AB is the diameter of a circle, centre O,

Solution:

We know that the area of minor segment of angle θ in a circle of radius r is,

$A=\left\{\frac{\pi \theta}{360^{\circ}}-\sin \frac{\theta}{2} \cos \frac{\theta}{2}\right\} r^{2}$

It is given that, $\angle B O C=\theta$

So, $\angle A O C=180^{\circ}-\theta$

Area, $A$ of minor segment cutoff by $A C$ at angle $\angle A O C=180-\theta$

$A=\left\{\frac{\pi\left(180^{\circ}-\theta\right)}{360^{\circ}}-\sin \frac{\left(180^{\circ}-\theta\right)}{2} \cos \frac{\left(180^{\circ}-\theta\right)}{2}\right\} r^{2}$

Now, since $\sin \left(90^{\circ}-\alpha\right)=\sin \alpha$ and $\cos \left(90^{\circ}-\alpha\right)=\cos \alpha$

$A=\left\{\frac{\pi\left(180^{\circ}-\theta\right)}{360^{\circ}}-\sin \frac{\theta}{2} \cos \frac{\theta}{2}\right\} r^{2}$

We know that the area of sector of a circle of radius r at an angle θ is

$A^{\prime}=\frac{\theta}{360^{\circ}} \times \pi r^{2}$

So, the area of sector $B O C, A^{\prime}=\frac{\theta}{360^{\circ}} \times \pi r^{2}$

It is given that,

Area of minor segment cutoff by $A C=2 \times$ Area of sector $B O C$

$\left\{\frac{\pi\left(180^{\circ}-\theta\right)}{360^{\circ}}-\cos \frac{\theta}{2} \sin \frac{\theta}{2}\right\} r^{2}=\frac{2 \theta}{360^{\circ}} \times \pi r^{2}$

$\frac{\pi\left(180^{\circ}-\theta\right)}{360^{\circ}}-\cos \frac{\theta}{2} \sin \frac{\theta}{2}=\frac{2 \pi \theta}{360^{\circ}}$

$\frac{\pi 180^{\circ}}{360^{\circ}}-\frac{\pi \theta}{360^{\circ}}-\cos \frac{\theta}{2} \sin \frac{\theta}{2}=\frac{2 \pi \theta}{360^{\circ}}$

$\cos \frac{\theta}{2} \sin \frac{\theta}{2}=\frac{\pi 180^{\circ}}{360^{\circ}}-\frac{\pi \theta}{360^{\circ}}-\frac{2 \pi \theta}{360^{\circ}}$

$\cos \frac{\theta}{2} \sin \frac{\theta}{2}=\frac{\pi}{2}-\frac{3 \pi \theta}{360^{\circ}}$

$\cos \frac{\theta}{2} \sin \frac{\theta}{2}=\frac{\pi}{2}-\frac{\pi \theta}{120^{\circ}}$

$\cos \frac{\theta}{2} \sin \frac{\theta}{2}=\pi\left(\frac{1}{2}-\frac{\theta}{120^{\circ}}\right)$