Inverse Trigonometric Functions

Inverse trigonometric functions allow us to determine an angle when we know the value of a trigonometric ratio.

In ordinary trigonometry, we begin with an angle and calculate a ratio. In inverse trigonometry, we reverse this process: we begin with a ratio and determine the angle.

For example:

sin 30° = 1/2

Using the inverse sine function:

sin⁻¹(1/2) = 30°

Inverse trigonometric functions are essential in geometry, navigation, physics, engineering, surveying, and calculus. They allow us to calculate angles from measurements, determine directions, and solve equations involving trigonometric relationships.

What Is an Inverse Trigonometric Function?

An inverse function reverses the operation of another function.

For example, consider:

y = sin θ

The sine function takes an angle θ and produces a numerical value.

The inverse sine function reverses this relationship:

θ = sin⁻¹

This means we are finding an angle whose sine equals y.

The three principal inverse trigonometric functions are:

• Inverse sine: sin⁻¹(x), also written arcsin(x)

• Inverse cosine: cos⁻¹(x), also written arccos(x)

• Inverse tangent: tan⁻¹(x), also written arctan(x)

These functions are sometimes called arc functions.

Example

We know:

cos 60° = 1/2

Therefore:

cos⁻¹(1/2) = 60°

Similarly:

tan 45° = 1

Therefore:

tan⁻¹(1) = 45°

Important Notation

The notation sin⁻¹(x) means inverse sine.

It does not mean 1/sin(x).

These are different operations:

• sin⁻¹(x) = inverse sine

• 1/sin(x) = cosecant, written csc(x)

For example:

sin⁻¹(1/2) = 30°

but:

1/sin 30° = 2

Why Trigonometric Functions Need Restricted Domains

An inverse function must produce exactly one output for each permitted input.

However, sine, cosine, and tangent are periodic functions. They repeat their values at different angles.

For example:

sin 30° = 1/2

sin 150° = 1/2

sin 390° = 1/2

If inverse sine returned every angle with sine equal to 1/2, it would produce infinitely many answers.

To create an inverse function, we restrict the original trigonometric function to an interval where it is one-to-one.

A function is one-to-one when different inputs produce different outputs.

The chosen interval determines the principal value returned by the inverse function.

Principal Values

A principal value is the single angle returned by an inverse trigonometric function.

The principal-value intervals are:

In radians:

Notice that inverse tangent never returns exactly −90° or 90°.

These angles are excluded because tangent is undefined there.

The Inverse Sine Function

The inverse sine function is:

y = sin⁻¹(x)

It answers the question:

What angle has a sine equal to x?

The domain is:

−1 ≤ x ≤ 1

The range is:

−π/2 ≤ y ≤ π/2

or:

−90° ≤ y ≤ 90°

Common Exact Values

Worked Example

Evaluate:

sin⁻¹(√3/2)

We know:

sin 60° = √3/2

Since 60° lies in the principal-value range:

sin⁻¹(√3/2) = 60° = π/3

Negative Values

Evaluate:

sin⁻¹(−1/2)

We know:

sin(−30°) = −1/2

Therefore:

sin⁻¹(−1/2) = −30° = −π/6

Domain Restrictions

The expression:

sin⁻¹(2)

is undefined over the real numbers because sine cannot produce a value greater than 1.

The Inverse Cosine Function

The inverse cosine function is:

y = cos⁻¹(x)

It answers:

What angle has a cosine equal to x?

Its domain is:

−1 ≤ x ≤ 1

Its range is:

0 ≤ y ≤ π

or:

0° ≤ y ≤ 180°

Common Exact Values

Worked Example

Evaluate:

cos⁻¹(−1/2)

We know:

cos 120° = −1/2

Therefore:

cos⁻¹(−1/2) = 120° = 2π/3

Why the Answer Is Not 240°

Cosine is also −1/2 at 240°.

However, inverse cosine returns only angles between 0° and 180°.

Therefore, 120° is the principal value.

The Inverse Tangent Function

The inverse tangent function is:

y = tan⁻¹(x)

It answers:

What angle has a tangent equal to x?

Its domain is:

All real numbers

Its range is:

−π/2 < y < π/2

or:

−90° < y < 90°

Common Exact Values

Worked Example

Evaluate:

tan⁻¹(√3)

Since:

tan 60° = √3

Therefore:

tan⁻¹(√3) = 60° = π/3

Worked Example

Evaluate:

tan⁻¹(−1)

Since:

tan(−45°) = −1

Therefore:

tan⁻¹(−1) = −45° = −π/4

Comparing the Three Principal Inverse Functions

The inverse sine and inverse cosine functions have restricted domains because sine and cosine values cannot exceed 1 or fall below −1.

