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Constructing
Inverse Trigonometric Functions

Learn why trig functions need restricted domains to have inverses, and how to build \(\sin^{-1}\), \(\cos^{-1}\), and \(\tan^{-1}\) from scratch.

Precalculus

What Is an Inverse Function?

A function \(f\) has an inverse \(f^{-1}\) if and only if every output comes from exactly one input. When that is true, we can "reverse" the function: if \(f(a) = b\), then \(f^{-1}(b) = a\).

Graphically, the inverse reflects the curve over the line \(y = x\). Algebraically, we swap \(x\) and \(y\) and solve for \(y\).

The Problem with Trig Functions

Trig functions like \(\sin x\) are periodic: they repeat the same outputs over and over. That means they are not one-to-one on their natural domains.

The Solution: Restricting the Domain

If we restrict the domain to an interval where the function is always increasing or always decreasing, we get a one-to-one piece. Then the inverse exists on that piece.

Always Increasing
As \(x\) moves right, \(f(x)\) always goes up. No horizontal line crosses twice.
Always Decreasing
As \(x\) moves right, \(f(x)\) always goes down. Still one-to-one.

The Invertibility Theorem

This is exactly the tool we use. For each trig function, we pick an interval where it is strictly monotonic, then define the inverse on that interval.

1
Choose a restricted domain where \(f\) is strictly increasing or decreasing.
2
Verify the range covers all possible output values.
3
Define \(f^{-1}\) by reversing inputs and outputs on that piece.

Restricting \(\sin x\) to Build \(\sin^{-1} x\)

The inverse is \(\sin^{-1} x\) (also written \(\arcsin x\)): domain \([-1,1]\), range \(\left[-\frac{\pi}{2},\,\frac{\pi}{2}\right]\).

Restricting \(\cos x\) to Build \(\cos^{-1} x\)

The inverse is \(\cos^{-1} x\) (also written \(\arccos x\)): domain \([-1,1]\), range \([0,\,\pi]\).

Restricting \(\tan x\) to Build \(\tan^{-1} x\)

The inverse is \(\tan^{-1} x\) (also written \(\arctan x\)): domain \((-\infty, \infty)\), range \(\left(-\frac{\pi}{2},\,\frac{\pi}{2}\right)\).

Domain vs. Range Swap

When you build an inverse, the domain and range switch roles. The restricted domain of the original becomes the range of the inverse, and vice versa.

Original (Restricted) Inverse
\(\sin\) / \(\sin^{-1}\) Domain: \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\), Range: \([-1,1]\) Domain: \([-1,1]\), Range: \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\)
\(\cos\) / \(\cos^{-1}\) Domain: \([0,\pi]\), Range: \([-1,1]\) Domain: \([-1,1]\), Range: \([0,\pi]\)
\(\tan\) / \(\tan^{-1}\) Domain: \(\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\), Range: \((-\infty,\infty)\) Domain: \((-\infty,\infty)\), Range: \(\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\)

Why These Specific Intervals?

1
Monotonic: the function is strictly increasing or strictly decreasing on the interval, so it passes the Horizontal Line Test.
2
Full range captured: the interval produces every output value in \([-1,1]\) (for sin/cos) or \((-\infty,\infty)\) (for tan).
3
Includes zero: the intervals include \(x = 0\), the most natural reference angle, keeping the inverse centered and intuitive.
4
Convention: mathematicians worldwide agree on these intervals so that inverse trig values are unambiguous.

How to Evaluate Inverse Trig

1
Identify which inverse trig function you need: \(\sin^{-1}\), \(\cos^{-1}\), or \(\tan^{-1}\).
2
Check that the input is in the correct domain (\([-1,1]\) for arcsin/arccos, all reals for arctan).
3
Ask: "What angle in the restricted range gives this trig value?"
4
Verify your answer is inside the correct range. If not, you picked the wrong quadrant.

Example 1: \(\sin^{-1}\!\left(-\frac{\sqrt{3}}{2}\right)\)

Question: Evaluate \(\sin^{-1}\!\left(-\frac{\sqrt{3}}{2}\right)\).

1
We need an angle \(\theta\) with \(\sin\theta = -\frac{\sqrt{3}}{2}\).
2
The reference angle is \(\frac{\pi}{3}\) because \(\sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}\).
3
Since the value is negative, \(\theta\) must be in \(\left[-\frac{\pi}{2},0\right)\). So \(\theta = -\frac{\pi}{3}\).
4
Check: \(-\frac{\pi}{3} \in \left[-\frac{\pi}{2},\frac{\pi}{2}\right]\). ✓

Example 2: \(\cos^{-1}\!\left(-\frac{1}{2}\right)\)

Question: Evaluate \(\cos^{-1}\!\left(-\frac{1}{2}\right)\).

1
We need \(\theta\) with \(\cos\theta = -\frac{1}{2}\) and \(\theta \in [0,\pi]\).
2
Reference angle: \(\frac{\pi}{3}\) (since \(\cos\frac{\pi}{3} = \frac{1}{2}\)).
3
Cosine is negative in quadrant II. In \([0,\pi]\), the Q II angle is \(\pi - \frac{\pi}{3} = \frac{2\pi}{3}\).

Example 3: \(\tan^{-1}(-1)\)

Question: Evaluate \(\tan^{-1}(-1)\).

1
We need \(\theta\) with \(\tan\theta = -1\) and \(\theta \in \left(-\frac{\pi}{2},\frac{\pi}{2}\right)\).
2
Reference angle: \(\frac{\pi}{4}\) (since \(\tan\frac{\pi}{4} = 1\)).
3
Tangent is negative in Q IV of the restricted interval: \(\theta = -\frac{\pi}{4}\).

Common Mistakes to Avoid

Wrong Quadrant
Returning \(\frac{5\pi}{6}\) for \(\sin^{-1}\!\left(\frac{1}{2}\right)\). That angle is outside \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\).
Negative Arccos
Claiming \(\cos^{-1}(-x)\) is negative. The range of \(\cos^{-1}\) is \([0,\pi]\): it is always non-negative.
Forgetting Restrictions
Using the unrestricted \(\sin x\) and listing multiple answers. The inverse must return exactly one value.

Summary of Inverse Trig Functions

Function Domain Range Monotonicity
\(\sin^{-1} x\) \([-1,1]\) \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\) Increasing
\(\cos^{-1} x\) \([-1,1]\) \([0,\pi]\) Decreasing
\(\tan^{-1} x\) \((-\infty,\infty)\) \(\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\) Increasing

Question 1 of 5

Why must we restrict the domain of \(\sin x\) before defining \(\sin^{-1} x\)?

Question 2 of 5

What is the range of \(\cos^{-1} x\)?

Question 3 of 5

Evaluate: \(\sin^{-1}\!\left(-\frac{\sqrt{2}}{2}\right)\)

Question 4 of 5

On the restricted domain \(\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\), the function \(\tan x\) is:

Question 5 of 5

A student says \(\cos^{-1}\!\left(-\frac{\sqrt{3}}{2}\right) = -\frac{5\pi}{6}\). What is the error?