Two simple moves on the unit circle explain a lot about sine and cosine: reflecting across the $x$-axis, and going all the way around. Both build on reference angles.
Idea clave
$$\sin(-\theta) = -\sin\theta \qquad \cos(-\theta) = \cos\theta \qquad \tan(-\theta) = -\tan\theta$$ and, for every $\theta$, $$\sin(\theta + 2\pi) = \sin\theta, \quad \cos(\theta + 2\pi) = \cos\theta, \quad \tan(\theta + \pi) = \tan\theta.$$
Por qué funciona
Symmetry. The angle $-\theta$ turns the same amount the other way, so its point is the reflection of $(x, y)$ across the $x$-axis: $(x, -y)$. The $x$-coordinate stays, so cosine is even; the $y$-coordinate flips, so sine is odd.
Periodicity. Adding $2\pi$ is one more full turn, which lands on the same point, so every value repeats. Adding only $\pi$ sends $(x, y)$ to the opposite point $(-x, -y)$; both signs flip, and the ratio $\frac{y}{x}$ does not change. That is why tangent repeats every $\pi$.
Periodicity. Adding $2\pi$ is one more full turn, which lands on the same point, so every value repeats. Adding only $\pi$ sends $(x, y)$ to the opposite point $(-x, -y)$; both signs flip, and the ratio $\frac{y}{x}$ does not change. That is why tangent repeats every $\pi$.
Worked examples
Ejemplo resuelto: An odd function
Find $\sin\left(-\frac{\pi}{3}\right)$.
- Sine is odd: $\sin\left(-\frac{\pi}{3}\right) = -\sin\frac{\pi}{3}$.
- $\sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}$.
Respuesta: $-\frac{\sqrt{3}}{2}$
Ejemplo resuelto: Using the period
Find $\cos\frac{17\pi}{6}$.
- $\frac{17\pi}{6} = \frac{5\pi}{6} + \frac{12\pi}{6} = \frac{5\pi}{6} + 2\pi$: one full turn past $\frac{5\pi}{6}$.
- So $\cos\frac{17\pi}{6} = \cos\frac{5\pi}{6}$.
- $\frac{5\pi}{6}$ is in Quadrant II with reference angle $\frac{\pi}{6}$: $\cos\frac{5\pi}{6} = -\frac{\sqrt{3}}{2}$.
Respuesta: $-\frac{\sqrt{3}}{2}$
Ejemplo resuelto: Without knowing the angle
If $\cos\theta = 0.6$, find $\cos(-\theta)$ and $\cos(\theta + \pi)$.
- Cosine is even: $\cos(-\theta) = 0.6$.
- A half turn sends the point to the opposite side of the circle: $\cos(\theta + \pi) = -0.6$.
Respuesta: $0.6$ and $-0.6$
Error común
Writing $\cos(-\theta) = -\cos\theta$, treating every trig function like sine. Check with $\theta = 0$: $\cos 0 = 1$, and $\cos(-0)$ is also $1$, not $-1$.
A function with $f(x + p) = f(x)$ for all $x$ is periodic, and the smallest such $p > 0$ is its period: $2\pi$ for sine and cosine, $\pi$ for tangent.