What will happen to $\theta$ when $\sin \theta =\frac{\sqrt {{2{\pm}\sqrt i}}}{2}$ and i >4? Some follow up questions  related to Two mysterious missing angles in the sine values of acute angle list?
.After finding two missing angles,   we have a list of 9 sine values for special angles between 0$^\circ$ to $90^\circ$: $\sin \theta =\frac{\sqrt {{2{\pm}\sqrt i}}}{2}, i=0,1,2,3,4 $. (A side note:This formula is easy to remember: start with
$\sin 45^\circ=$$\frac{\sqrt2}{2}$, let $\pm \sqrt i$ jump under the radical sign and follow  behind 2. Thanks @some guy to remind this trick!)
Now, I have some curious thoughts:

*

*Why is the formula symmetric about $\theta = 45^\circ$?

*One would naturally want to continue this list with i beyond 4,..  in which case |$\sin \theta$| will not be bounded by [-1,1]. What does this mean? What are those angles? I expect we step into complex numbers, but do not have a clear picture. Can you please help?

\begin{align}\sin 0^\circ =\frac{\sqrt {\color{green}{2-\sqrt {4}}}}{2},\sin 15^\circ =\frac{\sqrt {\color{green}{2-\sqrt3}}}{2},\sin 22.5^\circ =\frac{\sqrt {\color{green}{2-\sqrt2}}}{2} \end{align}
\begin{align}\sin 30^\circ =\frac{\sqrt {\color{green}{2-\sqrt{1}}}}{2}
, \sin 45^\circ =\frac{\sqrt {\color{green}{2-\sqrt{0}}}}{2}
, \sin 60^\circ =\frac{\sqrt {\color{green}{2+\sqrt{1}}}}{2}\end{align}
\begin{align}\sin 67.5^\circ =\frac{\sqrt {\color{green}{2+\sqrt2}}}{2},\sin 75^\circ =\frac{\sqrt {\color{green}{2+\sqrt 3}}}{2}, \sin 90^\circ =\frac{\sqrt {\color{green}{2+\sqrt{4}}}}{2}\end{align}
 A: The sine function takes all complex values on the complex numbers.  The definition is $$\sin z = \frac{e^{iz}-e^{-iz}}{2i},$$ so if we want to solve $\sin z =w$ we have to solve $$2iw=e^{iz}-e^{-iz}=x-\frac1x$$ where $x=e^{iz}$ and multiplying both sides by $x$, we have a quadratic in $x$.  Note that $e^{iz}$ never takes the value $0$, so there is never a division-by-zero problem.
Does this answer your question?
A: About your first question about why they are symmetric about $45^\circ$. Let's say that you have an angle that is $x^\circ$ from the $45^\circ$ angle to make an angle of $45-x^\circ$. Thus, the symmetric angle must be $45+x^\circ$. So, if you add these angles up, you get $90^\circ$, which means that they are complementary. You should know that cosine is equal to the sine of the complementary angle. So, if we take a pair of symmetric angles $45-x^\circ, 45+x^\circ$, their sines will be $sin(45-x),sin(45+x)$ or $sin(45-x),cos(45-x)$, if we use our identity. Now, we will use one final identity, and that is that $sin(\theta)^2+cos(\theta)^2=1$. So, we get that $sin(45-x)^2+cos(45-x)^2=1$. Now, we will prove that if $sin$ of one of the angles, $sin(45-x) = \frac{\sqrt{2 - \sqrt{m}}}2$, then the other angle should be $sin(45+x) = \frac{\sqrt{2 + \sqrt{m}}}2$. Using our cosine identity, we know that $sin(45+x) = cos(45-x) = \frac{\sqrt{2 + \sqrt{m}}}2$. If we plug in our values into $sin(45-x)^2+cos(45-x)^2=1$, we get $(\frac{\sqrt{2 - \sqrt{m}}}2)^2+(\frac{\sqrt{2 + \sqrt{m}}}2)=1$ and if you simplify, you can see that this is true. Thus, our original assumptions that if $sin(45-x) = \frac{\sqrt{2 - \sqrt{m}}}2$, then the other angle should be $sin(45+x) = \frac{\sqrt{2 + \sqrt{m}}}2$ is true. If there is any confusion, let me know.
