Let’s deduce formulas to calculate the trigonometric functions of $x$, with $x$ outside the 1st quadrant, relating $x$ with some number in the 1st quadrant. The goal is to know $\operatorname{sen} x$, $\cos x$ and $\operatorname{tg} x$ from a table that gives the circular functions of the reals between $0$ and $\frac{\pi}{2}$.

Drag $P$ across all quadrants. The point $P'$ of the 1st quadrant is the symmetric point of $P$; Compare the sine and cosine values ​​on the panel.

The story

The trigonometric tables always covered only a portion of the circle: Aryabhata’s went from $0^\circ$ to $90^\circ$, and for the other arcs the astronomer used the symmetries of the circle. What was missing was a rule for signs. While sine and cosine were segment lengths, they were always positive, and each quadrant required separate reasoning. The change came with the acceptance of negative numbers as something that has geometric meaning. In 1629, the Frenchman Albert Girard wrote that a negative solution, in geometry, “indicates a setback” — whoever walks $-3$ steps walks 3 steps backwards. It is exactly the idea that gives a signal to the coordinates of a point in the cycle and, with that, to the circular functions in each quadrant. Euler, in the following century, organized the rules of signs into the form we use.

  1. 628
    The Indian Brahmagupta gives rules for operating with “debts” (negatives) and “fortunes” (positives).
  2. 1545
    Cardano still calls negative roots numeri ficti, fictitious numbers.
  3. 1629
    Albert Girard: in geometry, the negative “indicates a setback”.
  4. 1637
    Descartes publishes Geometry, which gives rise to coordinates.
  5. 1748
    Euler organizes the signs of circular functions in the four quadrants.

Sources: Girard (MacTutor) · Trigonometric functions (MacTutor).

In everyday life

  • Calculators. To calculate $\operatorname{sen} 1000^\circ$, the machine first takes the turns ($1000^\circ - 720^\circ = 280^\circ$) and then reduces it to the 1st quadrant: $\operatorname{sen} 280^\circ = -\operatorname{sen} 80^\circ$.
  • Navigation. Walking 10 km heading $225^\circ$ (southwest, measured from east) moves $10\cos 225^\circ \approx -7{,}07$ km east and $10\operatorname{sen} 225^\circ \approx -7{,}07$ km north: the signs say “west” and “south”.
  • Games. A character facing $150^\circ$ walks in the direction $(\cos 150^\circ, \operatorname{sen} 150^\circ) = \left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$: to the left and a little up.
In a course in any quadrant, the cosine and sine signs tell whether the boat is heading east or west, north or south.

Theory

From the 2nd to the 1st quadrant

Given $x$ with $\frac{\pi}{2} < x < \pi$, let $P$ be its image in the cycle and $P'$ be the point of the cycle symmetric of $P$ with respect to the sine axis. Counterclockwise, $\overset{\frown}{AP} + \overset{\frown}{PA'} = \pi$; as $\overset{\frown}{AP'} = \overset{\frown}{PA'}$, comes $\overset{\frown}{AP} + \overset{\frown}{AP'} = \pi$, therefore $\overset{\frown}{AP'} = \pi - x$.

As $P$ and $P'$ have the same ordinate and opposite abscissa:

Propriedade2nd quadrant

$$\operatorname{sen} x = \operatorname{sen}(\pi - x) \qquad \cos x = -\cos(\pi - x)$$

and, through fundamental relationships,

$$\operatorname{tg} x = \frac{\operatorname{sen} x}{\cos x} = \frac{\operatorname{sen}(\pi - x)}{-\cos(\pi - x)} = -\operatorname{tg}(\pi - x)$$$$\operatorname{cotg} x = -\operatorname{cotg}(\pi - x) \qquad \sec x = -\sec(\pi - x) \qquad \operatorname{cossec} x = \operatorname{cossec}(\pi - x).$$

ExemploExamples

$$\operatorname{sen} 115^\circ = \operatorname{sen}(180^\circ - 115^\circ) = \operatorname{sen} 65^\circ \qquad \cos 130^\circ = -\cos(180^\circ - 130^\circ) = -\cos 50^\circ$$$$\operatorname{tg}\frac{2\pi}{3} = -\operatorname{tg}\left(\pi - \frac{2\pi}{3}\right) = -\operatorname{tg}\frac{\pi}{3} \qquad \operatorname{cotg}\frac{4\pi}{5} = -\operatorname{cotg}\left(\pi - \frac{4\pi}{5}\right) = -\operatorname{cotg}\frac{\pi}{5}$$

