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# Solving Complex Trig Inequalities by the Triple Unit Circle - Updated

advertisement ```SOLVING COMPLEX TRIG INEQUALITIES
BY THE TRIPLE UNIT CIRCLE - (Updated)
(Authored by Nghi H Nguyen - Updated March 28, 2021)
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 basic 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 more visual than the sign chart, and it shows the periodic
property of trig functions.
In this article, the author explains how does this new method work in solving complex
trig inequalities by using the number unit circle.
SOLVING BASIC INEQUALITIES BY THE NUMBER UNIT CIRCLE.
This new method 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. On the sin axis, take sin x = - 0.5
The unit circle gives 2 end points: x1 = 11Ꙥ/6 and x2 = 7Ꙥ/6. Plot these 2 end points
on the unit circle. They divide the circle into 2 arc lengths (7Ꙥ/6, 11Ꙥ/6) and
(-Ꙥ/6,7Ꙥ/6).
The function F(x) = sin x + 1/2 &lt; 0 when x is inside the arc length (7Ꙥ/6, 11Ꙥ/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 (3Ꙥ/2) to find the sign status of F(x) inside this
interval. We get: F(x) = sin (3Ꙥ/2) + 1/2 = - 1 + 1/2 &lt; 0. Color it blue.
F(x) &gt; 0 when x is inside the arc length (-Ꙥ/6, 7Ꙥ/6). Color it red. This way of coloring
will help later in solving complex trig inequalities.
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Figure 2
SOLVING TRIG INEQUALITIES BY THE BI-UNIT CIRCLE
We can use the bi-unit circle to solve complex trig inequalities in the form of F(x) that
can be transformed into 2 basic trig functions, or similar: 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 &lt; cos x + cos 2x + cos 3x
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 &lt; cos x + cos 3x + cos 2x
sin 2x(2cos x + 1) &lt; cos 2x (2cos x + 1)
F(x) = f(x).g(x) = (2cos x + 1)(sin 2x – cos 2x) &lt; 0
1. Solve f(x) = (2cos x + 1) = 0 --&gt; cos x = -1/2→ This gives 2 end points at: x = 2Ꙥ/3
and x = 4Ꙥ/3. Use check point (x = Ꙥ) --&gt; we get: f(x) = f(Ꙥ) = (- 2 + 1) = -1 &lt; 0.
Then, f(x) &lt; 0 when x is inside the interval (2Ꙥ/3, 4Ꙥ/3). Color it blue, and the other
arc length red.
2. Solve. g(x) = sin 2x – cos 2x = 0 → Divide by cos 2x (cos x ≠ Ꙥ/4, and x ≠ 3Ꙥ/4).
We get: tan 2x = 1. This gives 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 at: 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) &gt; 0. Then g(x) &gt; 0 inside interval
(Ꙥ/8, 5Ꙥ/8). Color it red and color the 3 other arc lengths, following the rule of end
points.
Figure 3
By superimposing, we see that the solution set of F(x) &lt; 0 are the 3 open intervals:
(5Ꙥ/8, 2Ꙥ/3) where f(x) &gt; 0 and g(x) &lt; 0, and
(9Ꙥ/8, 4Ꙥ/3) where f(x) &lt; 0 and g(x) &gt; 0
(13Ꙥ/8, Ꙥ/8 + 2Ꙥ) or (13Ꙥ/8, 17Ꙥ/8) where f(x) &gt; 0 and g(x) &lt; 0
SOLVING COMPLEX TRIG INEQUALITIES BY THE TRIPLE UNIT CIRCLE
Example 3. Solve
1 + cos 2x &gt; cos 4x + cos 6x
Transform both sides of the inequality. Use these 3 trig identities:
(1 + cos 2a = 2cos&sup2; a) and (cos a + cos b), and (cos a – cos b).
F(x) = 2cos&sup2; x - 2cos 5x.cos x &gt; 0
F(x) = - 2cos x(cos x – cos 5x) = - 2cos x.(- sin 3x.sin 2x) &gt; 0
F(x) = 4cos x.sin 3x.sin 2x = F(x) = f(x).g(x).h(x) &gt; 0
1. The function f(x) = cos x= 0 has 2 end points at: - Ꙥ/2 and Ꙥ/2. f(x) is positive (&gt; 0)
when x is inside the arc length (-Ꙥ/2, Ꙥ/2). Color it red and color the rest of the unit
circle blue.
