I've approached in a few different ways to prove it, but couldn't get any good results.
When :
$d = 2\,,$
$\text{radius}=\dfrac d2\;\rightarrow\;r=1\,.$
Since hypotenuse is the same length with $\,r\,,$
$\sin(\theta)=\dfrac{\text{opp}}{\text{hyp}(r)}$
But divided by $\,1\,$ is simply meaningless, so it should be:
$\sin(\theta)=\dfrac{\text{opp}}1$
$\sin(\theta)=\text{opp}\;\;$ (still unknown)
This time, I used Pythagorean Theorem to get the other lengths directly.
By using the formula:
$c=\sqrt{a^2+b^2}$
$c^2=a^2+b^2$
$a^2=c^2-b^2$
$a=\sqrt{c^2-b^2}$
Since $\,c\,$ represents $\,r\,,$
$a=\sqrt{r^2-b^2}$
$a=\sqrt{1^2-b^2}$
$a=\sqrt{1-b^2}\;\;$ (still unknown)
At the end, both ways couldn't find out what numbers of $\,o\,$ and $\,a\,$ are as long as hypotenuse is $\,1\,.$
But when I tap the calculator, $\,\sin\left(60\unicode{176}\right)\,$ throws $\,0.866025\ldots\,$ which doesn't make sense.
If I given only hypotenuse length, how can I prove that $\,\sin(\theta)=x\,?$
As already mentioned by @lulu, I would construct a $30-60-90$ triangle on the unit circle first. After we obtain all side lengths, we can use $\sin(\alpha) = \frac{\text{opposite}}{\text{hypotenuse}}$ to prove the value of $\sin(60°)$.
We first construct the triangle $\triangle ABC$, noting that the segment $AB$ has to be of length $1$, since that is the radius of the unit circle. By mirroring $\triangle ABC$ on the $x$-axis, we obtain $\triangle ACD$, noting that the segment $AD$ has to be of length $1$ too, for the same reason as above. We now have a larger triangle $\triangle ABD$.
Notice that all the angles of $\triangle ABD$ are $60°$, that means $\triangle ABD$ is equilateral. Thus, the side length of segment $BD$ is 1.
Since $BD$ is exactly twice the length of $BC$, we know that $BC$ is of length $\frac{1}{2}$. Also, since $AB$ is of length 1, we know that $AC = \sqrt{1^2 - \left(\frac{1}{2}\right)^2} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$.
Now, $\sin(60°) = \frac{AC}{1} = \frac{\sqrt{3}}{2}$.