SOLVING COMPLEX TRIG INEQUALITIES BY THE TRIPLE UNIT CIRCLE
(Authored by Nghi H Nguyen, Updated Nov. 24, 2020)
THE TRIG UNIT CIRCLE AND BASIC TRIG FUNCTIONS
It is a circle with radius R = 1, and with an origin O. The unit circle defines the main trig
functions of the variable arc x that varies and rotates counterclockwise on it.
On the unit circle, the value of the arc x is exactly equal to the corresponding angle x. It is
more convenient to use the arc x as the variable, instead of the angle x.
When the variable arc AM = x (in radians or degree) varies on the trig unit circle:
- The horizontal axis OAx defines the function f(x) = cos x. When the arc x varies from 0 to
2Ꙥ, the function f(x) = cos x varies from 1 to -1, then, back to 1.
- The vertical axis OBy defines the function f(x) = sin x. When the arc x varies from 0 to 2Ꙥ,
the function f(x) = sin x varies from 0 to 1, then to -1, then back to 0.
- The vertical axis AT defines the function f(x) = tan x. When x varies from -Ꙥ/2 to Ꙥ/2, the
function f(x) varies from - ∞ to +∞.
- The horizontal axis BU defines the function f(x) = cot x. When x varies from 0 to Ꙥ, the
function f(x) = cot x varies from +∞ to -∞.
Figure 1
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GENERALITIES ON SOLVING TRIG INEQUALITIES
Usually, to solve a complex trig inequality in the form of F(x) ≥ 0 or (≤ 0), we transform it into
an inequality in the form of
F(x) = f(x).g(x) ≥ 0 (or ≤ 0) or F(x) = f(x).g(x).h(x) ≥ 0 (or ≤ 0)
where f(x), g(x) and h(x) are simple trig functions, or similar.
To solve these types of trig inequalities, algebraically, we must create a Sign Chart that is not
easy to handle. This innovative number unit circle method offers students with a second
approach that is much more visual than the sign chart, and it shows the periodic characteristics
of the trig functions.
In this article, the author explains this new approach on solving complex trig inequalities by
using the number unit circle.
SOLVING BASIC INEQUALITIES BY THE NUMBER UNIT CIRCLE.
This new approach proceeds the same way as the number line check point method to solve
quadratic inequalities. The unit circle is numbered in radians or in degrees.
Example 1. Solve
Solution.
sin x ≤ 1/2
Transform the inequality into standard form: F(x) = sin x – 1/2 ≤ 0
First, solve F(x) = 0 → sin x = 1/2
There are 2 solution arcs: x1 = Ꙥ/6 and x2 = 5Ꙥ/6. Plot these 2 end points on the unit circle.
They divide the circle into 2 arc lengths (Ꙥ/6, 5Ꙥ/6) and (5Ꙥ/6, Ꙥ/6 + 2Ꙥ = 13Ꙥ/6).
F(x) = sin x - 1/2 < 0 when x is inside the arc length (5Ꙥ/6, 13Ꙥ/6).
The solution set of F(x) ≤ 0 is the closed interval [5Ꙥ/6, 13Ꙥ/6]. The 2 end points are included
in the solution set.
Note. We can use the check point (Ꙥ) to find the sign status of F(x) inside this arc length. We
get: F(x) = sin (Ꙥ) – 1/2 = 0 – 1/2 < 0. Color it blue.
F(x) > 0 when x is inside the arc length (Ꙥ/6, 5Ꙥ/6). Color it red. This way of coloring will help
later in solving complex trig inequalities.
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Fig. 2
SOLVING TRIG INEQUALITIES BY THE BI-UNIT CIRCLE
We can use the Bi-Unit Circle to solve trig inequalities in the form of F(x) that can be
transformed into 2 simple trig functions: F(x) = f(x).g(x) ≤ 0 (or ≥ 0)
This method separately figures the sign status (+ or -) of f(x) and (g(x) on two different
concentric number unit circles. It colors red for (+) arc lengths and blue for (-) arc lengths. By
superimposing, the combined solution set can be easily seen.
Example 2. Solve: sin x + sin 2x + sin 3x < cos x + cos 2x + cos 3x
(0 < x < 2Ꙥ)
Solution. Use Sum to Product Identities (cos a + cos b) and (sin a + sin b) to transform both
sides. Common Period 2Ꙥ
sin x + sin 3x + sin 2x < cos x + cos 3x + cos 2x
sin 2x(2cos x + 1) < cos 2x (2cos x + 1)
F(x) = f(x).g(x) = (2cos x + 1)(sin 2x – cos 2x) < 0
1. Solve f(x) = (2cos x + 1) = 0 --> cos x = -1/2→ 2 end points at: x = 2Ꙥ/3 and x = 4Ꙥ/3
Use check point x = Ꙥ --> we get: f(x) = f(Ꙥ) = (- 2 + 1) = -1 < 0.