2. Solve g(x) = sin 3x = 0 to find the end points. On the unit circle, sin 3x = 0 when:
3x = 0+ 2kꙤ. This gives: x = 0 + 2kꙤ/3
3x = Ꙥ+ 2kꙤ. This gives: x = Ꙥ/3 + 2kꙤ/3
3x = 2Ꙥ+ 2kꙤ. This gives: x = 2Ꙥ/3 + 2kꙤ/3
Page 4 of 16
For k = 0, k = 1, and k = 2, there are totally 6 end points at: (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 period (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 &gt; 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) --&gt; g(x) &lt; 0 (blue).
Arc length (2Ꙥ/3, Ꙥ) → g(x) &gt; 0 (red)
Arc length (Ꙥ, 4Ꙥ/3) → g(x) &lt; 0 (blue)
Arc length (4Ꙥ/3, 5Ꙥ/3) → g(x) &gt; 0 (red)
Arc length (5Ꙥ/3, 2Ꙥ) → g(x) &lt; 0 (blue)
3. Solve h(x) = sin 2x = 0 to find the end points. There are 4 end points: 0; Ꙥ/2; Ꙥ; and
3Ꙥ/2, and 4 equal arc lengths.
The function h(x) = sin 2x &gt; 0 when x is:
Inside the arc length (0, Ꙥ/2) → h(x) &gt; 0. (red)
Inside the arc length (Ꙥ/2, Ꙥ) → h(x) &lt; 0 (blue)
Inside the arc length (Ꙥ, 3Ꙥ/2) → h(x) &gt; 0 (red)
Inside the arc length (3Ꙥ/2, 2Ꙥ) → h(x) &lt; 0 (blue)
Figure 4
Report 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) &gt; 0 are the 2
open intervals (2Ꙥ/3, 4Ꙥ/3) and (5Ꙥ/3, Ꙥ/3 + 2Ꙥ = 7Ꙥ/3)
Page 5 of 16
Check.
Inequality: cos x.sin 3x.sin 2x &gt; 0
x = Ꙥ/6 →This gives F (Ꙥ/6) = cos (Ꙥ/6).sin(3Ꙥ/6).sin (2Ꙥ/6) =
= cos (Ꙥ/6).sin (Ꙥ/2).sin (Ꙥ/3) = (+)(+)(+) &gt; 0. Proved
x = 5Ꙥ/6 → This gives F(5Ꙥ/6) = cos (5Ꙥ/6).sin (15Ꙥ/6).sin (10Ꙥ/6) =
= cos (5Ꙥ/6).sin (5Ꙥ/2).sin (5Ꙥ/3) = (-)(+)(-) &gt; 0. Proved.
Example 4. Solve
sin 6x – sin 4x &lt; sin 2x
(0, 2Ꙥ)
Solution. Common period: 2Ꙥ. Use these trig identities to transform both sides:
(sin a – sin b) and sin 2a = 2sin a.cos a
2cos 5x.sin x &lt; 2sin x.cos x
F(x) = 2cos 5x.sin x – 2sin x.cos x &lt; 0
F(x) = 2sin x(cos 5x – cos x) &lt; 0
F(x) = 2sin x(- 2sin 3x.sin 2x) &lt; 0
F(x) = f(x).g(x).h(x) = 4sin 3x sin 2x.sin x &gt; 0
(side transposing)
Find the sign status of f(x), g(x), and h(x) within period (0, 2Ꙥ)
1. Solve f(x) = sin 3x = 0. See Example 3 for the sign status of h(x) = sin 3x. There are
6 end points and 6 equal arc lengths.
2. Solve g(x) = sin 2x = 0 → sin 2x = 0 when:
2x = 0 + 2kꙤ. This gives: x = 0 + kꙤ
2x = Ꙥ + 2kꙤ. This gives: x = Ꙥ/2 + kꙤ
2x = 2Ꙥ + 2kꙤ. This gives: x = Ꙥ + kꙤ
For k = 0, and k -1, there are 4 end points at: 0, Ꙥ/2, Ꙥ, and 3Ꙥ/2, and 4 equal arc
lengths.
g(x) = sin 2x &gt; 0 (red) when x varies inside the 2 intervals (0, Ꙥ/2) and (Ꙥ, 3Ꙥ,2)
g(x) = sin 2x &lt; 0 (blue) inside the 2 arc lengths: (Ꙥ/2, Ꙥ) and (3Ꙥ/2, 2Ꙥ)
3 f(x) = sin x &gt; 0 when x is inside the arc length (0, Ꙥ). Color it red. The rest of circle,
blue.
