Simple Area Trig ID's

The area of the shaded circle segment is equal the area of the circle sector with internal angle $\theta$ subtracts the area of isosceles triangle with the unequal angle $\theta$ . In formula:

$\begin{aligned} A_{\text{segment}} & = A_{\text{sector}} - \color{#3D99F6} A_\triangle & \small \color{#3D99F6} \text{Note that }A_\triangle = \frac 12 ab\sin \theta \\ & = \frac \theta{2\pi} \times \pi r^2 - \frac 12 r^2 \sin \theta \\ & = \frac 12 \theta - \frac 12 \sin \theta \\ & = \frac 12 \theta - 2 \left(\frac 14\right) \sin \theta \end{aligned}$

Therefore, $a+b = \frac 12 + \frac 14 = \frac 34 = \boxed{0.75}$ .

Draw the bisector of $\theta$ , which is also the perpendicular bisector of AB (I will not take the time to prove it here since it is a well known theorem). The area we are looking for is the Area of Sector AOB minus the Area of Triangle AOB . Since $\theta$ is in radians, the area of Sector AOB is $\frac{1}{2}$ $\theta$ $r^{2}$ . What is the area of Triangle AOB ? Well, it is $\frac{1}{2}$ ( AB )( OC ). By the law of cosines, $AB^{2}$ $=$ $r^{2}+r^{2}-2r^{2}\cos\theta$ $=$ $2-2cos\theta$ . Angle A and Angle B are equal. So $A=B=\frac{\pi-\theta}{2}$ . OC is then equal to $\sin\frac{\pi-\theta}{2}$ $=$ $\sin(\frac{\pi}{2}-\frac{1}{2}\theta)$ $=$ $\sin\frac{\pi}{2}\cos\frac{1}{2}\theta-\cos\frac{\pi}{2}\sin\frac{1}{2}\theta$ $=$ $\cos\frac{1}{2}\theta$ . So Triangle AOB is equal to $\frac{1}{2}(\cos\frac{1}{2}\theta\sqrt{2-2cos\theta})$ . Further reducing, $\cos\frac{1}{2}\theta$ $=$ $\sqrt{\frac{1+\cos\theta}{2}}$ . Then, $\sqrt{\frac{1+\cos\theta}{2}}$ $\sqrt{2-2\cos\theta}$ $=$ $\sqrt{\frac{2-2\cos^{2}\theta}{2}}$ $=$ $\sqrt{1-\cos^{2}\theta}$ $=$ $\sin\theta$ . Finally, the desired area is then $\frac{1}{2}\theta-\frac{1}{2}\sin\theta$ . So $a+b=$ $\boxed{\frac{3}{4}}$