Then, f(x) < 0 when x is inside the interval (2Ꙥ/3, 4Ꙥ/3). Color it blue.
The solution set for f(x) > 0 is colored red on unit circle.
2. Solve. g(x) = sin 2x – cos 2x = 0 → Divide by cos 2x
(Condition cos 2x ≠ 0 --> x ≠ Ꙥ/4, and x ≠ 3Ꙥ/4)
tan 2x = 1 → 2 solution arcs: 2x = Ꙥ/4 + kꙤ → that gives: x = Ꙥ/8 + kꙤ/2
and
2x = 5Ꙥ/4 + kꙤ → that gives: x = 5Ꙥ/8 + kꙤ/2
For k = 0, k =1 and k = 2, there are 4 end points: Ꙥ/8 ; 5Ꙥ/8 ; 9Ꙥ/8 and 13Ꙥ/8
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Use check point x = Ꙥ/2. We get g(Ꙥ/2) = 0 - (-1) > 0. Then g(x) > 0 inside interval (Ꙥ/8, 5Ꙥ/8)
Color it red and color the other arc lengths, following the rule for end points.
By superimposing, we see that the solution set are the 3 open intervals:
(5Ꙥ/8, 2Ꙥ/3) where f(x) > 0 and g(x) < 0, and
(9Ꙥ/8, 4Ꙥ/3) where f(x) < 0 and g(x) > 0
(13Ꙥ/8, Ꙥ/8 + 2Ꙥ) or (13Ꙥ/8, 17Ꙥ/8) where f(x) > 0 and g(x) < 0
SOLVING COMPLEX TRIG INEQUALITIES BY THE TRIPLE UNIT CIRCLE
Example 3. Solve
1 + cos 2x > cos 4x + cos 6x
Transform both sides of the inequality. Use these 3 trig identities:
(1 + cos 2a = 2cos² a) and (cos a + cos b), and (cos a – cos b).
F(x) = 2cos² x - 2cos 5x.cos x > 0
F(x) = - 2cos x(cos x – cos 5x) = - 2cos x.(- sin 3x.sin 2x) > 0
F(x) = 4cos x.sin 3x.sin 2x = F(x) = f(x).g(x).h(x) > 0
1. The function f(x) = cos x > 0 when x is inside the arc length (-Ꙥ/2, Ꙥ/2). Color it red and
color the rest of the unit circle green
2. Solve g(x) = sin 3x = 0 to find the end points. There are 3 solution arcs:
3x = 0 + 2kꙤ
3x = Ꙥ + 2kꙤ
3x = 2Ꙥ+ 2kꙤ
x = 2kꙤ/3
x = Ꙥ/3 + 2kꙤ/3
x = 2Ꙥ/3 + 2kꙤ/3
For k = 0, there are 3 end points:
x=0
x = Ꙥ/3
x = 2Ꙥ/3
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For k = 1, there are 3 end points:
x = 2Ꙥ/3
x=Ꙥ
x = 4Ꙥ/3
For k = 2, there are 3 end points
x = 4Ꙥ/3
x = 5Ꙥ/3
x = 2Ꙥ
There are totally 6 end points: (0), (Ꙥ/3), (2Ꙥ/3), (Ꙥ), (4Ꙥ/3), (5Ꙥ/3), and 6 equal arc lengths.
Use check point method to find the sign status of of g(x) = sin 3x for (0, 2Ꙥ)
Inside arc length (0, Ꙥ/3), select x = Ꙥ/6 as check point. We get
g(x) = sin (3*Ꙥ/6) = sin (Ꙥ/2) = 1 > 0. Color this arc length red.
By the property of end points, we get the sign status of g(x) inside the other arc lengths.
Arc length (Ꙥ/3, 2Ꙥ/3) --> g(x) < 0 ( green).
Arc length (2Ꙥ/3, Ꙥ) → g(x) > 0 (red)
Arc length (Ꙥ, 4Ꙥ/3) → g(x) < 0 (green)
Arc length (4Ꙥ/3, 5Ꙥ/3) → g(x) > 0 (red)
Arc length (5Ꙥ/3, 2Ꙥ) → g(x) < 0 (green)
3. Solve h(x) = sin 2x = 0 to find the end points. There are 4 end points: 0; Ꙥ/2; Ꙥ; and 3Ꙥ/2
The function h(x) = sin 2x > 0 when x is:
Inside the arc length (0, Ꙥ/2) → h(x) > 0. (red)
Inside the arc length (Ꙥ/2, Ꙥ) → h(x) < 0 (green)
Inside the arc length (Ꙥ, 3Ꙥ/2) → h(x) > 0 (red)
Inside the arc length (3Ꙥ/2, 2Ꙥ) → h(x) < 0 (green)
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Figure 4
Figure the sign status of f(x), g(x), and h(x) on 3 concentric number unit circles.
