Inverse Trigonometric Functions

Trigonometric functions allow us to find a ratio or coordinate when we know an angle. Inverse trigonometric functions allow us to work in the opposite direction: they help us find an angle when we know a trigonometric value.

For example:

sin(30°) = 0.5

Therefore:

sin⁻¹(0.5) = 30°

Similarly:

cos(60°) = 0.5

so:

cos⁻¹(0.5) = 60°

Inverse trigonometric functions are especially important in geometry, physics, engineering, navigation, surveying, waves, and mathematical modelling.

What Does "Inverse" Mean?

An inverse function reverses the action of another function.

Consider:

sin(30°) = 0.5

The sine function takes:

angle → ratio

The inverse sine function takes:

ratio → angle

So:

sin⁻¹(0.5) = 30°

This relationship can be thought of as:

30° → sin → 0.5

and:

0.5 → sin⁻¹ → 30°

The same idea applies to cosine and tangent.

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The Three Main Inverse Trigonometric Functions

The three most common 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)

The "arc" notation is often useful because it avoids confusion with exponents.

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Important: sin⁻¹(x) Does Not Mean 1/sin(x)

This is one of the most important distinctions in this topic.

sin⁻¹(x) means:

inverse sine

It does not mean:

1/sin(x)

The reciprocal of sine is:

csc(x) = 1/sin(x)

Similarly:

cos⁻¹(x) means inverse cosine.

But:

1/cos(x) = sec(x)

And:

tan⁻¹(x) means inverse tangent.

But:

1/tan(x) = cot(x)

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Why Do We Need Principal Values?

There is a problem with defining inverse trigonometric functions.

Trigonometric functions repeat.

For example:

sin(30°) = 0.5

but also:

sin(150°) = 0.5

and:

sin(390°) = 0.5

and infinitely many other angles have the same sine value.

So what should:

sin⁻¹(0.5)

equal?

An inverse function must give one output for each input.

Therefore, we restrict the output of inverse trigonometric functions to specific intervals called principal-value ranges.

Principal Values

The principal value is the single angle selected by an inverse trigonometric function.

The standard principal ranges are:

These ranges must be remembered or understood because they determine which answer an inverse function returns.

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Principal Range of arcsin

The inverse sine function returns angles in:

−π/2 ≤ y ≤ π/2

or:

−90° ≤ y ≤ 90°

This corresponds primarily to Quadrants I and IV, including the boundary angles.

For example:

arcsin(1/2) = 30°

not 150°.

Even though:

sin(150°) = 1/2

150° is outside the principal range of arcsin.

Therefore:

arcsin(1/2) = 30°

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Principal Range of arccos

The inverse cosine function returns angles in:

0 ≤ y ≤ π

or:

0° ≤ y ≤ 180°

This covers Quadrants I and II.

For example:

arccos(−1/2) = 120°

because:

cos(120°) = −1/2

and 120° lies within the principal range.

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Principal Range of arctan

The inverse tangent function returns angles in:

−π/2 < y < π/2

or:

−90° < y < 90°

For example:

arctan(1) = 45°

and:

arctan(−1) = −45°

Notice that ±90° are not included because tangent is undefined at these angles.

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Exact Principal Values

Some inverse trigonometric expressions can be evaluated exactly.

Inverse Sine

arcsin(0) = 0

arcsin(1/2) = π/6

arcsin(√2/2) = π/4

arcsin(√3/2) = π/3

arcsin(1) = π/2

Inverse Cosine

arccos(1) = 0

arccos(√3/2) = π/6

arccos(√2/2) = π/4

arccos(1/2) = π/3

arccos(0) = π/2

Inverse Tangent

arctan(0) = 0

arctan(√3/3) = π/6

arctan(1) = π/4

arctan(√3) = π/3

These values come directly from familiar trigonometric angles.

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Worked Example: Inverse Sine

Evaluate:

arcsin(√3/2)

Ask:

Which angle in the principal range has sine √3/2?

