Yes. It's known to be transcendental. The sequence of coefficients of your number is a variant of a *Sturmian* sequence. It has very low *complexity*. The definition of this: let the digit sequence be $a_1,a_2,a_3\ldots$ taking values in $ \lbrace 0,1\ldots,d-1 \rbrace ^{\mathbb N}$. A *subword* of length $k$ is a string $a_ia_{i+1}a_{i+2}\ldots a_{i+k-1}$. The complexity, $p(k)$, is a function from $\mathbb N$ to $\mathbb N$ taking $k$ to the number of subwords of the sequence of length $k$.

In a 2007 paper in the Annals of Mathematics (vol 165, p547--565), Adamczewski and Bugeaud (On the complexity of algebraic numbers. I. Expansions in integer bases) showed that if a number is algebraic, then its digit sequence in base $b$ has complexity satisfying $p(k)/k\to\infty$.

In your case, the complexity of the sequence of base 2 digits satisfies $p(k)=2k$. How to see this?
Define a map $f$ from $[0,1)$ to $\lbrace 0,1\rbrace$ by $f(x)=1$ if $x\in [0,1/2)$ and 0 otherwise.
The $n$th term of your sequence is $f(\alpha n\bmod 1)$, where $\alpha=1/(2\pi)$. Write $T$ for the transformation from $[0,1)$ to itself given by $T(x)=x+\alpha\bmod 1$. Then the $n$th term is just $f(T^n0)$. The sub-block of the digit sequence of length $k$ starting at the $j$th term is $f(T^j0)\ldots f(T^{j+k-1}0)$. Since the $T^j0$ are dense in $[0,1)$, we need to ask how many blocks $f(x)\ldots f(T^{k-1}x)$ are possible.

Consider taking $x=0$ and moving it around the circle (=$[0,1)$) once. As you move it, the $T^ix$ also each move around the circle one time. The sequence changes each time one of the $(T^ix)_{0\le i< k}$ crosses 0 or 1/2. This is a total of $2k$ changes. Hence the sequence takes on $2k$ values as $x$ moves around the circle, hence the estimate for the complexity.

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