$\tan \left( {{1 \over 4}{{\sin }^{ - 1}}{{\sqrt {63} } \over 8}} \right)$ ${\sin ^{ - 1}}\left( {{{\sqrt {63} } \over 8}} \right) = \theta$ $\sin \theta = {{\sqrt {63} } \over 8}$ $\cos \theta = {1 \over 8}$ $2{\cos ^2}{\theta \over 2} - 1 = {1 \over 8}$ ${\cos ^2}{\theta \over 2} = {9 \over {16}}$ $\cos {\theta \over 2} = {3 \over 4}$ ${{1 - {{\tan }^2}{\theta \over 4}} \over {1 + {{\tan }^2}{\theta \over 4}}} = {3 \over 4}$ $\tan {\theta \over 4} = {1 \over {\sqrt 7 }}$