Inverse tangent has no such domain restriction because tangent can produce any real value.

Interpreting Inverse Functions Graphically

The graph of an inverse function is the reflection of the original function's restricted graph across the line:

y = x

This happens because the input and output values exchange roles.

If the original function contains a point:

(a, b)

then its inverse contains:

(b, a)

Example: Sine and Inverse Sine

The sine function contains:

(π/6, 1/2)

The inverse sine function contains:

(1/2, π/6)

The coordinates are reversed.

Example: Cosine and Inverse Cosine

The cosine function contains:

(π/3, 1/2)

The inverse cosine function contains:

(1/2, π/3)

Again, the coordinates are reversed.

Example: Tangent and Inverse Tangent

The tangent function contains:

(π/4, 1)

The inverse tangent function contains:

(1, π/4)

The graphs reflect across y = x when tangent is restricted to its principal interval.

Important Graphical Features

Inverse Sine

The graph:

• Begins at (−1, −π/2).

• Passes through (0, 0).

• Ends at (1, π/2).

• Increases continuously.

Inverse Cosine

The graph:

• Begins at (−1, π).

• Passes through (0, π/2).

• Ends at (1, 0).

• Decreases continuously.

Inverse Tangent

The graph:

• Passes through (0, 0).

• Increases continuously.

• Approaches y = π/2 as x increases without bound.

• Approaches y = −π/2 as x decreases without bound.

The horizontal asymptotes are never reached.

Evaluating Inverse Trigonometric Expressions

Inverse trigonometric expressions may involve fractions, radicals, and combinations of functions.

Worked Example: sin⁻¹(1/2) + cos⁻¹(1/2)

Evaluate each function.

sin⁻¹(1/2) = π/6

cos⁻¹(1/2) = π/3

Add:

π/6 + π/3

= π/6 + 2π/6

= π/2

Worked Example: 2tan⁻¹(1)

tan⁻¹(1) = π/4

Therefore:

2tan⁻¹(1) = 2(π/4)

= π/2

Worked Example: cos⁻¹(−√2/2) − sin⁻¹(1/2)

cos⁻¹(−√2/2) = 3π/4

sin⁻¹(1/2) = π/6

Subtract:

3π/4 − π/6

= 9π/12 − 2π/12

= 7π/12

Compositions of Trigonometric and Inverse Functions

A composition involves applying one function to the output of another.

For example:

sin(sin⁻¹ x) = x

for −1 ≤ x ≤ 1.

Similarly:

cos(cos⁻¹ x) = x

for −1 ≤ x ≤ 1.

And:

tan(tan⁻¹ x) = x

for every real x.

These identities work because the inverse function reverses the original operation.

Worked Example

Evaluate:

sin[sin⁻¹(3/5)]

The inverse sine produces an angle whose sine is 3/5.

Applying sine returns the original value.

Answer = 3/5

Important Restriction

The reverse composition requires more care.

For example:

sin⁻¹(sin 150°)

We know:

sin 150° = 1/2

Therefore:

sin⁻¹(1/2) = 30°

sin⁻¹(sin 150°) = 30°

It is not 150° because inverse sine must return a principal value.

Another Example

Evaluate:

cos⁻¹(cos 240°)

cos 240° = −1/2

Therefore:

cos⁻¹(−1/2) = 120°

Answer = 120°

Using Right Triangles to Evaluate Compositions

Inverse trigonometric functions can be combined with other trigonometric functions.

For example:

cos[sin⁻¹(3/5)]

Let:

θ = sin⁻¹(3/5)

Then:

sin θ = 3/5

Since θ lies in the principal range of inverse sine and sine is positive, θ is in Quadrant I.

Construct a right triangle with:

• Opposite side = 3

• Hypotenuse = 5

Using the Pythagorean theorem:

Adjacent² = 5² − 3²

Adjacent² = 25 − 9

Adjacent = 4

Therefore:

cos θ = adjacent/hypotenuse

cos[sin⁻¹(3/5)] = 4/5

Worked Example

Evaluate:

sin[tan⁻¹(2)]

Let:

θ = tan⁻¹(2)

Then:

tan θ = 2 = 2/1

Construct a right triangle with:

• Opposite = 2

• Adjacent = 1

Find the hypotenuse:

r = √(2² + 1²)

r = √5

Therefore:

sin θ = 2/√5

Rationalise:

sin[tan⁻¹(2)] = 2√5/5

Solving Equations with Inverse Trigonometric Functions

Inverse trigonometric functions can be used to solve equations in which an unknown angle appears inside a trigonometric function.

