Trigonometric inequalities
An equation such as sin x = 0.6 picks out single angles. Where does the sine stay above 0.6, and how do you write down every one of those angles at once?
Where you are 4 in Solving trigonometric equations
Why this exists
Solving a trig equation finds the angles where a function hits a value exactly. Many questions ask something wider: for which angles does the quantity stay above a level, or below it? How long a Ferris wheel rider stays above a given height, or when a tide leaves a harbor deep enough, are questions of this kind, and those lessons answer them in context. This lesson sets out the method behind them.
The answer to such a question is not a list of angles but whole stretches of them, written in Interval and set-builder notation. The same task for polynomials is done in polynomial and rational inequalities: find where the expression equals zero or is undefined, cut the line there, and test each piece. Trig inequalities work the same way, with one new feature: the cut points repeat with the period, so the stretches repeat too.
Definition
A trigonometric inequality is an inequality whose unknown appears inside a trig function, such as 2\cos x + 1 < 0 or \tan x \ge 1. Its solution set is the set of every real x that makes it true, either inside a stated interval such as [0, 2\pi) or on the whole line. Angles are in radians unless degrees are named.
The equation made by putting = in place of the inequality sign is the matching equation. Its solutions are the crossings, the angles where the graph meets the level.
A strict inequality (< or >) leaves the crossings out, so a stretch ends there with an open end, a round bracket. A non-strict one (\le or \ge) keeps them, a closed end, a square bracket. A point where the function is undefined, such as an asymptote of tangent, is never a solution, so an end there is always open.
Intuition
The graph of y = \cos x is one unbroken wave. To pass from above the line \displaystyle y = -\frac{1}{2} to below it, the wave has to meet the line, so it can only change sides at a crossing. Between two neighboring crossings the wave stays on one side the whole way, and one test point tells which side.
The unit circle gives the same answer in a second picture. The angle x ends at the point (\cos x, \sin x), so \cos x \ge c asks which terminal points lie on or to the right of the vertical line at horizontal coordinate c, and \sin x > c asks which lie above the horizontal line at height c. Each answer is an arc, cut off by the line at the crossings.
Tangent is different in one way. Its graph is broken at the asymptotes \displaystyle x = \frac{\pi}{2} + \pi k: just left of \displaystyle \frac{\pi}{2} it is enormous and positive, just right of it enormous and negative. It changes sides there without crossing anything, so for tangent the asymptotes are cut points too.
Results
Cut points. For \sin x or \cos x compared with a number c, the inequality can change from true to false only at a solution of the matching equation. For \tan x it can also change at an asymptote, \displaystyle x = \frac{\pi}{2} + \pi k.
One test per stretch. Between two neighboring cut points the inequality is true at every point or at none.
Repetition. Every solution is a solution in one period plus an integer number of periods: 2\pi k for sine and cosine, \pi k for tangent, with k any integer.
For -1 < c < 1, with \alpha = \arcsin c and \beta = \arccos c, and for every c, with \gamma = \arctan c, the basic solution sets are, with k any integer:
- \sin x > c on the stretches (\alpha + 2\pi k, \ \pi - \alpha + 2\pi k);
- \cos x > c on the stretches (-\beta + 2\pi k, \ \beta + 2\pi k);
- \tan x > c on the stretches \displaystyle \left(\gamma + \pi k, \ \frac{\pi}{2} + \pi k\right).
The opposite inequality, <, holds on the rest of each period except the crossings (and, for tangent, the asymptotes); \ge and \le add the crossings.
Why it is true
As the angle grows, the terminal point slides around the unit circle without jumping, so its coordinates \cos x and \sin x change without jumps: their graphs are unbroken. Suppose \sin a < c and \sin b > c. An unbroken graph that starts below the line y = c and ends above it must meet the line somewhere between a and b, so there is a crossing between them. Read the other way: if there is no crossing between a and b, the inequality has the same truth value at both. That is the first two results for sine, and cosine is the same.
Tangent is \displaystyle \frac{\sin x}{\cos x}, a quotient of unbroken functions, and it is unbroken wherever its bottom \cos x is not zero. So the same argument works on each stretch between two asymptotes, but not across one.
