<s>[INST] Troubleshooting a trigonometry/geometry question - spot the (silly) mistake! I'm back again!
Again, another error - not sure if it's them or me this time...
Here is the question

In order to calculate the shaded region, I added some lines to the diagram. I realise that there are other ways to solve this (and indeed the solution give uses a different breakdown and there's seems correct).

I then calculated the area of the sector ADB - the area of the triangle ADB and added this to the area of the sector BOD. I then multiplied that value by 2 to get the area of the shaded region.
Here is my working (Let a = alpha and t = theta)
$Area = 2[\frac12(3r)^2(\frac{a}2)-\frac12(3r)^2\sin(\frac{a}2)+\frac12(2r)^2(\frac{t}2)]$
Giving me a final answer of $r^2[(\frac{9a}2)-9\sin(\frac{a}2)+2t]$
Can you see where I went wrong...?
EDIT:
Here is the solution they provided:

 [/INST] The two answers are equivalent.
You have: $r^2[(\frac{9 \alpha}2)-9\sin(\frac{\alpha}2)+2 \theta]$
They have: $ [\frac 9 2(\alpha -\sin \alpha) + 2(\theta - \sin \theta)]r^2$
The triangle $ABD$ is isosceles, so $\frac \alpha 2 + 2 \frac \theta 2=\pi$
$\frac \alpha 2 = \pi-\theta \Rightarrow \sin \frac \alpha 2=\sin(\pi-\theta)=\sin \theta$
That changes your answer to $r^2[9(\pi - \theta)-9\sin \theta+2 \theta]=r^2[9\pi - 9 \sin \theta -7 \theta]$
$\alpha +2\theta = 2\pi \Rightarrow \alpha = 2\pi - 2\theta$
$\sin \alpha = \sin (-2\theta)=-\sin (2\theta)=-2\sin \theta \cos \theta$
That changes their answer to $ [\frac 9 2(2\pi - 2\theta +2\sin \theta \cos \theta) + 2(\theta - \sin \theta)]r^2 = [9\pi - 9\theta +9\sin \theta \cos \theta + 2\theta - 2\sin \theta]r^2 $
$=r^2[9\pi +(9\cos \theta - 2)\sin \theta - 7\theta]$
These look more alike now they are both in terms of $\theta$ but there is still a difference.
We can apply the cosine rule to the triangle $ADB$ to get $\cos (\frac \theta 2)=\frac{2^2+3^2-3^2}{2.2.3}$
So $\cos  (\frac \theta 2)=\frac 1 3$
$\cos \theta = \cos (2 \frac \theta 2)=2 \cos^2 (\frac \theta 2) -1 = 2(\frac 1 3)^2-1=-\frac 7 9$
Then $9\cos \theta - 2= -7-2=-9$
which turns their answer into yours.
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