By superimposing, we can see that the combined solution set of F(x) > 0 are the 2 open
intervals (2Ꙥ/3, 4Ꙥ/3) and (5Ꙥ/3, Ꙥ/3 + 2Ꙥ = 7Ꙥ/3)
Check.
Inequality: cos x.sin 3x.sin2x > 0
x = Ꙥ/6 → F (Ꙥ/6) = cos (Ꙥ/6).sin(3Ꙥ/6).sin (2Ꙥ/6) = cos (Ꙥ/6).sin (Ꙥ/2).sin (Ꙥ/3) > 0. Proved
x = 5Ꙥ/6 → F(5Ꙥ/6) = cos (5Ꙥ/6).sin (15Ꙥ/6).sin (10Ꙥ/6) = cos (5Ꙥ/6).sin (Ꙥ/2).sin (2Ꙥ/3) < 0.
Proved.
Example 4. Solve
sin 6x – sin 4x < sin 2x
(0, 2Ꙥ)
Solution. Common period: 2Ꙥ. Use trig identities to transform both sides:
(sin a – sin b) and sin 2a = 2sin a.cos a
2cos 5x.sin x < 2sin x.cos x
2cos 5x.sin x – 2sin x.cos x < 0
2sin x(cos 5x – cos x) < 0
2sin x(- 2sin 3x.sin 2x) < 0
F(x) = f(x).g(x).h(x) = 4sin x sin 2x.sin 3x > 0
(side transposing)
Find the sign status of f(x), g(x), and h(x) within period (0, 2Ꙥ)
1. f(x) = sin x > 0 when x is inside the arc length (0, Ꙥ). Color it red. The rest of circle: green.
2. Solve g(x) = sin 2x = 0 → 2 solution arcs: x = Ꙥ/2 and x = 3Ꙥ/2
g(x) = sin 2x > 0 (red) when x varies inside the 2 intervals (0, Ꙥ/2) and (Ꙥ, 3Ꙥ,2)
g(x) = sin 2x < 0 (green) inside the 2 arc lengths: (Ꙥ/2, Ꙥ) and (3Ꙥ/2, 2Ꙥ)
3. Solve h(x) = sin 3x = 0. See Example 3 for the sign status of h(x) = sin 3x.
Figure separately the 3 sign status of f(x), g(x), and h(x) on 3 concentric unit circles.
By superimposing, we see that the combined solution set of F(x) > 0 are the 4 open intervals:
(0, Ꙥ/3) where the resulting sign is: F(x) = (+)(+)(+) > 0, and
(Ꙥ/2, 2Ꙥ/3) where → F(x) = (+)(-)(-) > 0, and
(Ꙥ, 4Ꙥ/3) where → F(x) = (-)(+)(-) > 0, and
(3Ꙥ/2, 5Ꙥ/3) where → F(x) = (-)(-)(+) > 0
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Figure 5
Check. Check the original trig inequality:
sin 6x – sin 4x < sin 2x
x = Ꙥ/6 → sin (Ꙥ) – sin (Ꙥ/3) = 0 – sin (Ꙥ/3) < sin (Ꙥ/3). Proved
x = 5Ꙥ/6 → sin (5Ꙥ) – sin (4Ꙥ/3) = sin (Ꙥ) - (-sin Ꙥ/3) = 0 + sin (Ꙥ/3) > sin (5Ꙥ/3). Proved
Note. Side transposing during transformation process doesn’t affect the original trig
inequality.
Example 5. Solve
tan 2x < tan x
(0, 360⁰)
Solution. Use the trig identity: tan a – tan b = sin (a – b)/(cos a.cos b) to transform the
inequality to:
F(x) = sin x/(cos x.cos 2x) = f(x)/[g(x).h(x)]
Common period 360⁰
1. f(x) = sin x > 0 when x is inside the half-circle (0, 180⁰) (red), and
f(x) < 0 inside (180⁰, 360⁰), (blue)
2. g(x) = cos x > 0 inside the interval (-90⁰, 90⁰) (red). Then, g(x) < 0 inside (90⁰, 270⁰) (blue)
3. h(x) = cos 2x = 0 → 2 end points: x = (45⁰) and (x = 135⁰)
h(x) = cos 2x < 0 inside the arc lengths (45⁰, 135⁰) and (225⁰, 315⁰) (blue)
h(x) = cos 2x > 0 inside the arc lengths (135⁰, 225⁰) and (315⁰, 45⁰ + 360⁰ = 405⁰). (red)
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Figure 6
By superimposing, we see that the combined solution set of F(x) < 0 are the 4 open intervals:
(45⁰, 90⁰) → F(x) = (+).(+).(-) < 0
(135⁰, 180⁰) → F(x) = (+).(-).(+) < 0
(225⁰, 270⁰) → F(x) = (-).(-).(-) < 0
(315⁰, 360⁰) → F(x) = (-).(+).(+) < 0
Check
inequality: tan 2x < tan x
Interval 1 → select x = 60⁰ → tan 2x = 120⁰ < tan 60⁰. Proved
Interval 2 → select x = 150⁰ → tan 2x = tan 300⁰ < tan 150⁰. Proved
Interval 3 → select x = 240⁰ → tan 2x = tan 480⁰ = tan 120⁰ < tan x = tan 240⁰. Proved
Interval 4 → select x = 330⁰ → tan 2x = tan 660⁰ = tan 300⁰ < tan x = tan 330⁰. Proved
Example 6. Solve
Solution.