Worked Example: sin θ = 0.6

Find the acute angle θ.

Apply inverse sine:

θ = sin⁻¹(0.6)

Using a calculator in degree mode:

θ ≈ 36.87°

Worked Example: cos θ = 0.25

For an acute angle:

θ = cos⁻¹(0.25)

θ ≈ 75.52°

Worked Example: tan θ = 1.5

For an acute angle:

θ = tan⁻¹(1.5)

θ ≈ 56.31°

These calculations are common in right-triangle problems.

Finding All Solutions to Trigonometric Equations

An inverse trigonometric function returns one principal angle.

However, a trigonometric equation may have several solutions over a specified interval.

Worked Example

Solve:

sin θ = 1/2

for:

0° ≤ θ < 360°

First find the principal angle:

θ = sin⁻¹(1/2)

θ = 30°

Sine is positive in Quadrants I and II.

The second angle is:

180° − 30° = 150°

Therefore:

θ = 30° or 150°

Worked Example

Solve:

cos θ = −1/2

for:

0° ≤ θ < 360°

The principal inverse cosine value is:

cos⁻¹(−1/2) = 120°

Cosine is negative in Quadrants II and III.

The second solution is:

360° − 120° = 240°

Therefore:

θ = 120° or 240°

Worked Example

Solve:

tan θ = 1

for:

0° ≤ θ < 360°

The principal angle is:

tan⁻¹(1) = 45°

Tangent repeats every 180°.

Therefore:

45° + 180° = 225°

θ = 45° or 225°

Solving Equations Containing Inverse Functions

Sometimes the inverse function itself appears in an equation.

Worked Example

Solve:

sin⁻¹(x) = π/6

Apply sine to both sides:

x = sin(π/6)

x = 1/2

This satisfies the domain of inverse sine.

Worked Example

Solve:

2cos⁻¹(x) = 2π/3

Divide by 2:

cos⁻¹(x) = π/3

Apply cosine:

x = cos(π/3)

x = 1/2

Worked Example

Solve:

tan⁻¹(2x − 1) = π/4

Apply tangent:

2x − 1 = tan(π/4)

2x − 1 = 1

2x = 2

x = 1

Inverse Trigonometry in Right-Triangle Problems

One of the most common applications of inverse trigonometric functions is finding unknown angles in right triangles.

Choosing the Correct Function

Use inverse sine when the opposite side and hypotenuse are known.

θ = sin⁻¹(opposite/hypotenuse)

Use inverse cosine when the adjacent side and hypotenuse are known.

θ = cos⁻¹(adjacent/hypotenuse)

Use inverse tangent when the opposite and adjacent sides are known.

θ = tan⁻¹(opposite/adjacent)

Worked Example: Finding an Angle

A right triangle has:

• Opposite side = 8 cm

• Adjacent side = 15 cm

Find θ.

Use tangent:

tan θ = 8/15

Therefore:

θ = tan⁻¹(8/15)

θ ≈ 28.07°

The unknown angle is approximately 28.1°.

Application: Angle of Elevation

An observer stands 40 m from the base of a building.

The top of the building is 30 m above the observer's eye level.

Find the angle of elevation.

Identify the Triangle

Opposite = 30 m

Adjacent = 40 m

Use tangent:

tan θ = 30/40

tan θ = 0.75

Apply inverse tangent:

θ = tan⁻¹(0.75)

θ ≈ 36.87°

The angle of elevation is approximately 36.9°.

Real-World Connection

Surveyors and engineers use similar calculations to determine heights and angles without directly measuring inaccessible distances.

Application: Navigation

A ship travels 12 km east and 5 km north.

Determine its direction relative to east.

Identify the Components

Horizontal displacement = 12 km

Vertical displacement = 5 km

The direction angle satisfies:

tan θ = 5/12

Therefore:

θ = tan⁻¹(5/12)

θ ≈ 22.62°

The ship's displacement is directed approximately:

22.6° north of east

The total displacement is:

d = √(12² + 5²)

d = √169

d = 13 km

Important Note About Direction

The ordinary inverse tangent function returns angles between −90° and 90°.

When finding a direction from horizontal and vertical components, the signs of both components must be considered to determine the correct quadrant.

For example, a vector pointing southwest requires an angle in Quadrant III, even though tan⁻¹(y/x) alone may return a positive acute angle.

