Maths Formulae Equations Trigonometric Inequation Sin

Sine Inequation – Solving Trigonometric Inequalities (sin)

Handle inequalities involving sin(x) using domain and identity rules. Ideal for trigonometry problem-solving.
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Definition

Trigonometric inequalities involving the sine function require finding all angle values (x) for which the sine of that angle satisfies a given condition (e.g., greater than, less than, or equal to a certain value 'a'). Unlike equations which often yield discrete solutions, inequalities typically result in continuous intervals or arcs of angles as solutions. The periodic nature of the sine function means these solution intervals repeat every 2π radians.

\[ \sin x \geq a \text{ or } \sin x \leq a \text{ or } \sin x > a \text{ or } \sin x < a \]
General Forms of Sine Inequalities
SymbolDescription
xThe angle variable, typically in radians.
a or mThe boundary value. For real solutions to exist, 'a' must be within the range of the sine function, [-1, 1].
k ∈ ℤAn integer used to represent all possible solutions due to the periodic nature of the sine function (period of 2π).
arcsin(a)The inverse sine function, which gives the principal value (an angle in the range [-π/2, π/2]) whose sine is 'a'.
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Key Formulas

\[ \sin x \geq a: \quad x \in [\arcsin(a) + 2\pi k, \pi - \arcsin(a) + 2\pi k] \]
Solution for sin(x) ≥ a
\[ \sin x \leq a: \quad x \in [\pi - \arcsin(a) + 2\pi k, 2\pi + \arcsin(a) + 2\pi k] \]
Solution for sin(x) ≤ a
\[ \sin x > a: \quad x \in (\arcsin(a) + 2\pi k, \pi - \arcsin(a) + 2\pi k) \]
Solution for sin(x) > a
\[ \sin x < a: \quad x \in (\pi - \arcsin(a) + 2\pi k, 2\pi + \arcsin(a) + 2\pi k) \]
Solution for sin(x) < a
\[ \sin x \geq 0: \quad x \in [2\pi k, \pi + 2\pi k] \]
Special Case: sin(x) is non-negative
\[ \sin x \leq 0: \quad x \in [\pi + 2\pi k, 2\pi + 2\pi k] \]
Special Case: sin(x) is non-positive
\[ \sin x \geq 1 \implies x = \frac{\pi}{2} + 2\pi k \]
Special Case: sin(x) at maximum
\[ \sin x \leq -1 \implies x = \frac{3\pi}{2} + 2\pi k \]
Special Case: sin(x) at minimum
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Diagram

c sin x > c x ∈ (arcsin(c)+2πk, π−arcsin(c)+2πk)
Sine inequation sin x > c: the solution is the arc where the sine curve is above y=c, between the two intersection points arcsin(c) and π−arcsin(c) per period.

A visual representation involves graphing the sine wave, y = sin(x), and a horizontal line, y = a. The solution to an inequality like sin(x) ≥ a consists of the intervals on the x-axis where the sine curve is on or above the horizontal line. These intervals repeat every 2π. Alternatively, on the unit circle, the solution corresponds to the arcs where the y-coordinate of points on the circle is greater than or equal to 'a'.

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Properties

Bounded Range: The sine function has a range of [-1, 1]. This means that for an inequality sin(x) ≥ a, if a > 1 there is no solution, and if a ≤ -1, the inequality is true for all real x.

Periodicity: The sine function is periodic with a period of 2π. If an interval [x₁, x₂] is a solution, then [x₁ + 2πk, x₂ + 2πk] is also a solution for any integer k.

Symmetry: The sine function has several key symmetries. It is an odd function, meaning sin(-x) = -sin(x). It also has a reflection property about the line x = π/2, given by sin(π - x) = sin(x). This supplementary angle identity is crucial for finding the second boundary point of a solution interval within one period.

\[ \sin(x + 2\pi k) = \sin x, \quad k \in \mathbb{Z} \]
Periodicity Property
\[ \sin(\pi - x) = \sin(x) \]
Supplementary Angle Identity
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Derivation Method

The solution to a sine inequality such as sin(x) ≥ a can be derived using the unit circle or the sine graph. Here we outline the graphical method.

\[ \text{Step 1: Graph } y = \sin x \text{ and the horizontal line } y = a \]
Visualize the functions

Draw the standard sine wave and a horizontal line at height 'a'. We assume -1 ≤ a ≤ 1, otherwise the solution is trivial.

\[ \text{Step 2: Find the intersection points in one period, e.g., } [0, 2\pi] \]
Solve the corresponding equation

Solve the equation sin(x) = a. The principal solution is x₁ = arcsin(a). Due to the symmetry sin(π - x) = sin(x), the second solution in the interval [0, 2π] is x₂ = π - arcsin(a).

\[ \text{Step 3: Identify the interval(s) where the inequality holds} \]
Analyze the graph

By observing the graph, the sine wave is above the line y = a between the two intersection points. Thus, for the interval [0, 2π], the solution is [x₁, x₂] or [arcsin(a), π - arcsin(a)].