Finally, if f has period p, then f(x + p) = f(x), so x satisfies the inequality exactly when x + p does. The solutions in one period, shifted by every multiple of p, are all the solutions.
Worked examples
A cosine below a level
Pacha needs every x in [0, 2\pi) with 2\cos x + 1 < 0.
SolutionSubtract 1 and divide by 2, a positive number, so the inequality keeps its direction: \displaystyle \cos x < -\frac{1}{2}. The matching equation \displaystyle \cos x = -\frac{1}{2} has reference angle \displaystyle \frac{\pi}{3}, and cosine is negative in Quadrants II and III, so the crossings are \displaystyle \frac{2\pi}{3} and \displaystyle \frac{4\pi}{3}. They cut [0, 2\pi) into three stretches, and one test point decides each.
| stretch | test point | \cos x | \displaystyle \cos x < -\frac{1}{2}? |
|---|---|---|---|
| \displaystyle \left[0, \frac{2\pi}{3}\right) | 0 | 1 | no |
| \displaystyle \left(\frac{2\pi}{3}, \frac{4\pi}{3}\right) | \pi | -1 | yes |
| \displaystyle \left(\frac{4\pi}{3}, 2\pi\right) | \displaystyle \frac{5\pi}{3} | \displaystyle \frac{1}{2} | no |
At the crossings \cos x equals \displaystyle -\frac{1}{2}, which is not less than \displaystyle -\frac{1}{2}, so the ends are open. The solution set is \displaystyle \left(\frac{2\pi}{3}, \frac{4\pi}{3}\right), the thick stretch in Figure 1. Over the whole line the same stretch comes back every turn:
An arc that wraps through zero
Solve \displaystyle \sqrt{2}\cos x \ge 1 in [0, 2\pi), then find all solutions.
SolutionDivide by \displaystyle \sqrt{2}: \displaystyle \cos x \ge \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}. On the unit circle, \cos x is the horizontal coordinate of the terminal point, so the solutions are the angles that end on or to the right of the vertical line at \displaystyle \frac{\sqrt{2}}{2}. That line meets the circle at the angles \displaystyle \frac{\pi}{4} and \displaystyle -\frac{\pi}{4}, and the part of the circle to its right is the arc from \displaystyle -\frac{\pi}{4} up through 0 to \displaystyle \frac{\pi}{4} (Figure 2). The inequality is non-strict, so both ends are closed.
Measured from 0 to 2\pi, that one arc is split by the start of the turn: the angles from 0 to \displaystyle \frac{\pi}{4}, and the angles from \displaystyle 2\pi - \frac{\pi}{4} = \frac{7\pi}{4} up to 2\pi. The end 0 is closed because 0 is in the interval and \cos 0 = 1 works; the end 2\pi is open only because the interval [0, 2\pi) stops short of it.
For all solutions the split disappears. Each turn holds one unbroken stretch, centered at a multiple of 2\pi:
Tangent, with an asymptote as an end
Onni wants all solutions of \tan x \ge 1, and then those in [0, 2\pi).
SolutionTangent has period \pi, so one period is enough: the branch on \displaystyle \left(-\frac{\pi}{2}, \frac{\pi}{2}\right), between two asymptotes. On it, \tan x = 1 only at \displaystyle \frac{\pi}{4}. The cut points are therefore \displaystyle -\frac{\pi}{2} (asymptote), \displaystyle \frac{\pi}{4} (crossing) and \displaystyle \frac{\pi}{2} (asymptote). Test each stretch: \tan 0 = 0, below 1, so \displaystyle \left(-\frac{\pi}{2}, \frac{\pi}{4}\right) fails; \displaystyle \tan\frac{\pi}{3} = \sqrt{3} \approx 1.73, so \displaystyle \left(\frac{\pi}{4}, \frac{\pi}{2}\right) works.