In computing, the function often called atan2(y, x) accounts for both coordinate signs and determines the appropriate direction.

Application: Inclined Planes

A ramp rises 1.5 m over a horizontal distance of 6 m.

Find the angle the ramp makes with the horizontal.

Identify the Ratio

Opposite = 1.5 m

Adjacent = 6 m

Therefore:

tan θ = 1.5/6

tan θ = 0.25

Apply inverse tangent:

θ = tan⁻¹(0.25)

θ ≈ 14.04°

The ramp makes an angle of approximately 14° with the horizontal.

This calculation is useful in architecture, accessibility design, and civil engineering.

Application: Forces and Vectors

A force has horizontal and vertical components:

Fₓ = 24 N

Fᵧ = 10 N

Determine the magnitude and direction of the force.

Find the Magnitude

F = √(Fₓ² + Fᵧ²)

F = √(24² + 10²)

F = √676

F = 26 N

Find the Direction

tan θ = Fᵧ/Fₓ

tan θ = 10/24

θ = tan⁻¹(10/24)

θ ≈ 22.62°

The force has magnitude 26 N and acts approximately 22.6° above the positive horizontal direction.

This method is widely used in mechanics, engineering, and vector analysis.

Application: Calculus

Inverse trigonometric functions also appear in differentiation and integration.

Their derivatives include:

d/dx[arcsin(x)] = 1/√(1 − x²)

for −1 < x < 1.

d/dx[arccos(x)] = −1/√(1 − x²)

for −1 < x < 1.

d/dx[arctan(x)] = 1/(1 + x²)

for all real x.

These formulas assume angles are measured in radians.

Example: Differentiating Inverse Tangent

Let:

f(x) = arctan(3x)

Using the chain rule:

f′(x) = 3/(1 + (3x)²)

Therefore:

f′(x) = 3/(1 + 9x²)

Connection to Integration

Inverse tangent appears when integrating expressions such as:

∫ 1/(1 + x²) dx

The result is:

arctan(x) + C

Inverse trigonometric functions are therefore important in advanced integration techniques.

Using Calculators Correctly

Most scientific calculators include inverse trigonometric functions.

They are commonly accessed using a secondary function key.

For example:

• sin⁻¹

• cos⁻¹

• tan⁻¹

Degree Mode

If a question asks for an angle in degrees, use degree mode.

For example:

tan⁻¹(1) = 45°

Radian Mode

If a question asks for radians, use radian mode.

For example:

tan⁻¹(1) = π/4

Approximately:

0.7854 radians

Checking Results

A useful way to verify an inverse trigonometric calculation is to apply the original trigonometric function.

For example:

θ = sin⁻¹(0.6) ≈ 36.87°

Check:

sin 36.87° ≈ 0.6

The result is consistent.

A Strategy for Inverse Trigonometric Problems

When solving a problem involving inverse trigonometric functions:

1. Identify the known trigonometric ratio or expression.

2. Determine which inverse function is appropriate.

3. Check that the input lies within the function's domain.

4. Find the principal value.

5. Determine whether additional angles are required.

6. Use quadrant information when necessary.

7. Convert between degrees and radians if required.

8. Check the result using the original trigonometric function.

9. Include units and interpret the result in context.

This process helps prevent errors involving principal values and quadrant restrictions.

Common Mistakes

Confusing inverse functions with reciprocals

sin⁻¹(x) is inverse sine, while 1/sin(x) is cosecant.

Ignoring principal-value ranges

Inverse trigonometric functions return values only within their defined principal intervals.

Assuming inverse sine returns every possible angle

Inverse sine returns one principal angle, not all solutions to a trigonometric equation.

Forgetting additional solutions

An equation such as sin θ = 1/2 has multiple solutions over a full revolution.

Using an input outside the domain

For real-valued inverse sine and inverse cosine, the input must be between −1 and 1.

Confusing degrees and radians

Calculator settings must match the units required by the problem.

Assuming sin⁻¹(sin θ) always equals θ

This is true only when θ lies within the principal-value range of inverse sine.

Ignoring quadrant information in vector problems

Inverse tangent alone may not determine the correct direction.

Using the wrong trigonometric ratio

Choose sine, cosine, or tangent based on the known sides of the right triangle.

Rounding too early

Keep sufficient precision during calculations and round only the final answer.

Forgetting units

Practical problems may require angles in degrees or radians and distances in metres or kilometres.

Key Terms

Inverse Function: A function that reverses the relationship between the inputs and outputs of another function.