We know:

sin(π/3) = √3/2

Therefore:

arcsin(√3/2) = π/3

or:

60°

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Worked Example: Inverse Cosine

Evaluate:

arccos(−√2/2)

We need an angle between:

0 and π

whose cosine is negative.

Cosine is negative in Quadrant II within this range.

The reference angle is:

π/4

Therefore:

arccos(−√2/2) = 3π/4

or:

135°

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Worked Example: Inverse Tangent

Evaluate:

arctan(−√3)

We need an angle between:

−π/2 and π/2

with tangent −√3.

The reference angle is:

π/3

Within the principal range, the negative angle is:

−π/3

Therefore:

arctan(−√3) = −π/3

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Domain of Inverse Sine

Since ordinary sine produces values only between −1 and 1:

−1 ≤ sin(x) ≤ 1

the input of arcsin must satisfy:

−1 ≤ x ≤ 1

Therefore, the domain of:

y = arcsin(x)

is:

[−1, 1]

For example:

arcsin(0.7)

exists.

But:

arcsin(2)

has no real value.

There is no real angle whose sine is 2.

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Domain of Inverse Cosine

Cosine also produces values only between −1 and 1.

Therefore:

−1 ≤ x ≤ 1

is the domain of:

arccos(x)

For example:

arccos(−0.4)

has a real value.

But:

arccos(1.5)

does not have a real value.

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Domain of Inverse Tangent

Tangent can produce any real number.

Therefore:

arctan(x)

has domain:

all real numbers

So expressions such as:

arctan(5)

arctan(100)

arctan(−250)

are all defined.

However, the resulting angles always remain between:

−π/2 and π/2

Why Restrictions Are Necessary

The ordinary sine function is not one-to-one over all real numbers.

For example:

sin(30°) = 1/2

sin(150°) = 1/2

Two different inputs produce the same output.

That prevents the unrestricted sine function from having an inverse that is itself a function.

To create arcsin, we restrict sine to:

−90° ≤ x ≤ 90°

Over this interval, sine is one-to-one.

Then we can reflect this restricted graph across:

y = x

to obtain the graph of arcsin.

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Inverse Functions as Reflections

A function and its inverse have graphs that are reflections of each other across the line:

y = x

For example, restricted:

y = sin(x)

and:

y = arcsin(x)

are reflections across y = x.

Coordinates switch positions.

If:

sin(a) = b

then:

arcsin(b) = a

So a point:

(a, b)

on the restricted sine graph corresponds to:

(b, a)

on the arcsin graph.

The Graph of arcsin(x)

The graph:

y = arcsin(x)

has:

Domain:

−1 ≤ x ≤ 1

Range:

−π/2 ≤ y ≤ π/2

It is increasing throughout its domain.

Important points include:

(−1, −π/2)

(0, 0)

(1, π/2)

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The Graph of arccos(x)

The graph:

y = arccos(x)

has:

Domain:

−1 ≤ x ≤ 1

Range:

0 ≤ y ≤ π

Unlike arcsin, arccos is decreasing.

Important points include:

(−1, π)

(0, π/2)

(1, 0)

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The Graph of arctan(x)

The graph:

y = arctan(x)

has:

Domain:

all real numbers

Range:

−π/2 < y < π/2

The graph approaches:

y = π/2

and:

y = −π/2

but never reaches them.

These are horizontal asymptotes.

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Calculator Mode Matters

When evaluating inverse trigonometric functions with a calculator, check whether the calculator is in:

• degree mode
• radian mode

For example:

arcsin(0.5)

in degree mode gives:

30°

In radian mode it gives approximately:

0.524

These are the same angle because:

π/6 ≈ 0.524 radians

A correct calculation in the wrong mode may appear to give the wrong answer.

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Using Inverse Trigonometry to Solve Equations

Suppose:

sin(x) = 0.7

We can apply inverse sine:

x = arcsin(0.7)

Using a calculator:

x ≈ 44.4°

But this is only the principal value.

