126 Calculate $\operatorname{sen} 15^\circ$, $\cos 15^\circ$ and $\operatorname{tg} 15^\circ$. See the resolution$\operatorname{sen} 15^\circ = \operatorname{sen}\dfrac{\pi}{12} = \dfrac{\ell_{12}}{2}$. Let us calculate $\ell_{12}$: let $Q$ be a vertex of the inscribed regular hexagon and $P$ be the midpoint of the arc on one side of the hexagon, so that $PQ = \ell_{12}$. The radius $OP$ cuts the side of the hexagon at its midpoint $R$, perpendicularly. Ative o JavaScript para ver esta figura interativa. In triangle $PQR$, rectangle in $R$: $$PQ = \ell_{12} \qquad QR = \frac{\ell_6}{2} = \frac{1}{2} \qquad RP = OP - OR = 1 - \frac{\sqrt{3}}{2}$$(since $OR$ is the height of equilateral triangle beside 1). Then $$\ell_{12}^2 = \left(\frac{1}{2}\right)^2 + \left(1 - \frac{\sqrt{3}}{2}\right)^2 = \frac{1}{4} + 1 - \sqrt{3} + \frac{3}{4} = 2 - \sqrt{3}$$$$\operatorname{sen}\frac{\pi}{12} = \frac{\ell_{12}}{2} = \frac{\sqrt{2 - \sqrt{3}}}{2} \qquad \cos\frac{\pi}{12} = \sqrt{1 - \frac{2 - \sqrt{3}}{4}} = \frac{\sqrt{2 + \sqrt{3}}}{2} \qquad \operatorname{tg}\frac{\pi}{12} = \frac{\sqrt{2 - \sqrt{3}}}{\sqrt{2 + \sqrt{3}}}.$$Note: in chapter VI we will see that these values are also written $\operatorname{sen} 15^\circ = \dfrac{\sqrt{6} - \sqrt{2}}{4}$, $\cos 15^\circ = \dfrac{\sqrt{6} + \sqrt{2}}{4}$ and $\operatorname{tg} 15^\circ = 2 - \sqrt{3}$.
We will verify that the circular functions of the reals $x = \dfrac{\pi}{n}$, with $n \in \mathbb{N}$ and $n \ge 3$, can be calculated from $\ell_n$, the side of the regular polygon with $n$ sides inscribed in the cycle. The story In book I of Almagest, Ptolemy constructed his table of chords exactly like this: he started from the sides of inscribed regular polygons — triangle, square, pentagon, hexagon and decagon —, which Euclid’s geometry knew how to calculate, and obtained the other arcs through formulas for sum, difference and half of arcs, which we will see in chapter VI. The chord of an arc $2\alpha$ in a circle of radius $R$ is $2R\operatorname{sen}\alpha$: the theorem in this chapter is the modern translation of this idea.