In the diagram, A is a point on the circumference of a circle with centre O and radius r. A circular arc with centre A meets the circumference at B and C. The angle OAB is $θ$ radians. The shaded region is bounded by the circumference of the circle and the arc with centre A joining B and C. The area of the shaded region is equal to half the area of the circle.
Show that $$\cos2θ = \frac{2\sin2θ – π}{4θ}$$
I had trouble showing this. I said that the area of the shaded region is equal to area of the BAC sector + $2*$ area of BOA sector – $2*$ area of BOA triangle and equated this to $\frac{1}{2}πr^2$, but I keep getting the same result of $2\sin2θ=\frac{1}{2}π-θ$, and I do not know how to continue.
Best Answer
The radius of the arc $BC$ is $|AB|=2r\cos\theta$, and $\measuredangle AOB=\pi-2\theta$. \begin{align} \text{Dashed area}&= \text{Area of sector }BAC+2(\text{Area of sector } BOA-\text{Area of triangle} BOA)\\ &=\frac{1}{2}|AB|^2(2\theta)+2\left[\frac{1}{2}r^2(\pi-2\theta)-\frac{1}{2}r^2\sin(\pi-2\theta)\right] \end{align} On the other hand he area of the dashed surface is $\frac{1}{2}\pi r^2$, then \begin{align} \frac{1}{2}|AB|^2(2\theta)+2\left[\frac{1}{2}r^2(\pi-2\theta)-\frac{1}{2}r^2\sin(\pi-2\theta)\right]&=\frac{1}{2}\pi r^2 \\ (2r\cos \theta)^2(\theta)+r^2\left[\pi-2\theta-\sin(\pi-2\theta)\right]&=\frac{1}{2}\pi r^2\\ 4\theta\cos^2\theta+\pi-2\theta-\sin2\theta&=\frac{1}{2}\pi \\ 2\theta(2\cos^2\theta-1)&=\sin 2\theta-\frac{1}{2}\pi \\ 2\theta\cos 2\theta&=\frac{2\sin 2\theta - \pi}{2}\\ \cos 2\theta&=\frac{2\sin 2\theta - \pi}{4\theta} \end{align}