Inverse Trigonometric Functions

arcsin, arccos, and arctan

SPEC 5.5 • INVERSE TRIGONOMETRIC FUNCTIONS

Why Do We Need Inverse Trig Functions?

In many problems, we need to find an angle when we know the value of a trigonometric ratio.

The Problem

If

sin θ = 0.5

, what is

θ

?

We need a function that "undoes" sine — this is the inverse sine function.

Applications

Finding angles in right-angled triangles

Solving trigonometric equations

Navigation and surveying calculations

Physics: resolving vectors, wave analysis

The Problem: Many-to-One Functions

For a function to have an inverse, it must be one-to-one (pass the horizontal line test).

The Issue

Trigonometric functions are many-to-one:

sin 30° = 0.5

sin 150° = 0.5

sin 390° = 0.5

Infinitely many angles give the same sine value!

The Solution

We restrict the domain of the original function to a portion where it IS one-to-one.

This gives us a unique output called the principal value.

Domains and Ranges

Each inverse trig function has a specific principal value range:

FunctionNotationDomainRange (Principal Values)

Inverse Sine

arcsin x

or

sin^{- 1} x

- 1 \leq x \leq 1

- \frac{π}{2} \leq y \leq \frac{π}{2}

Inverse Cosine

arccos x

or

cos^{- 1} x

- 1 \leq x \leq 1

0 \leq y \leq π

Inverse Tangent

arctan x

or

tan^{- 1} x

x \in R

- \frac{π}{2} < y < \frac{π}{2}

Key Observation

arcsin and arctan have the same range, but arctan uses strict inequalities (asymptotes at the boundaries).

Graph of y = arcsin x

Domain

- 1 \leq x \leq 1

Range

- \frac{π}{2} \leq y \leq \frac{π}{2}

Key Point Passes through

(0, 0)

Shape

S-shaped curve, increasing throughout. Endpoints at

(- 1, - \frac{π}{2})

and

(1, \frac{π}{2})

.

Graph of y = arccos x

Domain

- 1 \leq x \leq 1

Range

0 \leq y \leq π

Key Point Passes through

(0, \frac{π}{2})

Shape

Decreasing throughout. Endpoints at

(- 1, π)

and

(1, 0)

.

Graph of y = arctan x

Domain

x \in R

(all reals)

Range

- \frac{π}{2} < y < \frac{π}{2}

Key Point Passes through

(0, 0)

Shape

Horizontal asymptotes at

y = \pm \frac{π}{2}

. Increasing throughout, approaches but never reaches the asymptotes.

Worked Example01

Find the exact value of

arcsin (\frac{1}{2})

1

We need the angle

θ

where

sin θ = \frac{1}{2}

and

θ \in [- \frac{π}{2}, \frac{π}{2}]

2

From standard angles:

sin \frac{π}{6} = \frac{1}{2}

3

Since

\frac{π}{6}

is in the principal range:

arcsin (\frac{1}{2}) = \frac{π}{6}

Worked Example02

Find the exact value of

arccos (- \frac{3}{2})

1

We need

θ

where

cos θ = - \frac{3}{2}

and

θ \in [0, π]

2

We know

cos \frac{π}{6} = \frac{3}{2}

3

Cosine is negative in Q2, so the angle in

[0, π]

is:

4

θ = π - \frac{π}{6} = \frac{5 π}{6}

arccos (- \frac{3}{2}) = \frac{5 π}{6}

Worked Example03

Find the exact value of

arctan (- 1)

1

We need

θ

where

tan θ = - 1

and

θ \in (- \frac{π}{2}, \frac{π}{2})

2

We know

tan \frac{π}{4} = 1

3

Since tan is an odd function:

tan (- \frac{π}{4}) = - 1

4

- \frac{π}{4}

is in the principal range

(- \frac{π}{2}, \frac{π}{2})

arctan (- 1) = - \frac{π}{4}

Worked Example04

Find the exact value of

arcsin (- \frac{2}{2})

1

We need

θ

where

sin θ = - \frac{2}{2}

and

θ \in [- \frac{π}{2}, \frac{π}{2}]

2

We know

sin \frac{π}{4} = \frac{2}{2}

3

Since sine is an odd function:

sin (- \frac{π}{4}) = - \frac{2}{2}

4

- \frac{π}{4}

is in the principal range

arcsin (- \frac{2}{2}) = - \frac{π}{4}

Worked Example05

Solve the equation

arcsin (2 x - 1) = \frac{π}{6}

1

If

arcsin (2 x - 1) = \frac{π}{6}

, then:

2

2 x - 1 = sin \frac{π}{6} = \frac{1}{2}

3

2 x = 1 + \frac{1}{2} = \frac{3}{2}

4

x = \frac{3}{4}

Check:

arcsin (2 \times \frac{3}{4} - 1) = arcsin (\frac{1}{2}) = \frac{π}{6}

✓

x = \frac{3}{4}

Practice

Question 1: Find the exact value of

arccos (\frac{1}{2}) + arcsin (\frac{1}{2})

1

arccos \frac{1}{2} = \frac{π}{3}

(since

cos \frac{π}{3} = \frac{1}{2}

and

\frac{π}{3} \in [0, π]

)

2

arcsin \frac{1}{2} = \frac{π}{6}

(since

sin \frac{π}{6} = \frac{1}{2}

and

\frac{π}{6} \in [- \frac{π}{2}, \frac{π}{2}]

)

3

Sum =

\frac{π}{3} + \frac{π}{6} = \frac{2 π}{6} + \frac{π}{6} = \frac{3 π}{6} = \frac{π}{2}

\frac{π}{2}

Practice

Question 2: Find the exact value of

tan (arcsin \frac{5}{13})

1

Let

θ = arcsin \frac{5}{13}

, so

sin θ = \frac{5}{13}

and

θ \in [- \frac{π}{2}, \frac{π}{2}]

2

Using

sin^{2} θ + cos^{2} θ = 1

:

cos^{2} θ = 1 - \frac{25}{169} = \frac{144}{169}

3

Since

θ \in [- \frac{π}{2}, \frac{π}{2}]

,

cos θ > 0

, so

cos θ = \frac{12}{13}

4

tan θ = \frac{s i n θ}{c o s θ} = \frac{5/13}{12/13} = \frac{5}{12}

\frac{5}{12}

Summary

The Three Inverse Trig Functions

FunctionDomainRange

y = arcsin x

[- 1, 1]

[- \frac{π}{2}, \frac{π}{2}]

y = arccos x

[- 1, 1]

[0, π]

y = arctan x

R

(- \frac{π}{2}, \frac{π}{2})

Key Points to Remember

sin^{- 1} x

means arcsin x, NOT

\frac{1}{s i n x}

Always give answers within the principal range

arcsin and arctan return values in Q1 or Q4

arccos returns values in Q1 or Q2