Report the 3 sign status of f(x), g(x), and h(x) on 3 concentric unit circles. Figure 5.
By superimposing, we see that the combined solution set of F(x) &gt; 0 are the 4 open
intervals:
Page 6 of 16
(0, Ꙥ/3) where the resulting sign is: F(x) = (+)(+)(+) &gt; 0, and
(Ꙥ/2, 2Ꙥ/3) where → F(x) = (+)(-)(-) &gt; 0, and
(Ꙥ, 4Ꙥ/3) where → F(x) = (-)(+)(-) &gt; 0, and
(3Ꙥ/2, 5Ꙥ/3) where → F(x) = (-)(-)(+) &gt; 0
Figure 5
Check. Check the original trig inequality:
sin 6x – sin 4x &lt; sin 2x
x = Ꙥ/6 → sin (Ꙥ) – sin (Ꙥ/3) = 0 – sin (Ꙥ/3) &lt; sin (Ꙥ/3). Proved
x = 5Ꙥ/6 → sin (5Ꙥ) – sin (4Ꙥ/3) = sin (Ꙥ) - (-sin Ꙥ/3) = 0 + sin (Ꙥ/3) &gt; sin (5Ꙥ/3).
Proved
Example 5. Solve
tan 2x &lt; 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) = tan 2x – tan x = sin x/(cos x.cos 2x) = f(x)/[g(x).h(x)]
(Common period 360⁰)
1. Solve f(x) = sin x = 0. f(x) is positive when x is inside the half-circle (0, 180⁰) (red),
and f(x) &lt; 0 inside the interval (180⁰, 360⁰), (blue)
2. Solve g(x) = cos x = 0. g(x) is positive inside the arc length(-90⁰, 90⁰) (red), and
g(x) &lt; 0 inside the arc length (90⁰, 270⁰) (blue)
3. h(x) = cos 2x = 0 . There are 4 end points at: (45⁰), (135⁰), (225⁰), and (315⁰), and
4 equal arc lengths.
Page 7 of 16
h(x) = cos 2x &lt; 0 inside the arc lengths (45⁰, 135⁰) and (225⁰, 315⁰) (blue)
h(x) = cos 2x &gt; 0 inside arc lengths (135⁰, 225⁰) and (315⁰, 45⁰ + 360⁰ = 405⁰). (red)
Figure 6
3. h(x) = cos 2x = 0 . There are 4 end points at: (45⁰), (135⁰), (225⁰), and (315⁰), and
4 equal arc lengths. See Figure 6.
By superimposing, we see that the combined solution set of F(x) &lt; 0 are the 4 open
intervals:
(45⁰, 90⁰) → F(x) = (+).(+).(-) &lt; 0
(135⁰, 180⁰) → F(x) = (+).(-).(+) &lt; 0
(225⁰, 270⁰) → F(x) = (-).(-).(-) &lt; 0
(315⁰, 360⁰) → F(x) = (-).(+).(+) &lt; 0
Check
Inequality: tan 2x &lt; tan x
Interval 1 → select x = 60⁰ → tan 2x = 120⁰ &lt; tan 60⁰. Proved
Interval 2 → select x = 150⁰ → tan 2x = tan 300⁰ &lt; tan 150⁰. Proved
Interval 3 → select x = 240⁰ → tan 2x = tan 480⁰ = tan 120⁰ &lt; tan x = tan 240⁰. Proved
Interval 4 → select x = 330⁰ → tan 2x = tan 660⁰ = tan 300⁰ &lt; tan x = tan 330⁰. Proved
Page 8 of 16
Example 6. Solve
Solution.
tan x. tan 2x &lt; 0
F(x) = f(x).g(x) = tan x. tan 2x &lt; 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) &gt; 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) &gt; 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 &lt; 0
x = Ꙥ/3 → tan (Ꙥ/3).tan (2Ꙥ/3) → (+).(-) &lt; 0. Proved
x = (2Ꙥ/3) → tan (2Ꙥ/3).tan (4Ꙥ/3) → (-).(+) &lt; 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
Page 9 of 16
Example 7.
