You know the exact values in the first quadrant. Every other point on the unit circle is a mirror image of a first-quadrant point, so those few values cover the whole circle. You only need to know which mirror, and what it does to the signs.
Idea clave
Por qué funciona
| Quadrant | Reference angle | Positive there |
|---|---|---|
| I | $\theta$ | all three |
| II | $180^\circ - \theta$ or $\pi - \theta$ | $\sin$ (and $\csc$) |
| III | $\theta - 180^\circ$ or $\theta - \pi$ | $\tan$ (and $\cot$) |
| IV | $360^\circ - \theta$ or $2\pi - \theta$ | $\cos$ (and $\sec$) |
Worked examples
Ejemplo resuelto: Quadrant II
- $120^\circ$ is in Quadrant II. Its reference angle is $180^\circ - 120^\circ = 60^\circ$.
- $\cos 60^\circ = \frac{1}{2}$, so $|\cos 120^\circ| = \frac{1}{2}$.
- In Quadrant II, $x < 0$, so cosine is negative.
Respuesta: $\cos 120^\circ = -\frac{1}{2}$
Ejemplo resuelto: Quadrant IV, in radians
- $\frac{7\pi}{4}$ is just short of $2\pi$, in Quadrant IV. Reference angle: $2\pi - \frac{7\pi}{4} = \frac{\pi}{4}$.
- $\tan\frac{\pi}{4} = 1$.
- In Quadrant IV, $y < 0 < x$, so $\tan = \frac{y}{x}$ is negative.
Respuesta: $\tan\frac{7\pi}{4} = -1$
Ejemplo resuelto: A negative angle
- Add $2\pi$: $-\frac{5\pi}{6}$ is coterminal with $\frac{7\pi}{6}$, in Quadrant III.
- Reference angle: $\frac{7\pi}{6} - \pi = \frac{\pi}{6}$, and $\sin\frac{\pi}{6} = \frac{1}{2}$.
- In Quadrant III, $y < 0$, so sine is negative.
Respuesta: $-\frac{1}{2}$
Error común