Chapter 5 - Trigonometric Identities - Section 5.6 Half-Angle Identities - 5.6 Exercises - Page 244: 63a

$$\cos\frac{\theta}{2}=\frac{R-b}{R}$$

(The image is shown below) We see in the image that $BC$ is a circular curve where $OB=OC=R$. That means $OA=R$ as $A$ is a point in the circular curve. Thus, $OH=OA-AH=R-b$ As triangle $OHC$ is a right triangle, we can calculate $\cos\frac{\theta}{2}$ using the sides of the triangle. $$\cos\frac{\theta}{2}=\frac{OH}{OC}$$ $$\cos\frac{\theta}{2}=\frac{R-b}{R}$$