Solve
cos x – sin x.tan x + 1 &lt; 0
Solution. Replace tan x by (sin x/cos x), and 1 by cos x/cos x, then proceed the
transformation:
F(x) = [cos&sup2; x – sin&sup2; x + cos x]/cos x &lt; 0
F(x) = (cos 2x + cos x)/cos x &lt; 0
(Trig identity: cos&sup2; x – sin&sup2; x = cos 2x)
F(x) = [f(x).g(x)]/h(x) = (2cos 3x/2.cos x/2)/cos x &lt; 0
(Trig identity: cos a + cos b)
1. Solve f(x) = cos (3x)/2 = 0. The common period is 4Ꙥ. There are 2 solutions:
(3x)/2 = Ꙥ/2. This gives: 3x = Ꙥ + 4kꙤ --→ x = Ꙥ/3 + (4kꙤ)/3
(3x)/2 = 3Ꙥ/2. This gives: 3x = 3Ꙥ + 4kꙤ --→ x = Ꙥ + (4kꙤ)/3
For k = 0, k = 1, and k = 2, we get 6 equal arc lengths and 6 end points at:
(Ꙥ/3), (Ꙥ), (5Ꙥ/3), (7Ꙥ/3), (9Ꙥ/3), (11Ꙥ/3), inside the common period 4Ꙥ.
To find the sign status of f(x) = cos (3x)/2, select the check point (x = 0).
We get f(0) = 1 &gt; 0. So, f(x) is positive (&gt; 0) inside the arc length (11Ꙥ/3, Ꙥ/3) or
(-Ꙥ/3, Ꙥ/3). Color it red and color the 5 other arc lengths.
2. Solve g(x) = cos x/2 = 0. There are 2 solutions:
x/2 = Ꙥ/2 + 2kꙤ. This gives x = Ꙥ + 4kꙤ
x/2 = 3Ꙥ/2 + 2kꙤ. This gives x = 3Ꙥ + 4kꙤ.
There are 2 end points at: (Ꙥ) and (3Ꙥ) and 2 equal arc lengths for the period (0, 4Ꙥ).
Select the check point (x = 0). We get g(0) = cos 0 = 1 &gt; 0. Therefore, g(x) is positive
(&gt; 0) inside the half circle (3Ꙥ, 5Ꙥ). Color it red and color the other half circle blue.
3. Solve h(x) = cos x = 0. The function h(x) is positive (&gt; 0) inside the interval
(-Ꙥ/2, Ꙥ/2) and (3Ꙥ/2, 5Ꙥ/2) within the period 4Ꙥ. Color them red and the 2 others
blue. By superimposing, we see that the combined solution set of (F(x) &lt; 0 are the 4
open intervals:
(Ꙥ/3, Ꙥ/2) and (3Ꙥ/2, 5Ꙥ/3) and (7Ꙥ/3, 5Ꙥ/2), and (7Ꙥ/2, 11Ꙥ/3). See Figure 8.
Note. There are 4 discontinuities at (Ꙥ/2), (3Ꙥ2), (5Ꙥ/2), and (7Ꙥ/2) when cos x = 0
Check.
F(x) = (cos 2x + cos x)/cos x&lt; 0
x = Ꙥ/4 --→ F(Ꙥ/4) = 0 + 1/1 &gt; 0. Proved
x = 3Ꙥ/4 --→ F(3Ꙥ/4) = (0 + √2/2)√2/2 &gt; 0. Proved
x = 280⁰ --→ F(280) = (cos 560 + cos 280)cos 280 = (-0.94 + 017)/0.17 &lt; 0. Proved
Page 10 of 16
Figure 8
Example 8. Solve
Figure 9
5sin x + 2cos x &gt; 3
Solution. Divide both sides by 5, we get:
F(x) = sin x + (2/5)cos x – 0.6 &gt; 0
Call t that tan t = sin t/cos t = 2/5 = 0.4 --→ t = 21⁰80, and cos t = 0.93
F(x) = sin x + (sin t/cos t).cos x - 0.6.&gt; 0
F(x) = sin x.cos t + sin t.cos x – 0.56 &gt; 0
F(x) = sin (x + t) – 0.56 &gt; 0
(0.6*cos t= 0.56)
Solve F(x) = sin (x + 21⁰80) – 0.56 = 0. We get:
sin (x + 21⁰80) = 0.56 = sin (33⁰92) = sin (146⁰08). This leads to:
x + 21⁰80 = 33⁰92 --→ x = 33.92 – 21.80 = 12⁰12
x + 21.80 = 146.08 --→ x = 146.08 – 21.80 = 124⁰28.
