Inverse Trigonometric Functions

Inverse trigonometric functions help us answer a very practical question: if we know a trig value, can we find the angle that produced it? students, this lesson will show you how functions like $\sin^{-1}(x)$, $\cos^{-1}(x)$, and $\tan^{-1}(x)$ work, why their domains and ranges matter, and how they appear in geometry, measurement, and modeling 📐. By the end of this lesson, you should be able to explain the key terminology, use inverse trig functions to solve problems, and connect them to the wider Geometry and Trigonometry topic.

Why inverse trig functions matter

Regular trig functions take an angle and give a ratio. For example, if $\theta$ is an angle in a right triangle, then $\sin \theta$ is the ratio of the opposite side to the hypotenuse. Inverse trig functions do the reverse: they take a ratio or number and return an angle.

This is useful whenever you know side lengths but need an angle. For example, if a ramp rises $2$ m over a horizontal distance of $5$ m, you may want to find the angle of elevation. A trigonometric ratio gives the relationship, and an inverse trig function gives the angle. In real life, this appears in surveying, navigation, engineering, and physics 🔍.

The important idea is that inverse trig functions are not simply “undoing” trig functions on all possible angles. Trig functions repeat values, so they are not one-to-one unless their domains are restricted. That restriction is what makes an inverse function possible.

What “inverse” really means

For a function to have an inverse, each output must come from exactly one input. But the functions $\sin \theta$, $\cos \theta$, and $\tan \theta$ are periodic, so many different angles can produce the same value. For example, $\sin 30^\circ = \sin 150^\circ = \frac{1}{2}$.

To create inverse trig functions, we restrict the original trig functions to intervals where they are one-to-one:

$\sin x$ is restricted to $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$
$\cos x$ is restricted to $[0,\pi]$
$\tan x$ is restricted to $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$

These restrictions allow the inverse functions to give a unique angle.

The standard notations are:

$y=\sin^{-1}(x)$ or $y=\arcsin(x)$
$y=\cos^{-1}(x)$ or $y=\arccos(x)$
$y=\tan^{-1}(x)$ or $y=\arctan(x)$

Although the notation looks like a power, $\sin^{-1}(x)$ does not mean $\frac{1}{\sin(x)}$. It means the inverse sine function. The reciprocal of $\sin(x)$ is $\csc(x)$.

Domains, ranges, and principal values

Understanding domain and range is essential for inverse trig functions.

For the inverse sine function:

domain is $[-1,1]$
range is $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$

For the inverse cosine function:

domain is $[-1,1]$
range is $[0,\pi]$

For the inverse tangent function:

domain is all real numbers $\mathbb{R}$
range is $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$

The output of an inverse trig function is called the principal value. This is the unique angle chosen from the restricted range.

For example:

$\sin^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6}$ because $\frac{\pi}{6}$ is in the correct range.
$\cos^{-1}\left(-\frac{1}{2}\right)=\frac{2\pi}{3}$.
$\tan^{-1}(1)=\frac{\pi}{4}$.

It is important to check the range before answering. For instance, even though $\sin 150^\circ = \frac{1}{2}$, we do not write $\sin^{-1}\left(\frac{1}{2}\right)=150^\circ$, because $150^\circ$ is not in the principal range of $\sin^{-1}$.

Using inverse trig functions in right triangles

Inverse trig functions are especially useful in right-triangle problems. Suppose students you know two sides of a right triangle and want an angle.

Example 1

A right triangle has opposite side $7$ cm and adjacent side $10$ cm. Find the angle $\theta$.

Use tangent:

$$\tan \theta=\frac{7}{10}$$

Then apply inverse tangent:

$$\theta=\tan^{-1}\left(\frac{7}{10}\right)$$

This gives an angle of about $35.0^\circ$.

Example 2

A ladder $6$ m long reaches a wall, and the bottom of the ladder is $2$ m from the wall. Find the angle between the ladder and the ground.

The adjacent side is $2$ m and the hypotenuse is $6$ m, so:

$$\cos \theta=\frac{2}{6}=\frac{1}{3}$$

Then:

$$\theta=\cos^{-1}\left(\frac{1}{3}\right)$$

This angle is about $70.5^\circ$.

These kinds of problems often appear in geometry because they combine distance, angle, and shape. They also connect with trigonometric reasoning, which is a major IB theme.

Inverse trig functions on the unit circle

The unit circle helps explain why inverse trig functions work and why their ranges are chosen carefully.

On the unit circle, a point at angle $\theta$ has coordinates $(\cos \theta,\sin \theta)$. That means inverse trig functions can be thought of as recovering an angle from a coordinate value.

