How were trigonometrical functions of $\dfrac{2\pi}{17}$ calculated? I know they were calculated by Gauss, but how? Is there a method for calculating them?

Let $\omega = e^{2 \pi i/17}$.  Since $3$ is a primitive root mod $17$, i.e. a generator of the multiplicative group of nonzero integers mod $17$, write $R_j = \omega^{3^j}$ for $j = 0, 1, \ldots, 15$.  These and $1$ are the $17$'th roots of unity.
For $2^j \le i < 2^{j+1}$ let $x_i = \sum_{k \equiv i \mod 2^j} R_k$.  Then for $2^j \le i < 2^j + 2^{j-1}$, $x_i + x_{i+2^{j-1}}$ and $x_i x_{i+2^{j-1}}$ can be expressed in terms of the previous $x_k$'s, which allows $x_i$ and $x_{i+2^{j-1}}$ to be obtained by solving a quadratic equation: if $x + y = c$ and $xy = d \ne 0$, then $x$ and $y$ are the roots of $z^2 - c z + d$.
Thus:
$$x_1 = \sum_{j=0}^{15} R_j = -1$$
$x_2 = R_0 + R_2 + \ldots + R_{14}$ and $x_3 = R_1 + R_3 + \ldots R_{15}$ satisfy
$x_2 + x_3 = x_1 = -1$ and $x_2 x_3 = 4 x_1 = -4$. 
$x_4 = R_0 + R_4 + R_8 + R_{12}$ and $x_6 = R_2 + R_6 + R_{10} + R_{14}$ satisfy
$x_4 + x_6 = x_2$, $x_4 x_6 = -1$ 
$x_5 =  R_5 + R_9 + R_{13} + R_1$ and $x_7 = R_7 + R_{11} + R_{15} + R_3$ satisfy
$x_5 + x_7 = x_3$, $x_5 x_7 = -1$
$x_8 = R_0 + R_8$ and $x_{12} = R_4 + R_{12}$ satisfy
$x_8 + x_{12} = x_4$, $x_8 x_{12} = x_5$
$x_{16} = R_0 = \omega$ and $x_{24} = R_8$ satisfy $x_{16} + x_{24} = x_8$, $x_{16} x_{24} = 1$