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tan x. tan 2x < 0
F(x) = f(x).g(x) = tan x. tan 2x < 0
(common period: Ꙥ)
f(x) = tan x = 0 → two end points x = 0, and x = Ꙥ.
There is discontinuation at x = Ꙥ/2
Figure 7
f(x) > 0 when x varies inside interval (0, Ꙥ/2). Color it red, and the rest blue.
g(x) = tan 2x = 0 → 2 end points at x = 0, and x = Ꙥ
There are discontinuations at:
2x = Ꙥ/2 → x = Ꙥ/4, and 2x = 3Ꙥ/2 → x = 3Ꙥ/4
g(x) > 0 when x varies inside the interval (0, Ꙥ/4). Color it red and color the other intervals.
By superimposing, we see that the combined solution set are the open intervals:
(Ꙥ/4, Ꙥ/2) and (Ꙥ/2, 3Ꙥ/4)
Check.
Inequality:
tan x.tan 2x < 0
x = Ꙥ/3 → tan (Ꙥ/3).tan (2Ꙥ/3) → (+).(-) < 0. Proved
x = (2Ꙥ/3) → tan (2Ꙥ/3).tan (4Ꙥ/3) → (-).(+) < 0. Proved
x = Ꙥ/6 → tan (Ꙥ/6).tan (2Ꙥ/6) = tan (Ꙥ/6).tan (Ꙥ/3) → (+).(+). Not true
x = 5Ꙥ/6 → tan (5Ꙥ/6).tan (10Ꙥ/6) = tan (5Ꙥ/6).tan (5Ꙥ/3) → (-).(-). Not true
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REMARKS.
1. The unit circle can be numbered, in radians or degrees, depending on the common period
of the trig inequality.
For example:
F(x) = tan x.tan 2x < 0. The unit circle is numbered in interval (0, Ꙥ), (or from 0⁰ to 180⁰)
F(x) = sin (x/2).cos 2x < 0. The unit circle is numbered from 0 to 4Ꙥ, (or from 0⁰ to 720⁰)
F(x) = sin (x/3).cos x > 0. The unit circle is numbered from 0 to 6Ꙥ(or from 0⁰ to 1080⁰)
2. A few rules of endpoints and arc lengths to remember:
For f(x) = cos x & f(x) = sin x, or similar, there are 2 endpoints and 2 arc lengths (0, 2Ꙥ)
For f(x) = cos 2x & f(x) = sin 2x, or similar, there are 4 endpoints and 4 arc lengths (0, 2Ꙥ)
For f(x) = cos 3x & f(x) = sin 3x, or similar, there are 6 endpoints and 6 arc lengths (0, 2Ꙥ)
For f(x) = tan x & f(x) = cot x, There are one end point, one discontinuation, and 3 arc lengths.
3. On the trig unit circle, the values of the arc x and the corresponding angle x are exactly
equal.
- The french concept to select the arc x as variable, instead of the angle x, makes the whole
trigonometric study more convenient, concrete, visual, and less absurd.
- The popular expressions such as “Arc tan x”, and “Arc sin 3/4” make sense with this concept.
- About the trig functions tan x and cot x, when they go to infinities, the notion of infinity is
understandable as two axis lines going parallel to each other.
- The extended answer (30⁰ + 720⁰) can be visually understood as an arc length of 30⁰ plus 2
circumferences. However, the notion of a flat angle (30⁰ + 720⁰) may be confused as a flat area
- Solving such inequality (tan x < a), by using the tan axis AT is easy and simple. However,
solving the inequality (sin x/cos x < a) becomes complicated because it is dealing with 2
function variables. Solving (tan x < a) or (> a) by the tan function’s graph takes too much time
in creating the graph.
- In practice, the rules of end points and arc lengths, on the unit circle, make the solving process
more convenient than handling the sign chart.
- Finally, we can use the trig unit circle as a second approach to solve complex trig inequalities
by using the bi-unit circle, or triple unit circle.
(This math article is authored by Nghi H Nguyen, the author of the new Transforming Method
to solve quadratic equations – Updated Nov. 24, 2020)
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