From the 3rd to the 1st quadrant

Given $x$ with $\pi < x < \frac{3\pi}{2}$, let $P$ be its image and $P'$ be the point of the symmetric cycle of $P$ with respect to the center. Counterclockwise, $\overset{\frown}{AP} - \overset{\frown}{AP'} = \pi$, therefore $\overset{\frown}{AP'} = x - \pi$.

$P$ and $P'$ have opposite abscissa and ordinate:

Propriedade3rd quadrant

$$\operatorname{sen} x = -\operatorname{sen}(x - \pi) \qquad \cos x = -\cos(x - \pi)$$$$\operatorname{tg} x = \frac{-\operatorname{sen}(x - \pi)}{-\cos(x - \pi)} = \operatorname{tg}(x - \pi)$$$$\operatorname{cotg} x = \operatorname{cotg}(x - \pi) \qquad \sec x = -\sec(x - \pi) \qquad \operatorname{cossec} x = -\operatorname{cossec}(x - \pi).$$

ExemploExamples

$$\operatorname{sen} 210^\circ = -\operatorname{sen}(210^\circ - 180^\circ) = -\operatorname{sen} 30^\circ \qquad \cos 225^\circ = -\cos(225^\circ - 180^\circ) = -\cos 45^\circ$$$$\operatorname{tg}\frac{4\pi}{3} = \operatorname{tg}\left(\frac{4\pi}{3} - \pi\right) = \operatorname{tg}\frac{\pi}{3} \qquad \sec\frac{7\pi}{6} = -\sec\left(\frac{7\pi}{6} - \pi\right) = -\sec\frac{\pi}{6}$$

From the 4th to the 1st quadrant

Given $x$ with $\frac{3\pi}{2} < x < 2\pi$, let $P$ be its image and $P'$ be the point of the symmetric cycle of $P$ with respect to the cosine axis. Counterclockwise, $\overset{\frown}{AP} + \overset{\frown}{PA} = 2\pi$; as $\overset{\frown}{AP'} = \overset{\frown}{PA}$, comes $\overset{\frown}{AP} + \overset{\frown}{AP'} = 2\pi$, therefore $\overset{\frown}{AP'} = 2\pi - x$.

$P$ and $P'$ have the same abscissa and opposite ordinates:

Propriedade4th quadrant

$$\operatorname{sen} x = -\operatorname{sen}(2\pi - x) \qquad \cos x = \cos(2\pi - x)$$$$\operatorname{tg} x = \frac{-\operatorname{sen}(2\pi - x)}{\cos(2\pi - x)} = -\operatorname{tg}(2\pi - x)$$$$\operatorname{cotg} x = -\operatorname{cotg}(2\pi - x) \qquad \sec x = \sec(2\pi - x) \qquad \operatorname{cossec} x = -\operatorname{cossec}(2\pi - x).$$

ExemploExamples

$$\operatorname{sen} 280^\circ = -\operatorname{sen}(360^\circ - 280^\circ) = -\operatorname{sen} 80^\circ \qquad \cos 340^\circ = \cos(360^\circ - 340^\circ) = \cos 20^\circ$$$$\operatorname{tg}\frac{11\pi}{6} = -\operatorname{tg}\left(2\pi - \frac{11\pi}{6}\right) = -\operatorname{tg}\frac{\pi}{6} \qquad \operatorname{cossec}\frac{5\pi}{3} = -\operatorname{cossec}\left(2\pi - \frac{5\pi}{3}\right) = -\operatorname{cossec}\frac{\pi}{3}$$

ResumoRule of thumb

  1. If the arc has more than one turn, remove the entire turns (add or subtract multiples of $360^\circ$ or $2\pi$).
  2. Find the corresponding arc of the 1st quadrant: $180^\circ - x$ (2nd), $x - 180^\circ$ (3rd) or $360^\circ - x$ (4th).
  3. The absolute value of the function is the same; the sign is that of the function in the quadrant of $x$.

The exercise in this section is number 117 from the chapter list.