If we are solving a trigonometric equation over a larger interval, there may be additional solutions.

This distinction is extremely important.

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Principal Value vs All Solutions

Suppose:

sin(x) = 0.5

Then:

arcsin(0.5) = 30°

This gives the principal value.

But if the problem asks for all solutions between:

0° ≤ x < 360°

then sine is positive in Quadrants I and II.

Therefore:

x = 30°

or:

x = 150°

The inverse function gives a starting angle, not necessarily every solution to a periodic trigonometric equation.

Solving a Cosine Equation

Solve:

cos(x) = −0.6

for:

0° ≤ x < 360°

First find the principal angle:

x = arccos(−0.6)

≈ 126.9°

Cosine is negative in Quadrants II and III.

The second solution is:

360° − 126.9°

= 233.1°

Therefore:

x ≈ 126.9°, 233.1°

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Solving a Tangent Equation

Solve:

tan(x) = 2

for:

0° ≤ x < 360°

Find the principal value:

x = arctan(2)

≈ 63.4°

Tangent is positive in Quadrants I and III.

Therefore, the second solution is:

63.4° + 180°

= 243.4°

Solutions:

x ≈ 63.4°, 243.4°

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General Solutions

Because trigonometric functions are periodic, equations can have infinitely many solutions.

For:

sin(x) = a

if:

α = arcsin(a)

then solutions can be represented using:

x = α + 2πn

or:

x = π − α + 2πn

where n is any integer.

For cosine:

cos(x) = a

if:

α = arccos(a)

then:

x = ±α + 2πn

For tangent:

tan(x) = a

if:

α = arctan(a)

then:

x = α + πn

These forms describe infinitely many repeating solutions.

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Solving Equations with Coefficients

Consider:

2sin(x) = 1

First isolate the trigonometric function:

sin(x) = 1/2

Now apply inverse sine:

arcsin(1/2) = 30°

The principal angle is:

30°

If:

0° ≤ x < 360°

then the second solution is:

150°

Therefore:

x = 30°, 150°

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Solving an Equation with a Shift

Consider:

cos(x − 20°) = 0.5

Let:

u = x − 20°

Then:

cos(u) = 0.5

Over one full cycle:

u = 60°

or:

u = 300°

Now substitute back.

x − 20° = 60°

x = 80°

or:

x − 20° = 300°

x = 320°

Always check the requested interval when solving transformed trigonometric equations.

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Inverse Trigonometry and Right Triangles

One of the most common real-world uses of inverse trigonometric functions is finding an unknown angle in a right triangle.

Recall:

sin(θ) = opposite/hypotenuse

cos(θ) = adjacent/hypotenuse

tan(θ) = opposite/adjacent

If the side lengths are known, inverse functions allow us to determine θ.

Worked Example: Finding an Angle

A right triangle has:

opposite = 6 cm

adjacent = 8 cm

Use tangent:

tan(θ) = 6/8

= 0.75

Apply inverse tangent:

θ = arctan(0.75)

θ ≈ 36.9°

So the angle is approximately:

36.9°

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Choosing the Correct Inverse Function

The side information determines which function to use.

Opposite and Hypotenuse

Use:

θ = arcsin(opposite/hypotenuse)

Adjacent and Hypotenuse

Use:

θ = arccos(adjacent/hypotenuse)

Opposite and Adjacent

Use:

θ = arctan(opposite/adjacent)

A useful strategy is:

1. label the sides relative to the unknown angle
2. choose the appropriate ratio
3. calculate the ratio
4. apply the corresponding inverse function

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Real-World Example: Angle of Elevation

Suppose you stand 40 m from the base of a building.

The building is 30 m high.

Ignoring eye height, let θ be the angle of elevation to the top.

Then:

tan(θ) = 30/40

= 0.75

Therefore:

θ = arctan(0.75)

≈ 36.9°

Inverse tangent converts the measured ratio into an angle.

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Real-World Example: Ramp Angle

A ramp rises 1.2 m over a horizontal distance of 8 m.

