<s>[INST] Area enclosed by quarter circle arcs in a square 
The problem I have here asks to find the area $A$ enclosed by the two quarter circle arcs in terms of $a$. The first circle has radius $a/2$, and the second has radius $a$. This is what I did.

Firstly I determined that $$A=\frac{a^2}{8}(2\theta_1-\sin(2\theta_1))+\frac{a^2}{2}(2\theta_2-\sin(2\theta_2))$$ $$=\frac{a^2}{4}(\theta_1+4\theta_2-\sin\theta_1\cos\theta_1-4\sin\theta_2\cos\theta_2)$$
Then $$h_1+h_2=a\sqrt2=\sqrt{\frac{a^2}{4}-x^2}+\sqrt{a^2-x^2}$$
$$(a\sqrt2-\sqrt{a^2-x^2})^2=\frac{a^2}{4}-x^2$$
$$12a^2-8a\sqrt{2a^2-2x^2}=a^2$$
$$11a=8\sqrt{2a^2-2x^2}$$
$$121a^2=128a^2-128x^2$$
$$x=a\sqrt{\frac{7}{128}}=\frac{a\sqrt{14}}{16}$$
So
$$h_1=\sqrt{\frac{a^2}{4}-\frac{14a^2}{256}}=\frac{5a\sqrt2}{16}$$
$$h_2=\sqrt{a^2-\frac{14a^2}{256}}=\frac{11a\sqrt2}{16}$$
Therefore
$$\sin\theta_1\cos\theta_1=\frac{\sqrt{14}}{8}\times\frac{5\sqrt2}{8}=\frac{5\sqrt7}{32}$$
$$4\sin\theta_2\cos\theta_2=4\times\frac{\sqrt{14}}{16}\times\frac{11\sqrt2}{16}=\frac{11\sqrt7}{32}$$
So
$$A=\frac{a^2}{4}(\theta_1+4\theta_2-\frac{5\sqrt7}{32}-\frac{11\sqrt7}{32})$$
$$=\frac{a^2}{4}(\theta_1+4\theta_2-\frac{\sqrt7}{2})$$
Now I can find $\theta_1$ and $\theta_2$ with my calculator, but there must be a way to express them in terms of surds but I don't know how. Also, is what I did actually correct and is there another shorter/nicer way to solve this?
 [/INST] $\sin\theta_1=2\sin\theta_2$ 
$\frac a 2\cos\theta_1+a\cos\theta_2=a\sqrt 2 \implies\cos\theta_1+2\cos\theta_2=2\sqrt 2\implies\cos\theta_1=2\sqrt 2-2\cos\theta_2$
Then you can square both equations and add to each other
$$\sin^2\theta_1+\cos^2\theta_1=4\sin^2\theta_2+(2\sqrt2-2\cos\theta_2)^2$$ 
$$8\sqrt2 \cos\theta_2=11$$
$$\cos\theta_2=\frac{11}{8\sqrt2}$$
</s>