Two mysterious missing angles in the sine values of acute angle list?

Question

There is a well known list of 5 for trig values of special angles between 0$^o$ to $90^o$:

$\sin 0^o =\frac{\sqrt {\color{blue}{0}}}{2}$, $\sin 30^o =\frac{\sqrt {\color{blue}{1}}}{2}$ , $\sin 45^o =\frac{\sqrt {\color{blue}{2}}}{2}$ , $\sin 60^o =\frac{\sqrt {\color{blue}{3}}}{2}$, $\sin 90^o =\frac{\sqrt {\color{blue}{4}}}{2}$

If wee add 15$^o$ and 75$^o$ , we can make the angles evenly spaced by 15$^o$ and expand the list of 5 to a list of 7:

$\sin 0^o =\frac{\sqrt {\color{blue}{0}}}{2}$,$\sin 15^o =\frac{\sqrt {\color{red}{2-\sqrt3}}}{2}$, $\sin 30^o =\frac{\sqrt {\color{blue}{1}}}{2}$ , $\sin 45^o =\frac{\sqrt {\color{blue}{2}}}{2}$ , $\sin 60^o =\frac{\sqrt {\color{blue}{3}}}{2}$,$\sin 75^o =\frac{\sqrt {\color{red}{2+\sqrt 3}}}{2}$, $\sin 90^o =\frac{\sqrt {\color{blue}{4}}}{2}$

Reformat the list :

$\sin 0^o =\frac{\sqrt {\color{green}{2-\sqrt {4}}}}{2}$,$\sin 15^o =\frac{\sqrt {\color{green}{2-\sqrt3}}}{2}$, $\sin 30^o =\frac{\sqrt {\color{green}{2-\sqrt{1}}}}{2}$

, $\sin 45^o =\frac{\sqrt {\color{green}{2-\sqrt{0}}}}{2}$ , $\sin 60^o =\frac{\sqrt {\color{green}{2+\sqrt{1}}}}{2}$,$\sin 75^o =\frac{\sqrt {\color{green}{2+\sqrt 3}}}{2}$, $\sin 90^o =\frac{\sqrt {\color{green}{2+\sqrt{4}}}}{2}$

Now, we see a nice pattern of $\frac{\sqrt {{2\color{red}{\pm}\sqrt i}}}{2}$ with i=0,1,3,4.

But wait, why is 2 missing from this i list? Seems like we have two special angles lost. With them, we can expand this to list of 9. Where are these two mysterious angles? Can you find them and prove they fit into the pattern? Or, do you find ways to expand this list even further to a list of 11,13,...and find something amazing?

Answer 1

The missing angles are:

$$\sin 22.5º = \frac{\sqrt{2 - \sqrt2}}{2}$$ $$\sin 67.5º = \frac{\sqrt{2 + \sqrt2}}{2}$$

There are other values listed here on Wikipedia, but the pattern is not obvious. For instance, $\sin 18º = \frac{\sqrt5 - 1}{4}$, but $\arcsin \frac{\sqrt5 + 1}{4}$ is not $90º - 72º = 36º$ but $54º$. In fact, $\sin 36º = \frac{\sqrt{10 - \sqrt{20}}}{4}$ and $\sin 72º = \frac{\sqrt{10 + \sqrt{20}}}{4}$.

The nice numbers for the sines of multiples of $18º$ come from the fact that $\sin (3\pi/10) = \cos (2 \pi/10)$, and solving $\sin(3x) = \sin(2x)$ using the triple-angle and double-angle formulas gives a nice quadratic in disguise after substituting $u = x^2$. The formulas involving $\sin(4x)$ and so on do not reduce into quadratics, and so if they can be reduced into radicals at all, the forms will be more complicated.