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$ A B $ is the diameter of a circle, centre $ O . C $ is a point on the circumference such that $ \angle C O B=\theta . $ The area of the minor segment cut off by $ A C $ is equal to twice the area of the sector $ B O C $. Prove that $ \sin \frac{\theta}{2} \cos \frac{\theta}{2}=\pi\left(\frac{1}{2}-\frac{\theta}{120}\right) $
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Given:
\( A B \) is the diameter of a circle, centre \( O . C \) is a point on the circumference such that \( \angle C O B=\theta . \)
The area of the minor segment cut off by \( A C \) is equal to twice the area of the sector \( B O C \).
To do:
We have to prove that \( \sin \frac{\theta}{2} \cos \frac{\theta}{2}=\pi\left(\frac{1}{2}-\frac{\theta}{120}\right) \).
Solution:
Let $r$ be the radius of the circle.
From the figure,
$\angle \mathrm{BOC}=\theta$
This implies,
$\mathrm{AOC}=180^{\circ}-\theta$
Area of the sector $\mathrm{BOC}=\pi r^{2} \times \frac{\theta}{360^{\circ}}$
Area of the minor segment $\mathrm{AC}=2$ (Area of sector BOC)
$=2 \times \pi r^{2} \frac{\theta}{360^{\circ}}$
$=\frac{2 \pi r^{2} \theta}{360^{\circ}}$............(i)
Area of segment $=(\frac{\pi(180^{\circ}-\theta)}{360^{\circ}}-\sin \frac{180^{\circ}-\theta}{2} \cos \frac{180^{\circ}-\theta}{2}) r^{2}$.............(ii)
From (i) and (ii), we get,
$\frac{2 \pi r^{2} \theta}{360^{\circ}}=r^{2}(\frac{\pi(180^{\circ}-\theta)}{360^{\circ}}-\sin \frac{180^{\circ}-\theta}{2} \cos \frac{180^{\circ}-\theta}{2})$
$\Rightarrow \frac{2 \pi \theta}{360^{\circ}}=[\frac{\pi(180^{\circ}-\theta)}{360^{\circ}}-\sin (90^{\circ}-\frac{\theta}{2}) \cos (90^{\circ}-\frac{\theta}{2})]$
$\Rightarrow \frac{2 \pi \theta}{360^{\circ}}=\frac{\pi(180^{\circ}-\theta)}{360^{\circ}}-\cos \frac{\theta}{2} \sin \frac{\theta}{2}$
Therefore,
$\sin \frac{\theta}{2} \cos \frac{\theta}{2}=\frac{\pi(180^{\circ}-\theta)}{360^{\circ}}-\frac{2 \pi \theta}{360^{\circ}}$
$\sin \frac{\theta}{2} \cos \frac{\theta}{2}=\pi(\frac{180^{\circ}}{360^{\circ}}-\frac{\theta}{360^{\circ}}-\frac{2 \theta}{360^{\circ}})$
$\sin \frac{\theta}{2} \cos \frac{\theta}{2}=\pi(\frac{1}{2}-\frac{3 \theta}{360^{\circ}})$
$=\pi(\frac{1}{2}-\frac{\theta}{120})$
$\sin \frac{\theta}{2} \cos \frac{\theta}{2}=\pi(\frac{1}{2}-\frac{\theta}{120})$
Hence proved.
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