Okay so I wanted to calculate the exact irrational values of $\sin(\frac{π}{9})$ and $\cos(\frac{π}{9})$, but I keep running into a bit of a problem.
Let's take the sine as an example:
$\because \sin(3x) = 3\sin(x) - 4\sin^{3}(x)$
$\therefore \sin(x) = 3\sin(\frac{x}{3}) - 4\sin^{3}(\frac{x}{3})$
$\therefore -4\sin^{3}(\frac{x}{3}) + 3\sin(\frac{x}{3}) - \sin(x) = 0$
$\therefore \sin^{3}(\frac{x}{3}) - \frac{3\sin(\frac{x}{3})}{4} + \frac{\sin(x)}{4} = 0$
Plugging in $\frac{π}{3}$ for $x$, we get:
$\sin^{3}(\frac{π}{9}) - \frac{3\sin(\frac{π}{9})}{4} + \frac{\sin(\frac{π}{3})}{4} =$
$\sin^{3}(\frac{π}{9}) - \frac{3\sin(\frac{π}{9})}{4} + \frac{\sqrt{3}}{8} = 0$
Now we have a depressed cubic equation; We can solve depressed cubic equations with the formula:
$$x^{3} + px + q$$
$$x = \sqrt[3]{\frac{-q}{2} + \sqrt{\frac{q^{2}}{4} + \frac{p^{3}}{27}}} + \sqrt[3]{\frac{-q}{2} - \sqrt{\frac{q^{2}}{4} + \frac{p^{3}}{27}}}$$
Plugging in our values, we get:
$\sin(\frac{π}{9}) = \sqrt[3]
{\frac{-\sqrt{3}}{16} + \sqrt{\frac{3}{256} - \frac{1}{64}}} + \sqrt[3]{\frac{-\sqrt{3}}{16} - \sqrt{\frac{3}{256} - \frac{1}{64}}}$
Which ultimately simplifies into
$\frac{\sqrt[3]{-\sqrt{3} + i} + \sqrt[3]{-\sqrt{3} - i}}{2\sqrt[3]{2}}$
And that's where the problem lies.
Of course, I'll need to verify that this is between 0 and 1, and right now, I'm not even sure it's a real number. So I'll need to get rid of the $i$s if I can; I cannot do that while they're under cube root, so I'll have to solve the roots first.
In order to solve the roots, I'll have to convert my complex numbers into the polar form first, then get the cube root of the distance and the third of the angle, and that's where my predicament becomes more apparent.
$-\sqrt{3} + i$ and $-\sqrt{3} - i$ are $2\angle\frac{5π}{6}$ and $2\angle\frac{7π}{6}$ respectively. My problem is that those angles are multiples of $\frac{π}{6}$, which means that when I take the cube root, they'll be multiples of $\frac{π}{18}$, which I don't know the sine and cosine of, and for which I need to derive them for $\frac{π}{9}$ in the first place. Same thing happens with the cosine (but the angles there are multiple of $\frac{π}{3}$ instead).
Now, there's an obvious solution to this, figuring out a way to take the cube root of a complex number algebraically (without converting into the polar form), but I've looked it up and found nothing. There might also be a way different way entirely of deriving this, but there are none I know of. That's why I'm asking this question.
