Trigonometric inequalities

A trigonometric inequality looks like a trigonometric equation, except that instead of the equals sign it has <, >, ≤ or ≥. Take, for example, this one:

$$\sin x>\frac{1}{2}$$

We are looking for all the x for which sine is greater than one half. Part of the work is already done. We solved the equation $\sin x=\frac{1}{2}$ in the article on trigonometric equations, and within one turn we got two solutions: $\frac{\pi }{6}$ and $\frac{5\pi }{6}$. These two numbers act as boundaries. Somewhere between them sine is greater than a half, and elsewhere it is smaller. What remains is to find out where.

On the unit circle

The easiest way to see it is on the unit circle. The sine of an angle is the y-coordinate of the point on the circle, so the inequality $\sin x>\frac{1}{2}$ is really asking: which points of the circle lie above the line $y=\frac{1}{2}$?

The blue line meets the circle at two points, which correspond to the angles $\frac{\pi }{6}$ and $\frac{5\pi }{6}$. Only the red arc between them lies above the line. As long as the arm of the angle turns anticlockwise from $\frac{\pi }{6}$ to $\frac{5\pi }{6}$, sine is greater than a half. As soon as it leaves the arc, it drops below.

The end points do not belong to the solution, which is why they are open circles in the figure. There sine is exactly $\frac{1}{2}$, and we want it to be strictly greater. Within one turn we therefore get the open interval

$$(\frac{\pi }{6},\frac{5\pi }{6}).$$

But sine is a periodic function with period 2π: after every full turn everything repeats. So we add 2kπ to both ends and get the result

$$x\in (\frac{\pi }{6}+2k\pi ,\frac{5\pi }{6}+2k\pi ),$$

where k ∈ ℤ. For k = 0 this is the arc from the figure; for k = 1 it is the same arc shifted by one turn, that is, the interval $(\frac{13\pi }{6},\frac{17\pi }{6})$. The solution is infinitely many intervals, each always 2π away from the next.

Let’s check, just to be safe. Try $x=\frac{\pi }{2}$, which lies in the interval. We have $\sin \frac{\pi }{2}=1$, and one is indeed greater than a half. On the other hand, x = π does not lie in the interval and sin π = 0, which is not greater than a half. So the result works.

An interval through zero

Now for cosine:

$$\cos x>\frac{1}{2}$$

Cosine is the x-coordinate of the point on the circle, so we are looking for the points to the right of the vertical line that crosses the horizontal axis at $\frac{1}{2}$. The equation $\cos x=\frac{1}{2}$ also has two solutions within the first turn, $\frac{\pi }{3}$ and $\frac{5\pi }{3}$ ($\frac{\pi }{3}$ is 60°, from the exact values, and $\frac{5\pi }{3}$ is the symmetrical point below the x-axis). The arc to the right of the line starts at the bottom, at $\frac{5\pi }{3}$, passes through zero and ends at the top, at $\frac{\pi }{3}$.

And here there is a strong temptation to write

$$(\frac{5\pi }{3},\frac{\pi }{3}).$$

A trap. The left end of an interval must be smaller than the right end, but $\frac{5\pi }{3}$ is greater than $\frac{\pi }{3}$. Such an interval is empty: there is no number that is both greater than $\frac{5\pi }{3}$ and smaller than $\frac{\pi }{3}$.

The fix is simple. Instead of $\frac{5\pi }{3}$ we take the same point of the circle one turn back:

$$\frac{5\pi }{3}-2\pi =-\frac{\pi }{3}.$$

The angle $-\frac{\pi }{3}$ points to the same place on the circle as $\frac{5\pi }{3}$; we just get there by turning clockwise. The left end is now smaller than the right end, and the result is

$$x\in (-\frac{\pi }{3}+2k\pi ,\frac{\pi }{3}+2k\pi ),$$

where k ∈ ℤ. This is also easy to see on the graph of cosine:

The red section around zero corresponds exactly to the interval for k = 0. The section on the right, which starts at $\frac{5\pi }{3}$, is its copy for k = 1, that is, $(\frac{5\pi }{3},\frac{7\pi }{3})$. That is also a correct way to write it, just with a different k. The wrong one is the version in which the ends are swapped.

Tangent

One last example:

$$\tan x\ge 1$$

Tangent is equal to one for $x=\frac{\pi }{4}$. On each of its branches the graph of tangent is always increasing, so for a slightly larger x tangent is greater than one. But it only increases up to $\frac{\pi }{2}$, where it shoots off to infinity, and after that it starts again from minus infinity. On one branch of the graph, then, everything from $\frac{\pi }{4}$ up to $\frac{\pi }{2}$ satisfies the inequality.

There are two mistakes people commonly make here. The first is in the period. Tangent does not repeat every 2π like sine, but already every π. Anyone who adds 2kπ out of habit loses half of the solutions, for example $x=\frac{5\pi }{4}$, where tangent is also equal to one.

The second lies in wait at the right-hand end. The left end $\frac{\pi }{4}$ does belong to the solution, because the inequality is not strict and $\tan \frac{\pi }{4}=1$ satisfies it. But the right end $\frac{\pi }{2}$ never belongs to it, even for a non-strict inequality. Tangent is not defined there, so it can be neither greater than nor equal to one. It would be a mistake to write the closed interval $[\frac{\pi }{4},\frac{\pi }{2}]$ just because the question has ≥.

The correct result is therefore

$$x\in [\frac{\pi }{4}+k\pi ,\frac{\pi }{2}+k\pi ),$$

where k ∈ ℤ. To check, we substitute $x=\frac{\pi }{3}$, which lies in the interval for k = 0. We have $\tan \frac{\pi }{3}=\sqrt{3}\approx 1.73$, which is more than one.