Question

Someone please help me prove this.

Answer 1

I have worked out the length of AB= 2rcos(theta).
The area of sector OABC= (pi)(r^2) x (360-4(theta))/360
I don't know what to do next.

Answer 2

could you explain where you got AB ?
i'm not getting the same value.

Answer 3

Here: I got x=rcos(theta). AB=2x=2cos(theta)

Answer 4

ah, ok, that looks good to me.
(i was using law of cosine and getting something probably not in a simplified form)
so then it looks like we would have to find the area of the shaded region and area of the circle and compare the two (it is the main equality in our statement)

Answer 5

we could find the sector swept out from center O between OA and OB to find the segments in the corners
then the main sector with center A between AB and AC seems to make up the rest of the area, right?

(since that area above BOC looks pretty hard to compute alone after having sector OB through OC

Answer 6

Okay.
So I got the area of the segments \[(180-2\theta-\sin 2\theta)r^2\]
The area of the sector ABXC = \[\theta(4r^2\cos ^2 \theta)\]