Then:

tan(θ) = 1.2/8

= 0.15

Therefore:

θ = arctan(0.15)

≈ 8.5°

This kind of calculation is useful in construction and engineering.

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Real-World Example: Direction of Motion

Suppose an object moves:

12 m east

and:

5 m north.

Its direction relative to east can be found using:

tan(θ) = 5/12

Therefore:

θ = arctan(5/12)

≈ 22.6°

So the displacement direction is approximately:

22.6° north of east

Inverse trigonometry is extremely useful when working with vectors.

Inverse Trigonometry and Vectors

Suppose a vector has components:

Vₓ = 8

Vᵧ = 6

Its magnitude is:

V = √(8² + 6²)

= √100

= 10

Its direction can be found using:

tan(θ) = Vᵧ/Vₓ

= 6/8

Therefore:

θ = arctan(6/8)

≈ 36.9°

The inverse tangent converts component information into directional information.

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Quadrants Matter with Vectors

There is an important complication.

Suppose:

Vₓ = −8

Vᵧ = 6

If we simply calculate:

arctan(6/−8)

we obtain a negative principal angle.

But the vector actually lies in Quadrant II.

Therefore, when finding vector directions, we must examine the signs of the components and determine the correct quadrant.

Many programming languages and calculators provide a function often called atan2(y, x) that uses both component signs to determine the appropriate direction.

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Inverse Functions in Modelling

Inverse trigonometric functions appear whenever a model gives us a ratio and we need to recover an angle.

Applications include:

• surveying
• navigation
• astronomy
• robotics
• engineering
• computer graphics
• architecture
• physics
• satellite positioning
• slope measurement

In many cases, sensors measure distances or components first, and inverse trigonometry converts those measurements into angles.

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Example: Surveying

A surveyor measures:

horizontal distance = 120 m

vertical rise = 18 m

The angle of elevation is:

tan(θ) = 18/120

= 0.15

Therefore:

θ = arctan(0.15)

≈ 8.5°

This allows geometric measurements to be converted into useful directional information.

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Inverse Trigonometry and Slopes

Recall that:

slope = rise/run

For an inclined line:

tan(θ) = rise/run

Therefore:

tan(θ) = slope

and:

θ = arctan(slope)

This creates an important connection between algebra and trigonometry.

For example, if:

slope = 0.5

then:

θ = arctan(0.5)

≈ 26.6°

So the line forms an angle of approximately 26.6° with the positive horizontal axis, subject to the line's direction and quadrant.

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Example: Road Grade

Suppose a road has a grade of:

10%

This means approximately:

rise/run = 0.10

Therefore:

tan(θ) = 0.10

θ = arctan(0.10)

≈ 5.7°

A 10% grade does not mean the road is inclined at 10°.

This is an important real-world distinction.

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Compositions with Inverse Functions

Consider:

sin(arcsin(x))

For:

−1 ≤ x ≤ 1

the inverse sine produces an angle whose sine is x.

Therefore:

sin(arcsin(x)) = x

Similarly:

cos(arccos(x)) = x

for:

−1 ≤ x ≤ 1

And:

tan(arctan(x)) = x

for every real x.

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Reversing the Order Is More Complicated

It may seem that:

arcsin(sin(x)) = x

for every x.

But this is not true.

Remember that arcsin can only return values in:

−π/2 ≤ y ≤ π/2

For example:

sin(5π/6) = 1/2

Then:

arcsin(sin(5π/6))

= arcsin(1/2)

= π/6

not:

5π/6

The inverse function returns the principal value.

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Another Composition Example

Consider:

arccos(cos(4π/3))

First:

cos(4π/3) = −1/2

Then:

arccos(−1/2) = 2π/3

not 4π/3.

Why?

Because the principal range of arccos is:

0 ≤ y ≤ π

and 4π/3 lies outside this range.

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Graphical Interpretation of Principal Values

The restricted domains used to define inverse functions explain principal values.