Inverse Trigonometric Function: A function that determines an angle from a trigonometric ratio.

Inverse Sine: The function arcsin(x), which returns an angle whose sine equals x.

Inverse Cosine: The function arccos(x), which returns an angle whose cosine equals x.

Inverse Tangent: The function arctan(x), which returns an angle whose tangent equals x.

Principal Value: The unique angle returned by an inverse trigonometric function within its specified range.

Restricted Domain: A limited interval of input values chosen to make a function one-to-one.

One-to-One Function: A function in which different inputs produce different outputs.

Domain: The set of permitted input values.

Range: The set of possible output values.

Periodic Function: A function whose values repeat at regular intervals.

Composition of Functions: Applying one function to the output of another.

Inverse Relationship: A relationship in which two functions reverse each other's operations on appropriate domains.

Reflection Across y = x: The geometric relationship between the graph of a one-to-one function and its inverse.

Horizontal Asymptote: A horizontal line that a graph approaches as the input increases or decreases without bound.

Exact Value: A value expressed without rounding, often using fractions, radicals, or π.

Radian: A unit of angular measurement based on the ratio of arc length to radius.

Degree: A unit of angular measurement in which a complete revolution contains 360°.

Angle of Elevation: The angle measured upward from the horizontal to an object.

Angle of Depression: The angle measured downward from the horizontal to an object.

Vector Component: The part of a vector acting along a particular coordinate direction.

Direction Angle: An angle describing the orientation of a vector.

Quadrant: One of the four regions of the Cartesian coordinate plane.

Arctangent2: A two-input angle calculation that uses horizontal and vertical components to determine the correct quadrant.

Chain Rule: A differentiation rule used for composite functions.

Antiderivative: A function whose derivative equals a given function.

Check Your Understanding

1. Define an inverse trigonometric function and explain how it differs from a reciprocal trigonometric function.

2. State the domains and principal-value ranges of arcsin(x), arccos(x), and arctan(x).

3. Evaluate sin⁻¹(1/2) in degrees and radians.

4. Evaluate cos⁻¹(−√3/2).

5. Evaluate tan⁻¹(−1).

6. Explain why sin⁻¹(2) is undefined over the real numbers.

7. Evaluate sin⁻¹(sin 150°).

8. Evaluate cos⁻¹(cos 240°).

9. Simplify cos[sin⁻¹(5/13)].

10. Solve 2tan⁻¹(x) = π/2.

11. Solve sin θ = √3/2 for 0° ≤ θ < 360°.

12. Solve cos θ = −√2/2 for 0° ≤ θ < 360°.

13. A ladder reaches 6 m up a wall and its base is 2.5 m from the wall. Calculate the angle the ladder makes with the ground.

14. A displacement vector has components 8 km east and 15 km south. Determine its magnitude and direction relative to east.

15. A drone travels 120 m horizontally and rises 45 m vertically. Determine its angle of elevation and total displacement. Explain which inverse trigonometric function you used and why.

Key Takeaways

• Inverse trigonometric functions determine angles from known trigonometric ratios.

• The three principal inverse functions are arcsin(x), arccos(x), and arctan(x).

• Inverse trigonometric functions are different from reciprocal trigonometric functions.

• Trigonometric functions require restricted domains to have well-defined inverse functions.

• A principal value is the unique angle returned by an inverse trigonometric function.

• The domain of arcsin(x) is [−1, 1], and its range is [−π/2, π/2].

• The domain of arccos(x) is [−1, 1], and its range is [0, π].

• The domain of arctan(x) is all real numbers, and its range is (−π/2, π/2).

• Inverse sine and inverse tangent are increasing functions.

• Inverse cosine is a decreasing function.

• The graphs of inverse functions are reflections of their restricted original functions across y = x.

• The inverse tangent graph has horizontal asymptotes at y = ±π/2.

• Applying a trigonometric function to its corresponding inverse returns the original input when that input is in the inverse function's domain.

• Applying an inverse trigonometric function to a trigonometric function may return a principal angle different from the original angle.

• Solving trigonometric equations may require finding additional solutions beyond the principal value.

• Inverse trigonometric functions are useful for calculating unknown angles in right triangles.

• Inverse tangent is commonly used to calculate angles of elevation, slopes, and vector directions.

• Quadrant information is essential when determining directions from vector components.

• Calculator angle settings must match the required units.

• Inverse trigonometric functions also appear in differentiation and integration.

• Understanding inverse trigonometric functions provides a foundation for solving geometric, scientific, engineering, and advanced calculus problems involving angles and directions.