The story

The names even function and odd function come from the powers: $x^2$, $x^4$, $x^6$, … do not change when we change $x$ to $-x$, and $x$, $x^3$, $x^5$, … change signs. The terms appear in a work by the young Leonhard Euler on a curve problem proposed at the time, “reciprocal trajectories”, written around 1727. Euler also realized that the cosine behaves like even powers and the sine like odd powers, which is clear in the series he used: $\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \cdots$ only has even powers, and $\operatorname{sen} x = x - \frac{x^3}{6} + \frac{x^5}{120} - \cdots$, only odd ones.

The cosine (red) and the first sums of its series, $1 - \frac{x^2}{2}$ and $1 - \frac{x^2}{2} + \frac{x^4}{24}$: only even powers, so they are all symmetrical in relation to the vertical axis, like the cosine itself.

Sources: Euler (MacTutor) · Earliest Known Uses of Some of the Words of Mathematics: E (Jeff Miller).

In everyday life

  • Noise canceling headphones. They capture the noise and emit the same wave with the signal changed: as the sine is odd, the “inverted” wave is equivalent to a half-period shift, and the two cancel each other out.
  • Compressed photos. The JPEG format uses only cosines because it extends each block of the image symmetrically, like a mirror — and symmetric functions (evens) are written only with cosines.
  • Rotate both ways. Turning a handwheel $30^\circ$ to the right or left produces the same “forward” feed ($\cos$ is even) and opposite side deviations ($\operatorname{sen}$ is odd).
Noise cancellation: the phone emits the same wave with the switched signal.

Theory

The reductions apply to all x

By reducing to the 1st quadrant, we establish notable equalities. For example, if $\frac{\pi}{2} < x < \pi$, then $\operatorname{sen} x = \operatorname{sen}(\pi - x)$ and $\cos x = -\cos(\pi - x)$. Let us now extend these equalities to todo $x$ real.

TeoremaReduction Identities

For every real $x$ the following are valid:

  1. $\operatorname{sen} x = \operatorname{sen}(\pi - x)$ e $\cos x = -\cos(\pi - x)$;
  2. $\operatorname{sen} x = -\operatorname{sen}(x - \pi)$ e $\cos x = -\cos(x - \pi)$;
  3. $\operatorname{sen} x = -\operatorname{sen}(2\pi - x)$ e $\cos x = \cos(2\pi - x)$;
  4. $\operatorname{sen} x = \cos\left(\dfrac{\pi}{2} - x\right)$ e $\cos x = \operatorname{sen}\left(\dfrac{\pi}{2} - x\right)$.
View the demo

1) For all $x \in \mathbb{R}$ we have $x = x_0 + 2k\pi$, with $0 \le x_0 < 2\pi$ and $k \in \mathbb{Z}$. Thus, $\pi - x = (\pi - x_0) - 2k\pi$, which shows that $x$ and $\pi - x$ have images in the cycle symmetric with respect to the sine axis (regardless of the quadrant of $x_0$). As a result:

$$\operatorname{sen}(\pi - x) = \operatorname{sen} x \qquad \cos(\pi - x) = -\cos x \qquad \forall x \in \mathbb{R}.$$

2), 3) and 4) are proven analogously: $x - \pi$ has a symmetric image to that of $x$ in relation to the center; $2\pi - x$, in relation to the cosine axis; and $\frac{\pi}{2} - x$, in relation to angle bisector of the 1st quadrant. For example, $\frac{\pi}{2} - x = \left(\frac{\pi}{2} - x_0\right) - 2k\pi$, and the arcs $x_0$ and $\frac{\pi}{2} - x_0$ add up to $\frac{\pi}{2}$, so their images are symmetric in angle bisector $u = v$; this symmetry exchanges abscissa and ordinate, which gives $\operatorname{sen} x = \cos\left(\frac{\pi}{2} - x\right)$ and $\cos x = \operatorname{sen}\left(\frac{\pi}{2} - x\right)$.

ExemploArcs of type π/2 + x and 3π/2 ± x

Combining the identities, we obtain formulas for other arcs (exercise 119):

$$\operatorname{sen}\left(\frac{\pi}{2} + x\right) = \operatorname{sen}\left[\pi - \left(\frac{\pi}{2} + x\right)\right] = \operatorname{sen}\left(\frac{\pi}{2} - x\right) = \cos x$$$$\cos\left(\frac{\pi}{2} + x\right) = -\cos\left[\pi - \left(\frac{\pi}{2} + x\right)\right] = -\cos\left(\frac{\pi}{2} - x\right) = -\operatorname{sen} x$$

Even functions and odd functions

DefiniçãoPair function

A function $f\colon A \to B$ is par if and only if

$$f(x) = f(-x), \quad \forall x \in A,$$

that is, by giving symmetric values ​​to the variable, we obtain the same value for the function.

Examples: $f(x) = |x|$, because $|-x| = |x|$; $f(x) = x^2$, because $(-x)^2 = x^2$; $f(x) = \dfrac{1}{x^2}$, since $\dfrac{1}{(-x)^2} = \dfrac{1}{x^2}$ for all $x \in \mathbb{R}^*$.

From the definition it follows that the graph of an even function is symmetric with respect to the $y$ axis, because $(x, y) \in f \Rightarrow (-x, y) \in f$.

Even functions: $|x|$, $x^2$ and $\frac{1}{x^2}$ — symmetric graphs with respect to the $y$ axis.

DefiniçãoUnique function

A function $f\colon A \to B$ is odd if and only if

$$f(-x) = -f(x), \quad \forall x \in A,$$

that is, by giving symmetric values ​​to the variable, we obtain symmetric values ​​for the function.

Examples: $f(x) = 2x$, because $2(-x) = -2x$; $f(x) = x^3$, because $(-x)^3 = -x^3$; $f(x) = \dfrac{1}{x}$, since $\dfrac{1}{-x} = -\dfrac{1}{x}$ for all $x \in \mathbb{R}^*$.

From the definition it follows that the graph of an odd function is symmetric with respect to the origin of the Cartesian system, since $(x, y) \in f \Rightarrow (-x, -y) \in f$.

Odd functions: $2x$, $x^3$ and $\frac{1}{x}$ — symmetrical graphs with respect to the origin.

Parity of sine and cosine

The numbers $x$ and $-x$ have, in the cycle, images symmetric with respect to the cosine axis. As a result:

TeoremaParity

$$\operatorname{sen}(-x) = -\operatorname{sen} x \qquad \cos(-x) = \cos x \qquad \forall x \in \mathbb{R}.$$

Therefore, the sine function is odd and the cosine function is even. Hence, $\operatorname{tg}(-x) = \dfrac{-\operatorname{sen} x}{\cos x} = -\operatorname{tg} x$: the tangent is also odd (as are the cotangent and cosecant), and the secant is even.

Drag $P$: the point $P'$, image of $-x$, is the symmetric of $P$ on the cosine axis.

The exercises in this section are numbers 119 to 125 from chapter list.