For sine, we retain the section from:

−π/2 to π/2

For cosine:

0 to π

For tangent:

−π/2 to π/2

These restricted sections are one-to-one.

Their inverses therefore pass the vertical line test and behave as proper functions.

Checking an Inverse-Trig Answer

When evaluating an inverse trigonometric expression, use this process:

Check the Input

Is the input in the domain?

For arcsin and arccos:

−1 ≤ input ≤ 1

Determine the Reference Angle

Use known exact values or a calculator.

Check the Sign

Is the trigonometric value positive or negative?

Check the Principal Range

Does your angle lie in the correct output range?

Check the Calculator Mode

Should the answer be in degrees or radians?

This method prevents many common errors.

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Solving vs Evaluating

It is important to distinguish these two tasks.

Evaluate

arcsin(1/2)

asks for one principal value:

π/6

Solve

sin(x) = 1/2

may ask for multiple values of x.

For:

0 ≤ x < 2π

the solutions are:

x = π/6, 5π/6

So:

inverse function → one principal value

while:

trigonometric equation → potentially many solutions

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Common Mistakes

Thinking sin⁻¹(x) means 1/sin(x)

It means inverse sine.

The reciprocal is csc(x).

Giving every possible angle when evaluating an inverse function

An inverse function returns a principal value.

Giving only the principal value when solving an equation

A trigonometric equation may have additional solutions.

Forgetting principal ranges

arcsin, arccos, and arctan return angles from different intervals.

Using arcsin with an input greater than 1

For real numbers:

arcsin(1.5)

is undefined.

Using arccos with an input below −1

For real numbers:

arccos(−2)

is undefined.

Forgetting calculator mode

30° and approximately 0.524 radians represent the same angle.

Assuming arcsin(sin(x)) always equals x

This is only guaranteed when x lies within the appropriate principal range.

Ignoring quadrants when solving equations

The principal angle may not be the only solution.

Ignoring quadrants when finding vector directions

The signs of the vector components determine the actual quadrant.

Confusing grade with angle

A 10% slope does not mean a 10° angle.

Rounding too early

Keep several decimal places during calculations and round only at the end.

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Key Terms

Inverse function — A function that reverses the action of another function.

Inverse trigonometric function — A function used to determine an angle from a trigonometric value.

Inverse sine — The function arcsin(x) or sin⁻¹(x).

Inverse cosine — The function arccos(x) or cos⁻¹(x).

Inverse tangent — The function arctan(x) or tan⁻¹(x).

Principal value — The single output angle selected by an inverse trigonometric function.

Principal range — The restricted range from which an inverse trigonometric function returns its principal value.

arcsin — Alternative notation for inverse sine.

arccos — Alternative notation for inverse cosine.

arctan — Alternative notation for inverse tangent.

One-to-one function — A function in which different inputs produce different outputs.

Restricted domain — A limited part of a function's domain selected so that an inverse can be defined.

Domain — The set of permitted input values.

Range — The set of possible output values.

Reference angle — The positive acute angle associated with a trigonometric value.

Periodic function — A function whose values repeat at regular intervals.

General solution — An expression representing all solutions of an equation.

Angle of elevation — The angle measured upward from a horizontal line of sight.

Angle of depression — The angle measured downward from a horizontal line of sight.

Vector component — The part of a vector acting along a particular direction or axis.

Slope — Vertical change divided by horizontal change.

Road grade — Slope expressed as a percentage.

Radian — An angle unit based on the radius of a circle.

Degree — An angle unit in which a complete revolution contains 360°.

atan2 — A computational form of inverse tangent that uses two coordinates or components to identify the correct quadrant.

Horizontal asymptote — A horizontal line that a graph approaches without necessarily reaching.

