Question

AB is the diameter of a circle, center O.C is a point on circumference such that angle $COB=\theta$. The area of the minor segment cut off by AC is equal to twice the area of sector BOC. Prove that
$sin\frac{\theta }{2}cos\frac{\theta }{2}=\pi (\frac{1}{2}-\frac{\theta }{120})sin\frac{\theta }{2}cos\frac{\theta }{2}=\pi (\frac{1}{2}-\frac{\theta }{120})$

Answer 1

Let the radius be r
$\angle BOC=\theta$
$\therefore \angle AOC=1800-\theta$ [$\because$ AB is the diameter]
Area of minor segment cut by AC = Area of sector OADC - Area of
$\angle$ AOC
$\frac{180}{360}-\frac{\theta }{360}\pi r^2-r^{2}sin(90-\frac{\theta }{2}).cos(90-\frac{\theta }{2})$
$=r^2\left[(\frac{1}{2}-\frac{\theta }{360})\right]\pi -sin\left(\frac{\theta }{2}\right).cos\left(\frac{\theta }{2}\right)...\left(1\right)[\because sin(90-\theta )=cos\theta -2]$
$\left[cos\right(90-\theta \left)sin\theta \right]$
Area of sector BOC $=\frac{\theta }{360}\times \pi r^2...\left(2\right)$
$\Rightarrow r^2\left[(\frac{1}{2}-\frac{\theta }{360})\pi sin(\frac{\theta }{2}.cos\frac{\theta }{2})\right]=2\times \frac{\theta }{360}\times \pi r^2$
$\Rightarrow \frac{\pi }{2}-\frac{\theta \pi }{360}-\frac{\theta \pi }{180}=sin\frac{\theta }{2}.cos\frac{\theta }{2}$
$\Rightarrow \pi [\frac{1}{2}-\frac{\theta \pi -2\theta \pi }{360}]=sin\frac{\theta }{2}.cos\frac{\theta }{2}$
$\Rightarrow \pi [\frac{1}{2}-\frac{\theta \pi }{360}]=sin\frac{\theta }{2}.cos\frac{\theta }{2}$
$\Rightarrow \pi [\frac{1}{2}-\frac{\theta \pi }{120}]=sin\frac{\theta }{2}.cos\frac{\theta }{2}$