The crossing \displaystyle \frac{\pi}{4} is included, since \displaystyle \tan\frac{\pi}{4} = 1 and the sign is \ge. The asymptote \displaystyle \frac{\pi}{2} is not, since \displaystyle \tan\frac{\pi}{2} does not exist. In one period the solutions are \displaystyle \left[\frac{\pi}{4}, \frac{\pi}{2}\right), and all solutions are
In [0, 2\pi), k = 0 and k = 1 give \displaystyle \left[\frac{\pi}{4}, \frac{\pi}{2}\right) \cup \left[\frac{5\pi}{4}, \frac{3\pi}{2}\right); k = -1 ends before 0 and k = 2 starts after 2\pi.
Crossings that are not exact
Solve \sin x > 0.6 in [0, 2\pi) and in general, to four decimal places.
SolutionThe matching equation \sin x = 0.6 has no exact solution, so the inverse supplies one: \alpha = \arcsin 0.6 \approx 0.6435, in Quadrant I. The other crossing in the turn is its mirror across the y-axis, \pi - \alpha \approx 2.4981. Test the stretches: at \displaystyle \frac{\pi}{2}, between the crossings, \sin x = 1 > 0.6; at 0, before them, \sin 0 = 0; at \displaystyle \frac{3\pi}{2}, after them, \sin x = -1. Only the middle stretch works, and the sign is strict, so its ends are open.
In [0, 2\pi) the solution set is (0.6435, 2.4981). Over the whole line,
Exactly, the ends are \arcsin 0.6 + 2\pi k and \pi - \arcsin 0.6 + 2\pi k; the decimals only round them.
A multiple angle
Solve \displaystyle 2\sin 2x > \sqrt{2} in [0, 2\pi), and give all solutions.
SolutionDivide by 2: \displaystyle \sin 2x > \frac{\sqrt{2}}{2}. As in equations with a multiple angle, name the inside u = 2x. As x runs over [0, 2\pi), u runs over [0, 4\pi), two full turns. In one turn \displaystyle \sin u = \frac{\sqrt{2}}{2} at \displaystyle \frac{\pi}{4} and \displaystyle \frac{3\pi}{4}, and sine is above \displaystyle \frac{\sqrt{2}}{2} between them (at \displaystyle \frac{\pi}{2} it is 1). The second turn adds 2\pi:
Dividing by 2 keeps the directions, so x is in \displaystyle \left(\frac{\pi}{8}, \frac{3\pi}{8}\right) \cup \left(\frac{9\pi}{8}, \frac{11\pi}{8}\right). The two stretches are \pi apart, because \sin 2x has period \pi, so all solutions are \displaystyle \frac{\pi}{8} + \pi k < x < \frac{3\pi}{8} + \pi k. Check at \displaystyle x = \frac{\pi}{4}: \displaystyle 2\sin\frac{\pi}{2} = 2 > \sqrt{2}.
A product and a sign chart
Chaska needs every x in [0, 2\pi) with \cos x\,(2\sin x - 1) < 0.
SolutionNever divide by \cos x here: its sign changes along the interval, and it is zero at two points. A product is negative exactly when its two factors have opposite signs, so this is the sign chart of the polynomial lesson. Each factor changes sign only where it is zero. \cos x = 0 at \displaystyle \frac{\pi}{2} and \displaystyle \frac{3\pi}{2}; 2\sin x - 1 = 0 where \displaystyle \sin x = \frac{1}{2}, at \displaystyle \frac{\pi}{6} and \displaystyle \frac{5\pi}{6}. Cosine is positive in Quadrants I and IV, and \displaystyle \sin x > \frac{1}{2} exactly between \displaystyle \frac{\pi}{6} and \displaystyle \frac{5\pi}{6}.
| stretch | \cos x | 2\sin x - 1 | product |
|---|---|---|---|
| \displaystyle \left[0, \frac{\pi}{6}\right) | + | - | - |
| \displaystyle \left(\frac{\pi}{6}, \frac{\pi}{2}\right) | + | + | + |
| \displaystyle \left(\frac{\pi}{2}, \frac{5\pi}{6}\right) | - | + | - |
| \displaystyle \left(\frac{5\pi}{6}, \frac{3\pi}{2}\right) | - | - | + |
| \displaystyle \left(\frac{3\pi}{2}, 2\pi\right) | + | - | - |
At the four cut points the product is 0, not negative, so they are left out. At x = 0 it is 1 \cdot (-1) = -1, so 0 is in. The solution set is
Boundaries
- Moving terms. Adding a number to both sides, or multiplying by a positive number, keeps the direction of the inequality; multiplying or dividing by a negative number reverses it, so -2\sin x > 1 becomes \displaystyle \sin x < -\frac{1}{2}.