For example:

If $\sin \theta = \frac{1}{2}$, then the principal angle is $\theta=\frac{\pi}{6}$.
If $\cos \theta = -\frac{\sqrt{2}}{2}$, then the principal angle is $\theta=\frac{3\pi}{4}$.
If $\tan \theta = \sqrt{3}$, then the principal angle is $\theta=\frac{\pi}{3}$.

However, many angles can share the same trig value. For example, $\sin \theta = \frac{1}{2}$ at $\theta=\frac{\pi}{6}$ and $\theta=\frac{5\pi}{6}$, among others. The inverse function only gives the principal value. If a problem asks for all solutions, then you must use trig equations and periodicity, not just the inverse function.

Solving equations with inverse trig functions

Inverse trig functions are often the first step in solving equations.

Example 3

Solve $\sin x=\frac{3}{5}$ for $0\le x<2\pi$.

First find the principal angle:

$$x=\sin^{-1}\left(\frac{3}{5}\right)$$

This gives one solution in the first quadrant. Since sine is also positive in the second quadrant, the second solution is:

$$x=\pi-\sin^{-1}\left(\frac{3}{5}\right)$$

So the two solutions in $0\le x<2\pi$ are the principal angle and its supplementary angle.

Example 4

Solve $\cos x=-\frac{1}{2}$ for $0\le x<2\pi$.

The principal value is:

$$x=\cos^{-1}\left(-\frac{1}{2}\right)=\frac{2\pi}{3}$$

Cosine is negative in quadrants II and III, so the other solution is:

$$x=2\pi-\frac{2\pi}{3}=\frac{4\pi}{3}$$

This shows a key IB idea: inverse trig functions give a starting point, but the full solution set may require geometry of the unit circle.

Compositions and common identities

A useful way to test understanding is to compose trig and inverse trig functions.

For any $x$ in the domain of $\sin^{-1}$, we have:

$$\sin\left(\sin^{-1}(x)\right)=x$$

Similarly,

$$\cos\left(\cos^{-1}(x)\right)=x$$

and

$$\tan\left(\tan^{-1}(x)\right)=x$$

But the reverse direction is not always equal to the original angle, because inverse trig functions return principal values only.

For example:

$$\sin^{-1}(\sin \theta)\ne \theta$$

for all $\theta$.

Instead, the result depends on whether $\theta$ lies in the correct principal range.

A common example is:

$$\sin^{-1}(\sin \frac{5\pi}{6})=\frac{\pi}{6}$$

because $\frac{5\pi}{6}$ is not in the range of $\sin^{-1}$.

This idea is important in algebraic manipulation and in checking whether equations are valid for all values.

Real-world interpretation and IB connections

Inverse trig functions help convert ratios into angles, which is essential in geometry and measurement. Surveyors use them to determine heights and distances. Engineers use them to find slope angles. In physics, they help find direction from components. In vector geometry, angles between vectors can be found using dot products, and inverse cosine appears in the formula:

$$\cos \theta=\frac{\mathbf{a}\cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|}$$

$$\theta=\cos^{-1}\left(\frac{\mathbf{a}\cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|}\right)$$

This is a strong example of how inverse trig functions connect directly to 2D and 3D geometry.

Inverse trig also appears in calculus later in the course, especially when differentiating and integrating expressions involving inverse trig functions. That means this topic is not isolated; it supports later HL work across the syllabus.

Conclusion

Inverse trigonometric functions are the reverse process of trig ratios, giving a unique angle from a known value. students, the most important ideas are the restricted domains of the original trig functions, the principal value ranges of the inverse functions, and the difference between finding one angle and solving a full trig equation. These functions are central to Geometry and Trigonometry because they connect side lengths, angles, unit circle reasoning, and vectors. They are also useful in many real situations where angles must be measured from known ratios 📏.

Study Notes

$\sin^{-1}(x)$, $\cos^{-1}(x)$, and $\tan^{-1}(x)$ mean inverse trig functions, not reciprocals.
The reciprocal functions are $\csc(x)$, $\sec(x)$, and $\cot(x)$.
The inverse trig functions return principal values from restricted ranges.
$\sin^{-1}(x)$ has domain $[-1,1]$ and range $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$.
$\cos^{-1}(x)$ has domain $[-1,1]$ and range $[0,\pi]$.
$\tan^{-1}(x)$ has domain $\mathbb{R}$ and range $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$.
To solve a triangle or geometry problem, first choose the correct trig ratio, then apply an inverse trig function.
Inverse trig functions give one angle; solving trig equations may require more than one solution.
On the unit circle, inverse trig functions connect values like $\sin \theta$, $\cos \theta$, and $\tan \theta$ to angles.
In vector geometry, angle formulas often use $\cos^{-1}$ to recover the angle between vectors.
Always check whether your answer fits the required interval or principal range.