Sine inequalities

Let $f$ and $g$ be two trigonometric functions of the variable $x$. Solving the inequality $f(x) < g(x)$ is to obtain the set $S$ — the solution set or truth set — of numbers $r$ for which $f(r) < g(r)$ is a true sentence.

The story

The signals $<$ and $>$ appeared printed in 1631, in the book Artis analyticae praxis, published ten years after the death of the English mathematician Thomas Harriot, who was also an astronomer and navigator. The signs $\le$ and $\ge$ (with a dash under the sign) were introduced in 1734 by the French Pierre Bouguer, mathematician and hydrographer. Before that, inequalities were written out in full — “is greater than”, “is less than”.

Who was Thomas Harriot

Thomas Harriot (c. 1560–1621) began life as navigation teacher to Sir Walter Raleigh, the court adventurer of Elizabeth I. In 1585 he embarked on Raleigh’s expedition to Roanoke Island in present-day North Carolina; He learned the Algonquian language and wrote an account of the land and its inhabitants — the only book he published in his lifetime. Back in England, under the protection of the Earl of Northumberland, he dedicated himself to algebra, optics and astronomy: in July 1609, a few months before Galileo, he pointed a telescope at the Moon and drew what he saw. But Harriot published almost nothing, and thousands of pages of manuscripts remained forgotten for centuries. He died of nose cancer, perhaps caused by tobacco, a habit he helped bring from America.

1557
Robert Recorde invents the sign $=$: “two parallel straight lines, because nothing can be more equal”.
1631
The Harriot signals $<$ and $>$ are printed.
1734
Pierre Bouguer introduces $\le$ and $\ge$.

Sources: F. Cajori, A History of Mathematical Notations, vol. I (1928), §§ 484–485 · Harriot (MacTutor).

In everyday life

Ferris wheel. In the wheel of chapter VII, the cabin is above 29 m while $\operatorname{sen}\theta > \frac{1}{2}$, that is, during $\frac{1}{3}$ of the lap: 3 min 20 s every 10 minutes.
Solar energy. A horizontal panel only generates useful energy when the Sun is high enough; to require at least half the maximum intensity is to require $\operatorname{sen} h > \frac{1}{2}$, that is, Sun above $30^\circ$.
Crossing at low tide. The times at which a sandbank can be crossed on foot are the solutions to an inequality of the type $a + b\operatorname{sen}(ct + d) < m$.

How long does the cabin spend above the chosen height? The green band is the solution to the inequality.

Theory

Fundamental inequalities

Almost all trigonometric inequalities can be reduced to one of the following six types, where $m$ is a given real number:

$$\operatorname{sen} x > m \qquad \cos x > m \qquad \operatorname{tg} x > m$$$$\operatorname{sen} x < m \qquad \cos x < m \qquad \operatorname{tg} x < m$$

That’s why they are called fundamental inequalities, and you need to know how to solve them to solve the others. (With $\ge$ and $\le$ the reasoning is the same; only the extremes of the intervals change, which are now included.)

Resolution of sin x > m

ResumoRoad map

Mark the point $P_1$ on the sine axis such that $\overline{OP_1} = m$.
Draw through $P_1$ the line $r$ perpendicular to the axis. The images of the reals $x$ such that $\operatorname{sen} x > m$ are at the intersection of the cycle with the demiplane above $r$.
Describe the intervals to which $x$ can belong, starting from $A$ and going through the cycle counterclockwise until completing one revolution.

Exemplosin x > −√2/2

The line $r$ ($v = -\frac{\sqrt{2}}{2}$) cuts the cycle in the images of $\frac{5\pi}{4}$ and $\frac{7\pi}{4}$. The arc above it, traversed from $A$, gives two intervals:

$$2k\pi \le x < \frac{5\pi}{4} + 2k\pi \quad\text{ou}\quad \frac{7\pi}{4} + 2k\pi < x < 2\pi + 2k\pi.$$

AtençãoWhy start at A?

It would be wrong to write $\frac{7\pi}{4} + 2k\pi < x < \frac{5\pi}{4} + 2k\pi$: like $\frac{7\pi}{4} > \frac{5\pi}{4}$, there is no $x$ in that “range”. Starting from $A$, the arc that passes through $A$ is divided into two legitimate intervals. (Another way out is to use a negative determination: $-\frac{\pi}{4} + 2k\pi < x < \frac{5\pi}{4} + 2k\pi$.)