\[ \text{Step 4: Generalize the solution using the period } 2\pi \]
Account for periodicity

Since the sine function repeats every 2π, we add 2πk to the endpoints of the interval to represent all possible solutions. This gives the general solution.

\[ x \in [\arcsin(a) + 2\pi k, \pi - \arcsin(a) + 2\pi k], \quad k \in \mathbb{Z} \]
General Solution
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Worked Example

Find all solutions for the inequality \( \sin(x) \geq \frac{1}{2} \).
  1. Identify the boundary value a = 1/2. This is within the range [-1, 1].
  2. Solve the corresponding equation sin(x) = 1/2. The principal value is x₁ = arcsin(1/2) = π/6.
  3. Find the second solution in the interval [0, 2π] using the identity sin(π - x) = sin(x). So, x₂ = π - π/6 = 5π/6.
  4. Determine the interval. Since we need sin(x) ≥ 1/2, we look for where the sine wave is above y = 1/2. This occurs between the two solutions found: [π/6, 5π/6].
  5. Generalize the solution by adding the period 2πk to the endpoints to account for all cycles.
  6. The complete solution set is x ∈ [π/6 + 2πk, 5π/6 + 2πk], where k is any integer.
\[ x \in \left[ \frac{\pi}{6} + 2\pi k, \frac{5\pi}{6} + 2\pi k \right], \quad k \in \mathbb{Z} \]
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Try It

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Applications

Physics & Wave Mechanics: Sine inequalities are used to determine time intervals when the displacement, velocity, or energy of an oscillating system (like a pendulum or a spring) exceeds a certain threshold. In wave mechanics, they help define regions where wave amplitude is above a critical level.

Electrical Engineering: In AC circuits, voltage and current are modeled by sine functions. Inequalities help find the duration for which voltage or current is above a specific value, which is crucial for analyzing power delivery and component safety.

Signal Processing: Engineers use sine inequalities to design filters and detectors. For instance, determining when a signal's amplitude is within a certain range can be used for noise reduction or signal detection.

Climate Science: Seasonal patterns, such as temperature or daylight hours, can be approximated by sine waves. Inequalities can be used to predict the periods of a year when temperatures will be above a certain point (e.g., growing season) or below freezing.

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Real-World Examples

The height of a tide (in meters) above mean sea level is modeled by the function h(t) = 3 sin(πt/6), where t is the time in hours after midnight. For safety, a ship needs the water depth to be at least 1.5 meters. During which time intervals in a 12-hour cycle (t=0 to t=12) is it safe for the ship to be in the harbor?
  1. Set up the inequality: 3 sin(πt/6) ≥ 1.5.
  2. Simplify the inequality: sin(πt/6) ≥ 0.5.
  3. Let u = πt/6. Solve sin(u) ≥ 0.5. From the worked example, we know π/6 ≤ u ≤ 5π/6.
  4. Substitute back for t: π/6 ≤ πt/6 ≤ 5π/6.
  5. Multiply all parts by 6/π to solve for t: 1 ≤ t ≤ 5.
  6. The period of the function is 2π / (π/6) = 12 hours. The solution interval occurs once within this cycle.
It is safe for the ship to be in the harbor between 1:00 AM and 5:00 AM.
An AC generator produces a voltage described by V(t) = 170 sin(120πt), where V is in volts and t is in seconds. A sensitive component can be damaged if the voltage exceeds 120 volts. Find the total time duration per cycle that the voltage is in the danger zone (V > 120 V).
  1. Set up the inequality: 170 sin(120πt) > 120.
  2. Simplify: sin(120πt) > 120/170 ≈ 0.7059.
  3. Let u = 120πt. Solve sin(u) > 0.7059.
  4. Find the boundary angles: u₁ = arcsin(0.7059) ≈ 0.7825 rad and u₂ = π - u₁ ≈ 2.3591 rad.
  5. The inequality holds for u in the interval (0.7825, 2.3591).
  6. Substitute back: 0.7825 < 120πt < 2.3591.
  7. Solve for t: 0.7825 / (120π) < t < 2.3591 / (120π), which gives 0.00207 s < t < 0.00626 s.
  8. The duration is the difference between the endpoints: 0.00626 - 0.00207 = 0.00419 seconds.
  9. The period is T = 2π / (120π) = 1/60 ≈ 0.0167 s. The duration is per cycle.
The voltage is in the danger zone for approximately 4.19 milliseconds per cycle.
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Real-World Scenarios