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Key Takeaways

• Trigonometric functions convert angles into ratios or coordinates.
• Inverse trigonometric functions convert trigonometric values back into angles.
• The three main inverse functions are arcsin, arccos, and arctan.
• sin⁻¹(x) means inverse sine, not 1/sin(x).
• The reciprocal of sine is cosecant.
• Trigonometric functions repeat, so they must be restricted before inverse functions can be defined.
• Inverse functions return principal values.
• arcsin returns values from −π/2 to π/2.
• arccos returns values from 0 to π.
• arctan returns values between −π/2 and π/2.
• arcsin and arccos accept real inputs only between −1 and 1.
• arctan accepts every real number.
• A function and its inverse are reflections across y = x.
• Calculator mode must match the required angle unit.
• Evaluating an inverse function gives one principal value.
• Solving a trigonometric equation may produce multiple solutions.
• Periodicity explains why trigonometric equations can have infinitely many solutions.
• Inverse trigonometry is especially useful for finding unknown angles.
• Right-triangle ratios can be reversed using inverse functions.
• Inverse tangent can convert slope into an angle of inclination.
• Vector components can be used with inverse tangent to determine direction.
• Quadrant information must be considered when interpreting inverse tangent.
• sin(arcsin(x)) = x for −1 ≤ x ≤ 1.
• arcsin(sin(x)) does not necessarily equal x because arcsin must return a principal value.
• Inverse trigonometric functions are widely used in physics, engineering, surveying, navigation, robotics, and modelling.

A useful way to remember the basic purpose is:

Trig: angle → ratio

Inverse trig: ratio → angle

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Check Your Understanding

1. What does an inverse trigonometric function determine?

2. Explain why sin⁻¹(x) does not mean 1/sin(x).

3. Evaluate:

arcsin(1/2)

4. Evaluate:

arccos(1/2)

5. Evaluate:

arctan(1)

6. Evaluate:

arcsin(−1)

7. Evaluate:

arccos(−1)

8. Evaluate:

arctan(−√3)

9. State the principal range of arcsin in radians.

10. State the principal range of arccos in radians.

11. State the principal range of arctan in radians.

12. Explain why:

arcsin(1/2) = π/6

rather than 5π/6.

13. Does:

arcsin(1.4)

have a real value? Explain.

14. Does:

arctan(20)

have a real value? Explain.

15. Solve:

sin(x) = 1/2

for:

0° ≤ x < 360°

16. Solve:

cos(x) = −1/2

for:

0° ≤ x < 360°

17. Solve:

tan(x) = 1

for:

0° ≤ x < 360°

18. Explain the difference between evaluating:

arcsin(1/2)

and solving:

sin(x) = 1/2

19. A right triangle has opposite side 8 cm and adjacent side 15 cm. Calculate the angle between the adjacent side and hypotenuse.

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

21. A hill rises 35 m over a horizontal distance of 200 m. Calculate its angle of inclination.

22. A vector has components:

Vₓ = 12

Vᵧ = 5

Calculate its direction relative to the positive x-axis.

23. Explain why vector components must be checked for their signs when using inverse tangent.

24. A road has a 15% grade. Estimate its angle of inclination.

25. Simplify:

sin(arcsin(x))

and state the appropriate domain.

26. Evaluate:

arcsin(sin(5π/6))

27. Explain why the answer to Question 26 is not 5π/6.

28. Evaluate:

arccos(cos(4π/3))

29. Explain graphically why trigonometric functions require restricted domains before inverse functions can be defined.

30. Challenge: A drone is located 240 m horizontally from an observer and 180 m above the observer's level.

a. Draw or imagine the right triangle representing the situation.
b. Identify the opposite and adjacent sides relative to the angle of elevation.
c. Write a tangent equation for the angle θ.
d. Use inverse tangent to calculate θ.
e. Calculate the straight-line distance from the observer to the drone.
f. Explain why arctan is appropriate for determining the angle.
g. Explain what would happen to θ if the drone maintained the same height but moved farther horizontally away.
h. Explain what would happen to θ if the drone moved vertically upward while remaining at the same horizontal distance.
i. Describe how inverse trigonometry could be used by a tracking or navigation system.
j. Explain the difference between the principal value returned by arctan and the need to consider quadrants in a more general two-dimensional vector problem.