- No dividing by a trig function. \cos x or \sin x is positive in some places, negative in others and zero at some points, so dividing by it would reverse the inequality on part of the line only. Bring everything to one side, factor, and use a sign chart.
- Levels outside the range. Since -1 \le \sin x \le 1, the inequality \sin x > c has no solution when c \ge 1 and holds for every x when c < -1. At the edge, \sin x < 1 holds for every x except \displaystyle \frac{\pi}{2} + 2\pi k, and \sin x \ge 1 only at those points. Cosine behaves the same; tangent takes every value, so it has no such edge.
- Asymptotes. For tangent the cut points include the asymptotes, and an end at an asymptote is always open, even for \le or \ge.
- Wrapping. A stretch that passes through a multiple of 2\pi is one stretch on the whole line but falls into two pieces inside [0, 2\pi), as in Example 2. The brackets at 0 and 2\pi come from the given interval, not from the inequality.
- Multiple angles. With 2x inside, the period changes, and one turn of x holds two turns of the inside.
Where this goes next
The Ferris wheel and tide lessons use exactly this method on a fitted sinusoid: find the times when the height equals the level, then keep the stretch between them, as in the Ferris wheel model and tide models. When the matching equation is harder, such as 2\sin^2 x - \sin x - 1 = 0, factoring and identities find the crossings, and the sign chart of Example 6 finishes the inequality.
Exercises
One turn, exact crossings
Give each solution set in [0, 2\pi).
- \displaystyle \sin x < \frac{1}{2}
- \displaystyle 2\cos x - \sqrt{3} > 0
- \displaystyle \sqrt{2}\sin x + 1 \le 0
- \cos x \ge 0
- \displaystyle 4\sin x \ge 2\sin x + \sqrt{3}
- \displaystyle \tan x < \sqrt{3}
- \tan x + 1 > 0
All solutions
Give every solution, with k any integer.
- \sin x \ge 0
- \displaystyle 2\cos x < -\sqrt{2}
- \displaystyle \cos x > \frac{1}{2}
- \displaystyle \sqrt{3}\tan x \le -1
- \sin x < 1
- \cos x \le -1
Crossings from the inverse functions
Give radians to four decimal places and degrees to two.
- Solve \cos x < 0.4 in [0, 2\pi).
- Solve 5\sin x + 2 \ge 0 in [0, 2\pi).
- Find all solutions of \tan x > 3.
- Find every \theta with 0^\circ \le \theta < 360^\circ such that 3\cos\theta \ge 1.
Multiple angles and products
- Solve 2\cos 2x + 1 \le 0 in [0, \pi).
- Solve \sin x\,(2\cos x - 1) > 0 in [0, 2\pi).
- Solve \displaystyle \sin^2 x < \frac{1}{4} in [0, 2\pi).
Reasoning
- Explain why \sin x - c can change sign only where \sin x = c, and why the same reasoning does not cover \tan x - c on the whole line. Which extra points must be added for tangent?
- Onni solves \cos x > 0 and writes the answer in [0, 2\pi) as \displaystyle \left(\frac{3\pi}{2}, \frac{\pi}{2}\right). What is wrong with this, and what should he write, in [0, 2\pi) and for all solutions?
- Chaska finds that the only solutions of \tan x = 1 in [0, 2\pi) are \displaystyle \frac{\pi}{4} and \displaystyle \frac{5\pi}{4}. She tests x = \pi, finds \tan\pi = 0 < 1, and concludes that \tan x < 1 on all of \displaystyle \left(\frac{\pi}{4}, \frac{5\pi}{4}\right). Find her error and give the correct solution set of \tan x < 1 in that interval.
- For which numbers c does \sin x > c have no solution, and for which does it hold for every x? Explain with the unit circle.