There are 2 end points at (12⁰12) and (124⁰08), and 2 arc lengths. To find the sign
status of F(x), select the check point (x = 90⁰). We get: F(90⁰) = 5 + 0 – 3 = 2 &gt; 0.
Therefore, F(x) is positive (&gt; 0) inside the interval (12⁰12, 124⁰28). That is the solution
set of the trig inequality. See Figure 9.
If the answer must be in radians, we convert degrees to radians by the conversion
rule. The solution set is (in radians): (0.067Ꙥ, 0.69Ꙥ).
Page 11 of 16
NOTE. This new method helps solving complex trig inequalities whose common period
is different to 2Ꙥ by using the unit circle.
Example 9. Solve
sin x &lt; cos (x/3)
(Common period 6Ꙥ)
Solution. F(x) = sin x – cos (x/3) &lt; 0
Replace cos (x/3) by cos (Ꙥ/2 – x/3), we get
F(x) = sin x – sin (Ꙥ/2 - x/3) &lt; 0.
Use trig identity # 29 (sin a – sin b) to transform F(x)
F(x) = 2cos (x/3 + Ꙥ/4).sin (2x/3 – Ꙥ/4) &lt; 0
1. Solve f(x) = cos (x/3 + Ꙥ/4) = 0. There are 2 solutions for f(x) = 0
a. x/3 + Ꙥ/4 = Ꙥ/2. This gives: x/3 = Ꙥ/4 + 2kꙤ --→ x = 3Ꙥ/4 + 6kꙤ
b. x/3 + Ꙥ/4 = (3Ꙥ)/2. This gives: x/3 = (5Ꙥ)/4 + 2kꙤ --→ x = (15Ꙥ)/4 + 6kꙤ
It is easier to work with degrees. There are 2 arc lengths and 2 end points at:
(3Ꙥ/4) = 135⁰, and (15Ꙥ/4) = 675⁰ inside the common period 6Ꙥ = 1080⁰.
To find the sign status of f(x), select the check point (180⁰). We get:
f(180) = cos (60 + 45) = cos 105 &lt; 0. Therefore, f(x) is negative inside the arc length
(135⁰, 675⁰). Color it blue and color the rest red
2. Solve g(x) = sin (2x/3 – Ꙥ/4) = 0. We get g(x) = 0 when:
a. 2x/3 – 45⁰ = 0⁰. This gives: (2x/3) = 45⁰ + k360⁰ --→ x = 135⁰/2 = 67⁰5 + k540⁰
b. 2x/3 – 45 = 180⁰. This gives: (2x/3) = 225⁰ + k360⁰ --→ x= 337⁰5 + k540⁰
c. 2x/3 – 45⁰ = 360⁰. This gives (2x/3) = 765⁰ + k360⁰ --→ x = 607⁰5 + k540⁰
For k = 0 and k = 1, there are 4 equal arc lengths and 4 end points at:
(67⁰5), (337⁰5), (607⁰5) and (877⁰5).
To find the sign status of g(x), select check point (x = 180⁰).
We get: g(180) = sin (120 – 45) = sin (75⁰) &gt; 0. Therefore, g(x) is positive inside the
arc length (67⁰5, 337⁰5). Color it red and color the 3 other arc lengths.
By superimposing, the solution set of F(x) &lt; 0 are the 3 open intervals:
(135, 337⁰5) and (607⁰5, 675⁰), and [877⁰5, (67⁰5 + 1080⁰)]. See Figure 10
Page 12 of 16
Figure 10
Check by calculator.
F(x) = sin x – cos (x/3)
x = 270⁰ --→ F(270) = sin 270 – cos 90 = - 1 – 0 = -1 &lt; 0. Proved
x = 540 --→ F(540) = sin 540 – cos 180 = 0 - (-1) = 1 &gt; 0. Proved.
