$$2\text{ days ago}$$
$2$ days ago, I asked this question asking about how to find the exact value of $\sin(\def\a#1{#1\unicode{xB0}}\a{10})$, which was surprisingly well received question for me. In this question, I used the knowledge of $\sin\left(\a{\dfrac x2}\right)$ being equal to $\pm\sqrt{\dfrac12(1-\cos(x))}$ to generate a formula to determine a formula to get the exact value of $\sin\left(\a{\dfrac x3}\right)$, however, I am wondering now if I could have just plugged in $2\a0$ for $x$ in the formula to determine the exact value of $\sin\left(\a{\dfrac x2}\right)$ to get the exact value of $\sin(1\a0)$. Here is my reasoning:
$$\sin(1\a0)=\pm\sqrt{\dfrac12(1-\cos(20))}$$$$=\pm\sqrt{0.5-\dfrac{\cos\left(\dfrac\pi9\right)}2}$$$$=\pm\sqrt{\dfrac{1-\cos\left(\dfrac\pi9\right)}2}$$Which, if we wanted to simplify this even further, we would get$$\pm\dfrac12\cdot\sqrt2\sqrt{1-\cos\left(\dfrac\pi9\right)}$$Setting this equal to the answer posted by @Robert Israel:$$\pm\dfrac12\cdot\sqrt2\sqrt{1-\cos\left(\dfrac\pi9\right)}=\frac{\omega - \overline{\omega}}{2i} = -\frac{i}{2} \left( \left(\frac{\sqrt{3}+i}{2}\right)^{1/3} - \left(\frac{\sqrt{3}-i}{2}\right)^{1/3}\right)$$And when we plug this into Wolfram Alpha, it turns out that the two equations are only equal to each other when $\sin(2\a0)$ is positive:
And when $\sin(2\a0)$ is negative:
Imgur links for reference if they are not loading:
Test for when the sine of x over 2 yields a positive result
Test for when the sine of x over 2 yields a negative result
My question
Why is it, that when we achieve the solution for the exact solution for $\sin(1\a0)$ using the formula for finding the exact value of $\sin\left(\a{\dfrac x2}\right)$, it is only equal to the exact value of $\sin(1\a0)$ that the answer to my last question received only when the formula for $\sin\left(\a{\dfrac x2}\right)$ provides a positive result?
To clarify
Note: If my title seems a bit long, you may edit it. However, and I know this may not be a valid point, I just feel like it is important to mention that this is an addendum to my previous question so it doesn't get marked as a duplicate of my previous question. It also might seem a bit long in the sense that it is a total of $134$ characters long at the time of posting this question. Feel free to edit this post in any way if you think that that part of the question might need to be edited for any reason. However, if you do decide to edit this, please be careful to not edit any defined functions, as that might break the entire post.