Answers
- \displaystyle \left[0, \frac{\pi}{6}\right) \cup \left(\frac{5\pi}{6}, 2\pi\right). The crossings are \displaystyle \frac{\pi}{6} and \displaystyle \frac{5\pi}{6}; sine is above \displaystyle \frac{1}{2} between them and below outside.
- \displaystyle \left[0, \frac{\pi}{6}\right) \cup \left(\frac{11\pi}{6}, 2\pi\right). \displaystyle \cos x > \frac{\sqrt{3}}{2}: the arc right of the line at \displaystyle \frac{\sqrt{3}}{2} runs from \displaystyle -\frac{\pi}{6} to \displaystyle \frac{\pi}{6} and wraps through 0.
- \displaystyle \left[\frac{5\pi}{4}, \frac{7\pi}{4}\right]. \displaystyle \sin x \le -\frac{\sqrt{2}}{2}, the arc below the line at height \displaystyle -\frac{\sqrt{2}}{2}, ends included.
- \displaystyle \left[0, \frac{\pi}{2}\right] \cup \left[\frac{3\pi}{2}, 2\pi\right). Cosine is 0 at \displaystyle \frac{\pi}{2} and \displaystyle \frac{3\pi}{2} and positive in Quadrants I and IV.
- \displaystyle \left[\frac{\pi}{3}, \frac{2\pi}{3}\right]. Subtract 2\sin x: \displaystyle 2\sin x \ge \sqrt{3}, so \displaystyle \sin x \ge \frac{\sqrt{3}}{2}, true from \displaystyle \frac{\pi}{3} to \displaystyle \frac{2\pi}{3}.
- \displaystyle \left[0, \frac{\pi}{3}\right) \cup \left(\frac{\pi}{2}, \frac{4\pi}{3}\right) \cup \left(\frac{3\pi}{2}, 2\pi\right). Cut points: crossings \displaystyle \frac{\pi}{3}, \displaystyle \frac{4\pi}{3} and asymptotes \displaystyle \frac{\pi}{2}, \displaystyle \frac{3\pi}{2}. Each branch is below \displaystyle \sqrt{3} up to its crossing; on \displaystyle \left(\frac{3\pi}{2}, 2\pi\right) tangent is negative.
- \displaystyle \left[0, \frac{\pi}{2}\right) \cup \left(\frac{3\pi}{4}, \frac{3\pi}{2}\right) \cup \left(\frac{7\pi}{4}, 2\pi\right). \tan x > -1; the crossings are \displaystyle \frac{3\pi}{4} and \displaystyle \frac{7\pi}{4}, and each branch is above -1 from its crossing to its right asymptote.
- 2\pi k \le x \le \pi + 2\pi k. Sine is 0 at 0 and \pi and positive in between, the upper half of the circle.
- \displaystyle \frac{3\pi}{4} + 2\pi k < x < \frac{5\pi}{4} + 2\pi k. \displaystyle \cos x < -\frac{\sqrt{2}}{2}, the arc left of the line at \displaystyle -\frac{\sqrt{2}}{2}.
- \displaystyle -\frac{\pi}{3} + 2\pi k < x < \frac{\pi}{3} + 2\pi k. The arc right of the line at \displaystyle \frac{1}{2}, centered at angle 0.
- \displaystyle -\frac{\pi}{2} + \pi k < x \le -\frac{\pi}{6} + \pi k. \displaystyle \tan x \le -\frac{\sqrt{3}}{3}; on the branch \displaystyle \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) the crossing is \displaystyle -\frac{\pi}{6}, and tangent is smaller to its left, down to the asymptote.
- Every x except \displaystyle \frac{\pi}{2} + 2\pi k, that is, \displaystyle \frac{\pi}{2} + 2\pi k < x < \frac{5\pi}{2} + 2\pi k. Sine reaches 1 only at the top of the circle.
- x = \pi + 2\pi k. Cosine is never below -1, so only the equality can hold, at (-1, 0).
- (1.1593, 5.1239). \arccos 0.4 \approx 1.1593 and 2\pi - 1.1593 \approx 5.1239; cosine is below 0.4 between them (at \pi it is -1).