X = 650⁰ --→ F(650) = sin 650 – cos 216.7 = - 0.94 - (- 0.80) = - 0.14 &lt; 0. Proved
x = 900 --→ F(900) = sin 900 – cos 300 = 0 – 0.5 = - 0.5 &lt; 0. Proved.
We may convert to radians, if required.
(135, 337.5) and (607.5, 675) and (877.5, 67.5)
(3Ꙥ/4, 3Ꙥ/8), and (27Ꙥ/8, 15Ꙥ/4), and (39Ꙥ/8, 3Ꙥ/8 + 6Ꙥ)
Note. Solving this trig inequality by the sign chart is unadvised because the sign chart
gets stuffy and confusing.
Example 10. Solve
sin x + sin 3x + sin 5x + sin 7x &gt; 0
Solution. Use trig identity #28: (sin a + sin b) to transform the inequality:
F(x) = (sin x + sin 7x) + (sin 3x + sin 5x) &gt; 0
F(x) = 2sin 4x.cos 3x + 2sin 4x.cos x = 2sin 4x(cos 3x + cos x) &gt; 0
Using trig identity #10 (sin 2a = 2sin a.cos a) and #26 (cos a + cos b), we get:
F(x) = 4 sin 2x.cos 2x(2cos 2x.cos x) = 8cos x.sin 2x.cos&sup2; 2x &gt; 0
1. Solve cos x = 0. There are 2 end points at (-Ꙥ/2) and (Ꙥ/2). The function f(x) = cos
x is positive (&gt; 0) inside the arc length (-Ꙥ/2, Ꙥ/2). Color it red and color the rest blue.
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2. Solve g(x) = sin 2x = 0. There are 4 end points at (0), (Ꙥ/2), (Ꙥ) and (3Ꙥ/2).
g(x) is positive (&gt; 0) inside the arc length (0, Ꙥ/2). Color it red and color the 3 other arc
lengths.
3. Solve h(x) = cos&sup2; 2x = 0. This function is always positive regardless of x. Therefor,
the sign status of F(x) is the sign status of the product f(x).g(x).
By superimposing, the solution set of F(x) &gt; 0 is the open interval (0, Ꙥ). Figure 11.
Figure 11
Check.
Figure 12
f(x).g(x) = cos x.sin 2x
x = Ꙥ/4. This gives: f(Ꙥ/4).g(Ꙥ/2) = (√2/2).(1) &gt; 0. Proved
x = 3Ꙥ/4. This gives: f(3Ꙥ/4).g(3Ꙥ/2) = (-√2/2)(-1) &gt; 0. Proved
x = 5Ꙥ/4. This gives: f(5Ꙥ/4).g(5Ꙥ/2) = (-√2/2)(1) &lt; 0. Proved
Example 11. Solve
sin 3x + sin 2x &lt; cos 2x – cos 3x
Solution. F(x) = (sin 3x + sin 2x) + (cos 3x – cos 2x) &lt; 0
Use trig identities (sin a + sin b) and (cos a – cos b) to transform f(x).
F(x) = 2sin (5x/2).cos (x/2) – 2sin (5x/2).sin (x/2) &lt; 0
F(x) = 2sin (5x/2)(cos x/2 – sin x/2) = f(x).g(x) &lt; 0
1. Solve f(x) = sin (5x/2) = 0. This gives 3 solutions:
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a. 5x/2 = 0. This gives: 5x/2 = 0 + 2kꙤ --→ x = 0 + (4kꙤ)/5
b. 5x/2 = Ꙥ. This gives: 5x/2 = Ꙥ + 2kꙤ --→ x = (2Ꙥ/5) + (4kꙤ)/5
c. 5x/2 = 2Ꙥ. This gives 5x/2 = 2Ꙥ + 2kꙤ --→ x = (4Ꙥ/5) + (4kꙤ)/5
For k = 0, k= 1, k = 2, and k = 3, k = 4, there are total 10 end points at: (2Ꙥ/5), (4Ꙥ/5),
(6Ꙥ/5), (8Ꙥ/5), (10Ꙥ/5), (12Ꙥ/5), (14Ꙥ/5), (16Ꙥ/5), (18Ꙥ/5), and (4Ꙥ).
There are 10 equal arc lengths inside the common period 4Ꙥ.
To find the sign status of f(x), select the check point x = (Ꙥ/10).
