jasperE3 wrote: Determine all real numbers $\alpha$ with the property that all numbers in the sequence $\cos\alpha,\cos2\alpha,\cos2^2\alpha,\ldots,\cos2^n\alpha,\ldots$ are negative. WLOG $\alpha\in[0,2\pi)$ Write $u=\frac{\alpha}{2\pi}$ in binary representation $\overline{0.xxxxxxxxxx...}_2$ If two consecutive digits are $00$, then $\exists n$ such that $\{2^nu\}=0.00xxxxxx\in[0,\frac 14)$ So $2^nu=k+v$ with $v\in[0,\frac 14)$ $\implies$ $2^n\alpha=2k\pi+v'$ with $v'\in[0,\frac{\pi}2)$ And so $\cos 2^n\alpha>0$, not suitable If two consecutive digits are $11$, then $\exists n$ such that $\{2^nu\}=0.11xxxxxx\in[\frac 34,1)$ So $2^nu=k+v$ with $v\in[\frac 34,1)$ $\implies$ $2^n\alpha=2k\pi+v'$ with $v'\in[\frac{3\pi}2,2\pi)$ And so $\cos 2^n\alpha\ge 0$, not suitable So $u$ can not have two equal consecutive digits and so only two possibilities : $u=\overline{0.010101010101...}_2)=\frac 13$ and $u=\overline{0.101010101010...}_2)=\frac 23$ And so $\boxed{\alpha=\frac{2\pi}3+2k\pi\text{ or }\alpha=\frac{4\pi}3+2k\pi}$, which indeed both fit (all cosinus are $-\frac 12$)