Resolution of sin x < m

ResumoRoad map

Mark the point $P_1$ on the sine axis such that $\overline{OP_1} = m$ and draw the straight line $r$ perpendicular to the axis through it.
The images of the reals $x$ such that $\operatorname{sen} x < m$ are at the intersection of the cycle with the demiplane below $r$.
Starting from $A$ and going through the cycle counterclockwise until completing one revolution, describe the appropriate intervals.

Exemplosin x < 1/2

The line $v = \frac{1}{2}$ cuts the cycle in the images of $\frac{\pi}{6}$ and $\frac{5\pi}{6}$:

$$2k\pi \le x < \frac{\pi}{6} + 2k\pi \quad\text{ou}\quad \frac{5\pi}{6} + 2k\pi < x < 2\pi + 2k\pi.$$

PropriedadeEdge cases

If $m \ge 1$: $\operatorname{sen} x > m$ has no solution and $\operatorname{sen} x < m$ is valid for all $x$ (except $x = \frac{\pi}{2} + 2k\pi$, when $m = 1$).
If $m < -1$: $\operatorname{sen} x > m$ is valid for all $x$ and $\operatorname{sen} x < m$ has no solution.

To go further: $\operatorname{sen} x < x < \operatorname{tg} x$ and the fundamental limit

One of the most important trigonometric inequalities is not solved in the cycle, but by comparing areas. For $0 < x < \frac{\pi}{2}$, in the center cycle $O$, let $A = (1, 0)$, $P$ be the image of $x$ and $T$ be the point on the tangent axis on the line $OP$.

The triangle $OAP$ (area $\frac{1}{2}\operatorname{sen}x$) is inside the sector $OAP$ (area $\frac{x}{2}$), which is inside the triangle $OAT$ (area $\frac{1}{2}\operatorname{tg}x$).

$$\frac{1}{2}\operatorname{sen} x \;<\; \frac{x}{2} \;<\; \frac{1}{2}\operatorname{tg} x \qquad\Longrightarrow\qquad \operatorname{sen} x < x < \operatorname{tg} x.$$

Dividing everything by $\operatorname{sen} x > 0$ and inverting: $\cos x < \dfrac{\operatorname{sen} x}{x} < 1$. As $x$ approaches $0$, $\cos x$ approaches $1$, and the ratio is squeezed between the two:

$$\lim_{x \to 0} \frac{\operatorname{sen} x}{x} = 1.$$

(For $x < 0$ the same applies, because $\frac{\operatorname{sen}(-x)}{-x} = \frac{\operatorname{sen} x}{x}$.) In practice: for small arcs, measured in radians, $\operatorname{sen} x \approx x$ — that was exactly what Aryabhata took advantage of when using radius 3438.

The graph of $\frac{\operatorname{sen}x}{x}$ (red) goes between $\cos x$ (dashed) and $1$ close to zero, and goes to $1$ when $x \to 0$.

ExemploTwo limits

$$\lim_{x \to 0} \frac{\operatorname{sen} 2x}{x} = \lim_{x \to 0} 2\cdot\frac{\operatorname{sen} 2x}{2x} = 2 \cdot 1 = 2, \qquad \lim_{x \to 0} \frac{\operatorname{sen} x}{3x} = \frac{1}{3}\cdot 1 = \frac{1}{3}.$$

See: the derivative of sine

The rate of change of the sine at point $x$ is the limit of $\dfrac{\operatorname{sen}(x + h) - \operatorname{sen} x}{h}$ when $h \to 0$. By the sine formula of the sum (chapter VI),

$$\frac{\operatorname{sen}(x + h) - \operatorname{sen} x}{h} = \cos x \cdot \frac{\operatorname{sen} h}{h} - \operatorname{sen} x \cdot \frac{1 - \cos h}{h}.$$

The first quotient tends to $1$. The second tends to $0$, because $\dfrac{1 - \cos h}{h} = \dfrac{2\operatorname{sen}^2\frac{h}{2}}{h} = \operatorname{sen}\frac{h}{2}\cdot\dfrac{\operatorname{sen}\frac{h}{2}}{\frac{h}{2}} \to 0 \cdot 1$. Therefore, the derivative of the sine is cosine. Likewise, the cosine is $-\operatorname{sen} x$.

Derivating twice, the sine returns with the changed sign: $(\operatorname{sen} x)'' = -\operatorname{sen} x$. This is why sines and cosines describe springs, in which the acceleration is proportional to the displacement, with opposite sign — see Euler and the spring.

The exercises in this section are numbers 271 to 285 from chapter list.

Euler's Eyes