We get: f(2Ꙥ/10) = sin (5x/2) = sin (Ꙥ/2) = 1 &gt; 0. Therefor, f(x) is positive (&gt; 0) inside
the interval (0, 2Ꙥ/5). Color it red and color the 9 other arc lengths.
2. Solve g(x) = (cos x/2 – sin x/2) = 0.
Use trig identity (sin a – cos a) = - √2 cos (a + Ꙥ/4).
We get: g(x) = - (sin x/2 – cos x/2) = √2cos (x/2 + Ꙥ/4) = 0. There are 2 solutions:
a. (x/2 + Ꙥ/4) = Ꙥ/2. This gives: x/2 = Ꙥ/4 + 2kꙤ --→ x = Ꙥ/2 + 4kꙤ
b. (x/2 + Ꙥ/4) = 3Ꙥ/2. This gives: x/2 = 5Ꙥ/4 + 2kꙤ --→ x= 10Ꙥ/2 + 4kꙤ
Select the check point (Ꙥ). We get g(Ꙥ) = (cos Ꙥ/2 – sin Ꙥ/2) = 0 – 1 &lt; 0. Therefor,
g(x) is negative in the interval (Ꙥ/2, 5Ꙥ/2). Color it blue and the other half circle red.
The solution set of F(x) &lt; 0 are the 6 open intervals: (2Ꙥ/5, Ꙥ/2) and (4Ꙥ/5, 6Ꙥ/5)
and (8Ꙥ/5, 2Ꙥ) and (12Ꙥ/5, 10Ꙥ/4) and (14Ꙥ/5, 16Ꙥ/5) and (18Ꙥ/5, 4Ꙥ).
Check.
F(x) = (sin 3x + sin 2x) + (cos 3x – cos 2x) &lt; 0
x = Ꙥ/4 --→ (sin 3Ꙥ/4 + sin Ꙥ/2) + (cos 3Ꙥ/4 – cos Ꙥ/2) = √2/2 + 1 - √2/2 – 0 &gt; 0. OK
x = 80⁰ --→ sin 240 + sin 160 + cos 240 – cos 160 = -0.87 + 0.34 - 0.5 – 0.94 &lt; 0. OK
x = 3Ꙥ --→ sin 3Ꙥ + sin 2Ꙥ + cos 3Ꙥ – cos 2Ꙥ = 0 + 0 – 1 – 1 = -2 &lt; 0. Proved
x = Ꙥ --→ sin 3Ꙥ + sin 2Ꙥ + cos 3Ꙥ – cos 2Ꙥ = 0 + 0 - 1 – 1 = - 2 &lt; 0. Proved
x = 700⁰ -→ sin 2100 + sin 1400 + cos 2100 – cos 1400 = - 0.87 – 0.64 + 0.5 – 0.77 &lt;
0. Proved
REMARKS.
1. If the answers must be in radians, we convert them from degrees to radians by
using the conversion rule: x Radians = x Degrees.(Ꙥ/180)
2. A few rules of endpoints and arc lengths to remember:
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For f(x) = cos x and f(x) = sin x, or similar, there are 2 endpoints and 2 arc lengths for
(0, 2Ꙥ)
For f(x) = cos 2x and f(x) = sin 2x, or similar, there are 4 endpoints and 4 arc lengths
for (0, 2Ꙥ)Page 13 of 14
For f(x) = cos 3x and f(x) = sin 3x, or similar, there are 6 endpoints and 6 arc lengths
for (0, 2Ꙥ)
For f(x) = tan x and 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, 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.
- Solving such inequality (tan x &lt; a), by using the tan axis AT is easy and simple.
However, solving the inequality (sin x/cos x &lt; a) becomes complicated because it is
dealing with 2 function variables.
- 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.
- On the sign chart, the 2 extremities (0 or 2Ꙥ) are separated. On the unit circle, the 2
extremities joint together and show the periodic property of trig functions.
- On the sign chart, all x-values are places on the first line. In case of complex trig
inequalities such as F(x) = f(x).g(x).h(x) ≤ 0 (or ≥ 0), the high number of x-values
makes the chart stuffy and confusing. In the new method, number of x-values depends
on, in each step, the solving of only one basic trig inequality.
(This math article is authored by Nghi H Nguyen – Updated, March 28, 2021)
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