\infty x_n = \prod_{n=0}^\infty \cos(s/2^n) = \dfrac{\sin(2s)}{2s}$$ In particular you're taking the case $s = \pi/3$ so $x_0 = 1/2$, and then the product is $\sin(2 \pi/3)/(2\pi/3) = 3 \sqrt{3}/(4 \pi)$.

If $x_0 = \cos(s)$, $0 < s < \pi$, and $x_{n+1} = \sqrt{\dfrac{1+x_n}{2}}$, then $x_{n} = \cos(s/2^n)$. Now $$ \prod_{n=0}^\infty x_n = \prod_{n=0}^\infty \cos(s/2^n) = \dfrac{\sin(2s)}{2s}$$ In particular you're taking the case $s = \pi/3$ so $x_0 = 1/2$, and then the product is $\sin(2 \pi/3)/(2\pi/3) = 3 \sqrt{3}/(4 \pi)$. Of course if $s$ is any rational multiple of $\pi$, $\sin(2s)$ is algebraic, and then (unless $\sin(2s)=0$) the product is transcendental.

If $x_0 = 1/2$ and $x_{n+1} = \sqrt{\dfrac{1+x_n}{2}}$, it appears that $$ \prod_{n=0}^\infty x_n = \dfrac{3 \sqrt{3}}{4 \pi}$$ (verified numerically to $300$ decimal places, but I don't have a proof)

If $x_0 = \cos(s)$, $0 < s < \pi$, and $x_{n+1} = \sqrt{\dfrac{1+x_n}{2}}$, then $x_{n} = \cos(s/2^n)$. Now $$ \prod_{n=0}^\infty x_n = \prod_{n=0}^\infty \cos(s/2^n) = \dfrac{\sin(2s)}{2s}$$ In particular you're taking the case $s = \pi/3$ so $x_0 = 1/2$, and then the product is $\sin(2 \pi/3)/(2\pi/3) = 3 \sqrt{3}/(4 \pi)$.