- [0, 3.5531] \cup [5.8717, 2\pi). \sin x \ge -0.4; \arcsin(-0.4) \approx -0.4115, so the crossings are \pi + 0.4115 \approx 3.5531 and 2\pi - 0.4115 \approx 5.8717, and sine is below -0.4 only between them.
- \displaystyle 1.2490 + \pi k < x < \frac{\pi}{2} + \pi k. \arctan 3 \approx 1.2490, and each branch is above 3 from its crossing to its right asymptote.
- [0^\circ, 70.53^\circ] \cup [289.47^\circ, 360^\circ). \displaystyle \cos\theta \ge \frac{1}{3}; \displaystyle \arccos\frac{1}{3} \approx 70.53^\circ and 360^\circ - 70.53^\circ = 289.47^\circ.
- \displaystyle \left[\frac{\pi}{3}, \frac{2\pi}{3}\right]. With u = 2x in [0, 2\pi), \displaystyle \cos u \le -\frac{1}{2} for \displaystyle \frac{2\pi}{3} \le u \le \frac{4\pi}{3}; divide by 2.
- \displaystyle \left(0, \frac{\pi}{3}\right) \cup \left(\pi, \frac{5\pi}{3}\right). Cut points 0, \displaystyle \frac{\pi}{3}, \pi, \displaystyle \frac{5\pi}{3}. Both factors are positive on \displaystyle \left(0, \frac{\pi}{3}\right) and both negative on \displaystyle \left(\pi, \frac{5\pi}{3}\right); at 0 the product is 0, so 0 is out.
- \displaystyle \left[0, \frac{\pi}{6}\right) \cup \left(\frac{5\pi}{6}, \frac{7\pi}{6}\right) \cup \left(\frac{11\pi}{6}, 2\pi\right). \displaystyle \sin^2 x < \frac{1}{4} means \displaystyle -\frac{1}{2} < \sin x < \frac{1}{2}, the angles ending strictly between the lines at heights \displaystyle -\frac{1}{2} and \displaystyle \frac{1}{2}.
- The graph of y = \sin x is unbroken, because the terminal point moves around the circle without jumping. If \sin x - c were negative at a and positive at b, the graph would go from below the line y = c to above it, and an unbroken graph cannot do that without meeting the line, so \sin x = c somewhere between. Tangent’s graph is broken at \displaystyle x = \frac{\pi}{2} + \pi k, where it jumps from very large positive values to very large negative ones without passing through c. So the asymptotes \displaystyle \frac{\pi}{2} + \pi k must be added to the cut points.
- An interval is written from its smaller end to its larger, so \displaystyle \left(\frac{3\pi}{2}, \frac{\pi}{2}\right) names nothing. The stretch he means wraps through 0. In [0, 2\pi) it is \displaystyle \left[0, \frac{\pi}{2}\right) \cup \left(\frac{3\pi}{2}, 2\pi\right); for all solutions it is the single stretch \displaystyle -\frac{\pi}{2} + 2\pi k < x < \frac{\pi}{2} + 2\pi k.
- She tested only one point, but \displaystyle \left(\frac{\pi}{4}, \frac{5\pi}{4}\right) contains the asymptote \displaystyle \frac{\pi}{2}, which is a cut point. On \displaystyle \left(\frac{\pi}{4}, \frac{\pi}{2}\right) tangent is above 1 (for example \displaystyle \tan\frac{\pi}{3} = \sqrt{3}), and only after the asymptote does it come in from below. The solution set of \tan x < 1 in \displaystyle \left(\frac{\pi}{4}, \frac{5\pi}{4}\right) is \displaystyle \left(\frac{\pi}{2}, \frac{5\pi}{4}\right).
- No solution when c \ge 1; every x when c < -1. The heights of the points of the unit circle run from -1 to 1. When c \ge 1, no point lies strictly above the line y = c. When c < -1, every point does, even the bottom one at height -1. For c = -1 the bottom point \left(0, -1\right) fails, so \displaystyle x = \frac{3\pi}{2} + 2\pi k is excluded.