For the tangent, the border is no longer a straight line parallel to an axis, but the straight line $OT$ that passes through the center: it cuts the cycle into two diametrically opposite points, and the solutions always come in pairs of arcs symmetric with respect to $O$ — the reflection of the tangent having a period $\pi$.

The story

The name tangent comes from the axis we use to measure it: the segment $\overline{AT}$ is on the tangent line to the cycle at $A$. The word, in Latin tangens, was introduced by the Danish mathematician Thomas Fincke in the book Geometria rotundi (1583), together with secans. Before, the Arabs already calculated this magnitude as the “shadow” of a vertical rod illuminated by the Sun, in a study linked to sundials (see chapter II).

Sources: Fincke (MacTutor) · Trigonometric functions — History (MacTutor).

In everyday life

  • Accessibility. The Brazilian standard limits ramps to 8,33% of inclination: $\operatorname{tg}\alpha \le 0{,}0833$, that is, $\alpha \le 4{,}76^\circ$.
  • Roofs. Ceramic tiles require a minimum slope, around 30% depending on the model, for water to drain: $\operatorname{tg}\alpha \ge 0{,}30$, that is, $\alpha \ge 16{,}7^\circ$.
  • Roads. “Sharp slope” signs warn of sections where the tangent of the slope exceeds a certain value, which is dangerous for heavy trucks.
The accessibility norm is a tangent inequality: $\operatorname{tg}\alpha \le 0{,}0833$.

Theory

Resolution of tg x > m

ResumoRoad map

  1. Mark the point $T$ on the tangent axis such that $\overline{AT} = m$ and draw the straight line $r = \overleftrightarrow{OT}$.
  2. The images of the real $x$ such that $\operatorname{tg} x > m$ are at the intersection of the cycle with the angle $r\hat{O}v$ (between the straight line $r$ and the axis $v$, on the side of $T$) and with its opposite at the vertex.
  3. Describe the appropriate intervals.

Exemplotg x > 1

Ative o JavaScript para ver esta figura interativa.

$$\frac{\pi}{4} + 2k\pi < x < \frac{\pi}{2} + 2k\pi \quad\text{ou}\quad \frac{5\pi}{4} + 2k\pi < x < \frac{3\pi}{2} + 2k\pi,$$

which are summarized in $\frac{\pi}{4} + k\pi < x < \frac{\pi}{2} + k\pi$.

Resolution of tg x < m

ResumoRoad map

  1. Mark $T$ with $\overline{AT} = m$ and trace $r = \overleftrightarrow{OT}$.
  2. The images of the reals $x$ such that $\operatorname{tg} x < m$ are at the intersection of the cycle with the angle $v\hat{O}r$ on the other side of $r$ (the one containing $A$) and with its opposite at the vertex.
  3. Describe the ranges that suit you, remembering that $\frac{\pi}{2} + k\pi$ is never a solution.

Exemplotg x < √3

Ative o JavaScript para ver esta figura interativa.

$$2k\pi \le x < \frac{\pi}{3} + 2k\pi \;\text{ ou }\; \frac{\pi}{2} + 2k\pi < x < \frac{4\pi}{3} + 2k\pi \;\text{ ou }\; \frac{3\pi}{2} + 2k\pi < x < 2\pi + 2k\pi,$$

that is, $-\frac{\pi}{2} + k\pi < x < \frac{\pi}{3} + k\pi$.

The exercises in this section are numbers 311 to 320 from chapter list.