Question

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

Answer 1

Given, $\angle BOC=\theta$

$\Rightarrow \angle AOC={180}^o-\theta$

Now, area of sector BOC $=\pi r^2\times \frac{\theta }{{360}^o}$

and area of minor segment cut-off by AC $=\pi r^2\times \frac{{180}^o-\theta }{{360}^o}-r^{2}sin\left(\frac{{180}^o-\theta }{2}\right)cos\left(\frac{{180}^o-\theta }{2}\right)$

According to the question, $2\times \pi r^2\times \frac{\theta }{{360}^o}=\pi r^2\times \frac{{180}^o-\theta }{{360}^o}-r^{2}sin\left(\frac{{180}^o-\theta }{2}\right)cos\left(\frac{{180}^o-\theta }{2}\right)$

$\Rightarrow \frac{\pi \theta }{{180}^o}=\pi (\frac{1}{2}-\frac{\theta }{{360}^o})-sin({90}^o-\frac{\theta }{2})cos({90}^o-\frac{\theta }{2})$

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

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

$\Rightarrow sin\frac{\theta }{2}cos\frac{\theta }{2}=\pi (\frac{1}{2}-\frac{\theta }{{120}^o})$

Hence, proved.