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Constructing
Inverse Trigonometric
Functions

Discover how restricting a domain where a trig function is always increasing or decreasing allows its inverse to be constructed.

Geometry

What Is an Inverse Function?

An inverse function reverses the original: if \( f(a) = b \), then \( f^{-1}(b) = a \). Every input-output pair gets swapped.

Graphically, the inverse is the reflection of the original across the line \( y = x \). But if the original fails the Horizontal Line Test, its reflection will fail the Vertical Line Test: not a function.

The Problem with Trig Functions

Consider \( y = \sin x \). Because sine is periodic (repeats every \( 2\pi \)), infinitely many \( x \)-values give the same \( y \)-value.

The Solution: Domain Restriction

If we restrict a trig function to a carefully chosen interval where it is strictly monotonic (always increasing or always decreasing), the restricted version passes the Horizontal Line Test.

Strictly Increasing
Output rises as input grows: each output has exactly one input.
Strictly Decreasing
Output falls as input grows: again, every output is unique.

Restricting \( \sin x \) to Build \( \arcsin x \)

Restricting \( \cos x \) to Build \( \arccos x \)

Restricting \( \tan x \) to Build \( \arctan x \)

Increasing vs. Decreasing Restrictions

Both strategies produce valid inverses, but the mathematical community has adopted standard restricted domains for each trig function:

\( \sin x \)
Restricted to \( [-\frac{\pi}{2}, \frac{\pi}{2}] \)
Increasing
\( \cos x \)
Restricted to \( [0, \pi] \)
Decreasing
\( \tan x \)
Restricted to \( (-\frac{\pi}{2}, \frac{\pi}{2}) \)
Increasing

The Core Principle

1
Monotonic on D: \( f \) is strictly increasing or strictly decreasing throughout \( D \).
2
Surjective onto R: \( f \) hits every value in its full range on \( D \).
3
Inverse exists: \( f^{-1}(y) \) returns the unique \( x \in D \) with \( f(x) = y \).

How to Construct an Inverse

1
Identify the trig function: Which function do you need to invert (\( \sin \), \( \cos \), or \( \tan \))?
2
Choose a monotonic interval: Find an interval where the function is strictly increasing or strictly decreasing.
3
Verify full range coverage: Confirm the function achieves every value in its range on that interval.
4
Define the inverse: The inverse maps each range value back to the unique domain value. State the domain and range of the inverse.
5
Verify: Check that \( f(f^{-1}(x)) = x \) and \( f^{-1}(f(x)) = x \) on the restricted domain.

Example 1: Evaluating \( \arcsin\!\bigl(\tfrac{1}{2}\bigr) \)

Question: Find \( \arcsin\!\bigl(\tfrac{1}{2}\bigr) \).

1
We need \( x \) in \( \bigl[-\frac{\pi}{2},\,\frac{\pi}{2}\bigr] \) such that \( \sin x = \frac{1}{2} \).
2
Recall: \( \sin\!\bigl(\frac{\pi}{6}\bigr) = \frac{1}{2} \), and \( \frac{\pi}{6} \in \bigl[-\frac{\pi}{2},\,\frac{\pi}{2}\bigr] \). ✓
✓
Answer: \( \arcsin\!\bigl(\tfrac{1}{2}\bigr) = \dfrac{\pi}{6} \).

Example 2: Evaluating \( \arccos(-1) \)

Question: Find \( \arccos(-1) \).

1
We need \( x \in [0,\,\pi] \) such that \( \cos x = -1 \).
2
\( \cos(\pi) = -1 \), and \( \pi \in [0,\,\pi] \). ✓
✓
Answer: \( \arccos(-1) = \pi \).

Example 3: Evaluating \( \arctan(-1) \)

Question: Find \( \arctan(-1) \).

1
We need \( x \in \bigl(-\frac{\pi}{2},\,\frac{\pi}{2}\bigr) \) such that \( \tan x = -1 \).
2
\( \tan\!\bigl(-\frac{\pi}{4}\bigr) = -1 \), and \( -\frac{\pi}{4} \in \bigl(-\frac{\pi}{2},\,\frac{\pi}{2}\bigr) \). ✓
✓
Answer: \( \arctan(-1) = -\dfrac{\pi}{4} \).

Why These Specific Intervals?

Many monotonic intervals exist for each trig function. The standard choices are preferred because they:

1
Include the origin (or are centered near it), making computations and graph interpretations simpler.
2
Cover the full range of the original function, so the inverse is defined for every possible output.
3
Are contiguous (no gaps), forming a single connected interval rather than scattered pieces.

Common Mistakes to Avoid

✕ Ignoring the Restriction
Saying \( \arcsin\!\bigl(\tfrac{1}{2}\bigr) = \frac{5\pi}{6} \) because \( \sin\frac{5\pi}{6} = \frac{1}{2} \). That angle is outside \( [-\frac{\pi}{2}, \frac{\pi}{2}] \).
✕ Confusing Notation
Thinking \( \sin^{-1}(x) = \frac{1}{\sin x} \). The \( {-1} \) means inverse function, not reciprocal. The reciprocal is \( \csc x \).
✕ Wrong Range for Arccos
Giving a negative answer for \( \arccos \). The range is \( [0, \pi] \), so the output is always \( \geq 0 \).

Summary Table

Function Restricted Domain Monotonicity Inverse Range
\( \sin x \) \( [-\frac{\pi}{2}, \frac{\pi}{2}] \) Increasing \( [-\frac{\pi}{2}, \frac{\pi}{2}] \)
\( \cos x \) \( [0, \pi] \) Decreasing \( [0, \pi] \)
\( \tan x \) \( (-\frac{\pi}{2}, \frac{\pi}{2}) \) Increasing \( (-\frac{\pi}{2}, \frac{\pi}{2}) \)

Question 1 of 5

Why must we restrict the domain of \( \sin x \) before constructing \( \arcsin x \)?

Question 2 of 5

What is the standard restricted domain for \( \cos x \) when defining \( \arccos x \)?

Question 3 of 5

What is the value of \( \arcsin\!\bigl(-\frac{\sqrt{2}}{2}\bigr) \)?

Question 4 of 5

On the restricted domain \( \bigl(-\frac{\pi}{2}, \frac{\pi}{2}\bigr) \), the tangent function is:

Question 5 of 5

A student claims \( \arccos\!\bigl(\tfrac{1}{2}\bigr) = -\frac{\pi}{3} \). What is wrong with this answer?