<s>[INST] Derivation of $\sin(15^\circ)$ geometrically This is my attempt:-
Let us consider a right $\triangle ABC$ such that angle  $A$ is $15^\circ$ and $C$ is $75^\circ$.
On the line $AB$, let us assume a point $D$ such that $\frac{BC}{BD} =\frac{ 1}{\sqrt{3}}$ (Without Loss Of Generality). So $\angle BDC$ becomes $30^\circ$ and $\angle BCD$ becomes $60^\circ$. Then $\angle DCA$ becomes equal to $\angle BAC$,  that is $15^\circ$; so $CD = DA$.

$CD$ will be $2$ times $BC$ (angle $BDC = 30^\circ$; $\sin 30^\circ$). On adding $BD$ and $AD$ we get $$AB = BC(2+\sqrt{3})$$
$$BC^2 + AB^2 = AC^2$$
$$\therefore AC = 2BC\sqrt{2 + \sqrt{3}}$$
$$\sin(15^\circ) = \frac{BC}{AC}$$
$$\sin(15^\circ) = \frac{1}{\sqrt{2 + \sqrt{3}}}$$
Rationalising the denominator 2 times we get:-
$$\sin(15^\circ) = \frac{(4-2\sqrt{3})(\sqrt{2+\sqrt{3}})}{4}$$
Further simplifying:-
$$\sin(15^\circ) = \frac{(2-\sqrt{3})(\sqrt{2+\sqrt{3}})}{2}$$
$$\sin(15^\circ) = \frac{(\sqrt{2-\sqrt{3}})(\sqrt{2-\sqrt{3}}) (\sqrt{2+\sqrt{3}})}{2}$$
$$\sin(15^\circ) = \frac{\sqrt{2-\sqrt{3}}}{2}$$
Is my answer correct? And is there any other method or way to get the value of $\sin(15^\circ)$ geometrically?
 [/INST] There is another method that starts with a fun and challenging geometry problem. Not sure if you've seen it before, but it is fairly famous (I don't know if it has a specific name though):

Now, once you've solved that purely with elementary geometry, no trig, you can then get some nice trig results from that. $M$ and $N$ are the midpoints of $AB$ and $CD$ respectively:
Let the square have side $2$. You've already shown $\triangle AOB$ is equilateral, so $\angle OAB = 60^{\circ}$. Which gives $OM = AM\tan 60^{\circ} = \sqrt 3$. Then $ON = 2 - \sqrt 3$. You can get $OD$ by Pythagoras': $OD = \sqrt{(2 - \sqrt 3)^2 + 1^2} = 2\sqrt{2 - \sqrt 3}$. So $\sin 15^{\circ} = \frac{ON}{OD} = \frac 12\sqrt{2-\sqrt 3}$.
</s>