General formula giving $ p-1 $ sign of $ - $ and $ q $ sign of $ + $ with a period of $ p + q - 1 $ \\ latex-code. we will consider the two following cases: \\ case of $ p = 2k + 1 $ odd \begin{align}  2\sin\left(\dfrac{(2^{p+q-1}+2^{q+1}-3)\pi}{3\times 2^{p+q}-6}\right)&=2\cos\left(\dfrac{(2^{p+q}-2^{q+1})\pi}{3\times 2^{p+q}-6}\right)\\&  =\sqrt{2-\sqrt{\cdots-\sqrt{2+\sqrt{\cdots+\sqrt{2}}}}} \end{align} case of $ p = 2k $ peer \begin{align}  2\sin\left(\dfrac{(2^{p+q-1}-2^{q+1}+3)\pi}{3\times 2^{p+q}+6}\right)&=2\cos\left(\dfrac{(2^{p+q}+2^{q+1})\pi}{3\times 2^{p+q}+6}\right)\\&  =\sqrt{2-\sqrt{\cdots-\sqrt{2+\sqrt{\cdots+\sqrt{2}}}}} \end{align} We arbitrarily choose p and q, For example, if we want (04) sign of (-) and (05) sign of (+), we will have: p=4+1=5 ; q=5 ; \begin{align} 2\sin\left(\dfrac{(2^{5+5-1}+2^{5+1}-3)\pi}{3\times 2^{5+5}-6}\right)&=2\cos\left(\dfrac{(2^{5+5}-2^{5+1})\pi}{3\times 2^{5+5}-6}\right)\\&  =\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2-\cdots}}}}}}}}}} \end{align} we are getting: \begin{align} 2\sin(\dfrac{191\pi}{1022})=2\cos(\dfrac{320\pi}{1022})=\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2-\cdots}}}}}}}}}} \end{align} An approximate evaluation under maxima gives: 2*sin((191*Pi)/(1022))=sqrt(2-sqrt(2-sqrt(2-sqrt(2-sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2-sqrt(2-sqrt(2-sqrt(2-sqrt(2))))))))))))));