A chord of a circle systems …

Given
Area of minor segment cut off by $A B = \frac { 1 } { 8 } \times$ Area of circle
$\Rightarrow \quad \frac { \theta } { 360 ^ { \circ } } \times \pi r ^ { 2 } - \frac { 1 } { 2 } r ^ { 2 } \sin \theta = \frac { 1 } { 8 } \times \pi r ^ { 2 }$
$\Rightarrow \quad \frac { \theta } { 360 ^ { \circ } } \times \pi r ^ { 2 } = \frac { 1 } { 8 } \pi r ^ { 2 } + \frac { 1 } { 2 } r ^ { 2 } \sin \theta$
$\Rightarrow \quad \frac { \theta } { 360 ^ { \circ } } \times \pi = \frac { \pi } { 8 } + \frac { 1 } { 2 } \sin \theta \quad \left[ \text { Divide by } r ^ { 2 } \right]$
$\Rightarrow \quad \frac { \pi \theta } { 45 ^ { \circ } } = \pi + 4 \sin \theta \quad$ [Multiply by 8]
$\Rightarrow \quad \frac { \pi \theta } { 45 ^ { \circ } } = \pi + 4 \times 2 \sin \frac { \theta } { 2 } \cos \frac { \theta } { 2 } \quad [ \sin 2 \theta = 2 \sin \theta \cos \theta ]$
$\Rightarrow \quad \frac { \pi \theta } { 45 ^ { \circ } } = \pi + 8 \sin \frac { \theta } { 2 } \cos \frac { \theta } { 2 }$

Hence Proved