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?
 A: You obtained the correct answer.  However, not every step you wrote down is correct.
You found that
$$AC = 2BC\sqrt{2 + \sqrt{3}}$$
from which it follows that
$$\sin(15^\circ) = \frac{BC}{AC} = \frac{1}{2\sqrt{2 + \sqrt{3}}}$$
Multiplying the numerator and denominator by $\sqrt{2 - \sqrt{3}}$ yields
$$\sin(15^\circ) = \frac{\sqrt{2 - \sqrt{3}}}{2\sqrt{4 - 3}} = \frac{\sqrt{2 - \sqrt{3}}}{2}$$
As for an alternative method, consider the diagram below.

Since $\angle AEC$ is a right angle, $\triangle ACE$ is a right triangle.  Since the acute angles of a right triangle are complementary and $m\angle CAE = 45^\circ$, $m\angle ACE = 45^\circ$.  Since $CE = 1$ and the ratio of the side lengths of a $45^\circ, 45^\circ, 90^\circ$ right triangle is $1: 1: \sqrt{2}$, $AE = 1$ and $AC = \sqrt{2}$.
Since $\angle AEC$ is a right angle, $\overline{AD} \perp \overline{CE}$.  Thus, $\angle CED$ is also a right angle.  Therefore, $\triangle CDE$ is a right triangle.  Since the acute angles of a right triangle are complementary and $m\angle CDE = 30^\circ$, $m\angle ECD = 60^\circ$.  Since the ratio of the side lengths in a $30^\circ, 60^\circ, 90^\circ$ right triangle is $1: \sqrt{3}: 2$ and $CE = 1$, $DE = \sqrt{3}$ and $CD = 2$.
By the Segment Addition Postulate, $AD = AE + ED = 1 + \sqrt{3}$.
Since $\angle ABD$ is a right angle, $\triangle ABD$ is a right triangle.  Since the acute angles of a right triangle are complementary and $m\angle BDA = 30^\circ$, $m\angle BAD = 60^\circ$.  Since the ratio of the side lengths in a $30^\circ, 60^\circ, 90^\circ$ right triangle is $1: \sqrt{3}: 2$ and $AD = 1 + \sqrt{3}$, $AB = \dfrac{1 + \sqrt{3}}{2}$ and $BD = \dfrac{3 + \sqrt{3}}{4}$.
By the Angle Addition Postulate, $m\angle BAD = m\angle BAC + m\angle CAE$.  Hence, $m\angle BAC = m\angle BAD - m\angle CAE = 60^\circ - 45^\circ = 15^\circ$.
Since $\angle ABC$ is a right angle, $\triangle ABC$ is a right triangle.  Since the acute angles of a right triangle are complementary and $m\angle BAC = 15^\circ$, $m\angle ACB = 75^\circ$.
By the Segment Addition Postulate, $BC + CD = BD$.  Hence, $BC = BD - CD = \dfrac{3 + \sqrt{3}}{4} - 2 = \dfrac{-1 + \sqrt{3}}{2}$.
We summarize these results in the diagram below.

\begin{align*}
\sin(15^\circ) & = \frac{BC}{AC}\\
               & = \frac{\frac{-1 + \sqrt{3}}{2}}{\sqrt{2}}\\
               & = \frac{-1 + \sqrt{3}}{2\sqrt{2}}\\
               & = \frac{-1 + \sqrt{3}}{2\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}\\
               & = \frac{-\sqrt{2} + \sqrt{6}}{4}\\
               & = \frac{\sqrt{6} - \sqrt{2}}{4}
\end{align*}
which is equivalent to your answer since
\begin{align*}
\frac{\sqrt{2 - \sqrt{3}}}{2} & = \frac{\sqrt{2 - \sqrt{3}}}{2} \cdot \frac{2}{2}\\
                              & = \frac{\sqrt{8 - 4\sqrt{3}}}{4}\\
                              & = \frac{\sqrt{6 - 4\sqrt{3} + 2}}{4}\\
                              & = \frac{\sqrt{(\sqrt{6} - \sqrt{2})^2}}{4}\\
                              & = \frac{\sqrt{6} - \sqrt{2}}{4}
\end{align*}
A: 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 $\sin 15^{\circ} = \frac{ON}{OD} = \frac 12\sqrt{2-\sqrt 3}$.