sin > c: arc window
Flood Risk Window in River Levels
A river's seasonal level follows h(t) = 2·sin(2πt/12) + 3 metres. Flood risk occurs when h > 4.5 m: sin(2πt/12) > 0.75, giving a two-month arc per year around peak season. Flood risk engineers solve this sine inequation to set warning thresholds and calculate the number of days per year when levee gates must be closed — enabling cost-benefit analysis of flood defence investments.
sin > 0.6 top arc: valid angles
Safe Knee Flexion Range in Physiotherapy
During rehabilitation, a patient's knee angle θ must stay within the range where sin θ > 0.6 (ensuring sufficient muscle activation). This gives θ ∈ (arcsin(0.6), π−arcsin(0.6)) ≈ (36.9°, 143.1°). Physiotherapy devices and motion-capture systems use this sine inequation to trigger real-time feedback when the joint moves outside the therapeutic range, preventing re-injury.
|sin x| < c: both halves
Vibration Amplitude Safety in Structures
A bridge deck vibrates as A(t) = A₀·sin(ωt). Safety requires |A(t)| < A_max: |sin(ωt)| < A_max/A₀, giving ωt outside arcs near π/2 and 3π/2. Structural engineers solve this sine inequation to identify resonance windows to avoid and to certify that wind and traffic loads won't push vibration amplitudes past fatigue limits — critical in the Tacoma Narrows Bridge failure analysis.

Seasonal Analysis
Ecologists and farmers model daylight hours using a sine function. They use sine inequalities to determine the specific date ranges when the length of the day exceeds a certain number of hours, which is critical for predicting plant growth cycles, animal behavior, and planning agricultural activities.

Structural Engineering
When analyzing the sway of a skyscraper or a bridge in the wind, engineers model the periodic displacement with sine functions. They use inequalities to ensure that the maximum displacement remains within safe structural limits, preventing material fatigue and ensuring public safety.

Audio Engineering
A sound engineer uses an audio compressor to manage the volume of a track. This device acts based on sine inequalities: when the amplitude of the sound wave (modeled as a sum of sines) exceeds a set threshold, the compressor reduces the volume to prevent distortion and create a more balanced sound.

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Classification of Solutions

The nature of the solution to an inequality like sin(x) ≥ a depends entirely on the value of 'a'.

Case (Value of 'a')Nature of Solution for sin(x) ≥ a
a > 1No solution (∅). The sine function never exceeds 1.
a = 1Discrete points. Solution is x = π/2 + 2πk.
-1 < a < 1A set of intervals. Solution is x ∈ [arcsin(a) + 2πk, π - arcsin(a) + 2πk].
a = -1All real numbers (ℝ). The sine function is always greater than or equal to -1.
a < -1All real numbers (ℝ). The sine function is always greater than any number less than -1.
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Common Mistakes

⚠️ Forgetting Periodicity: A common error is finding the solution only within the interval [0, 2π] and forgetting to add the '+ 2πk' term. This omits the infinite other solution intervals.
💡 Mixing up Intervals for ≤ and ≥: Students often confuse the solution interval for sin(x) ≥ a with that for sin(x) ≤ a. Always visualize the unit circle or sine graph. For '≥ a', you want the arc/curve *above* the line y=a; for '≤ a', you want the arc/curve *below* it.
⚠️ Incorrectly Handling Negative Coefficients: When solving an inequality like -2sin(x) > 1, you must divide by -2 and flip the inequality sign to get sin(x) < -1/2. Forgetting to flip the sign is a frequent mistake.
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Study Strategy

1 📚 Grasp the Core Concepts
  • Review the definition of the sine function and its graphical representation as a periodic wave.
  • Understand how an inequation like sin(x) > c defines a range of angles, not just specific points.
  • Use the unit circle diagram to visualize the intervals where the y-coordinate (sine value) is positive, negative, or within a certain range.
  • Study the property of periodicity (2π) and how it leads to an infinite number of solution intervals for a sine inequation.
2 🧠 Memorize Key Solution Patterns
  • Memorize the principal values of sin(x) for standard angles (e.g., 0, π/6, π/4, π/3, π/2).
  • Learn the general solution formula for the boundary equation sin(x) = c, which is x = nπ + (-1)^n * arcsin(c).
  • Internalize the method for identifying the solution interval within one period [0, 2π] for sin(x) > c and sin(x) < c.
  • Commit to memory how to express the final general solution by adding the period multiple (2nπ) to the interval endpoints.
3 ✍️ Solve and Analyze Problems
  • Deconstruct the provided worked example, focusing on how the boundary points are found and how the correct interval is selected.
  • Practice solving basic inequations like sin(x) ≥ 1/2 and sin(x) < -√2/2, sketching the graph to verify your answer.
  • Work through problems with transformed arguments, such as sin(2x) > 0 or sin(x - π/4) ≤ 0, noting how the period and phase shift affect the solution.
  • Analyze the 'Common Mistakes' section and attempt problems that test for errors like incorrect interval selection or forgetting the general solution.
4 🌍 Connect to Real-World Scenarios
  • Analyze the real-world examples, such as determining when an AC voltage is above a certain threshold.
  • Use a sine inequation to model and find the time intervals when the height of a tide is above a specific safety level.
  • Solve a problem involving oscillations, like finding when a swinging pendulum's displacement from the center is greater than a certain distance.
  • Formulate your own simple scenario, such as daily temperature variation, and write a sine inequation to find when the temperature is below freezing.
By systematically building from core concepts to real-world applications, you will master the logic behind solving any trigonometric sine inequation.

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