In short

An inverse trigonometric function takes a ratio as input and returns an angle as output. Because trig functions are periodic and hit each value many times, you must restrict their domains to make them one-to-one before inverting. The restricted version is called the principal value branch, and it gives a unique angle for each input. The six inverse trig functions — \sin^{-1}, \cos^{-1}, \tan^{-1}, \cot^{-1}, \sec^{-1}, \csc^{-1} — each have their own domain, range, and graph.

You know that \sin 30° = \frac{1}{2}. Now turn the question around: which angle has a sine of \frac{1}{2}?

If you are working with trigonometric equations, the answer is "infinitely many angles" — 30°, 150°, 390°, -210°, and so on. Every angle of the form n \cdot 180° + (-1)^n \cdot 30° works. That is the general solution.

But sometimes you do not want infinitely many answers. Sometimes you want the answer — a single, definite angle. When you type \sin^{-1}(0.5) into a calculator, it does not return a list of infinitely many numbers. It returns one number: 30° (or \frac{\pi}{6} in radian mode). The calculator has made a choice.

That choice — which one angle to return — is the entire subject of this article. The function that makes this choice is called the inverse sine function, written \sin^{-1} or \arcsin. It takes a number (a ratio) as input and returns an angle. But to make that function well-defined, you have to be very careful about which angle it is allowed to return. Get that wrong, and you get contradictions. Get it right, and you have a precise tool that converts ratios back into angles.

Why you cannot just "invert" sine directly

Here is the problem in its sharpest form.

A function has an inverse only if it is one-to-one: each output comes from exactly one input. Consider f(x) = x^2. It is not one-to-one because f(3) = 9 and f(-3) = 9 — two different inputs give the same output. So "f^{-1}(9)" is ambiguous: is it 3 or -3?

The sine function has the same problem, but worse. Not just two inputs give the same output — infinitely many do. The function \sin x takes the value \frac{1}{2} at x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{13\pi}{6}, -\frac{7\pi}{6}, \ldots

The sine function hits $y = \frac{1}{2}$ infinitely many times. Each intersection is a valid answer to "which angle has sine $\frac{1}{2}$?" — but a function must give only one answer. To create an inverse, you must pick a piece of the sine curve where each $y$-value appears exactly once.

If you try to define \sin^{-1}\left(\frac{1}{2}\right) as "the angle whose sine is \frac{1}{2}," the definition is ambiguous — there are infinitely many such angles. A function cannot return infinitely many values. A function must return exactly one.

The solution: restrict the domain of sine to an interval where it is one-to-one, and then invert only that restricted version.

Choosing the right restriction: the principal value branch

Which interval should you pick? You need a piece of the sine curve where:

  1. Sine is one-to-one (no two inputs give the same output — the curve does not double back on itself).
  2. The piece covers the entire range [-1, 1] (so that \sin^{-1}(y) is defined for every y between -1 and +1).

For sine, the natural choice is \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]. On this interval, the sine curve goes from \sin(-\frac{\pi}{2}) = -1 up to \sin(\frac{\pi}{2}) = 1, hitting every value between -1 and 1 exactly once. It is strictly increasing, never doubling back.

The sine function with the principal branch highlightedThe graph of y equals sine x from negative 2 pi to 2 pi. The portion from negative pi over 2 to pi over 2 is drawn with a thick solid line, while the rest is drawn with a thin dashed line. The thick portion is the principal branch — the one-to-one piece that gets inverted. x y −π/2 π/2 −π π −3π/2 3π/2 1 −1 principal branch
The sine function with the principal branch (from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$) drawn in red. The dashed portions are the parts of sine that get discarded for the purpose of defining the inverse. On the red portion, every height from $-1$ to $1$ is reached exactly once — that is what makes it invertible.

This restricted piece is called the principal value branch of sine. When you invert it, you get the inverse sine function.

The three main inverse trig functions

Inverse sine ($\sin^{-1}$ or $\arcsin$)

y = \sin^{-1} x \quad \Longleftrightarrow \quad \sin y = x \;\text{ and }\; y \in \left[-\frac{\pi}{2},\, \frac{\pi}{2}\right]

Domain: [-1, 1]. Range: \left[-\frac{\pi}{2}, \frac{\pi}{2}\right].

Inverse cosine ($\cos^{-1}$ or $\arccos$)

y = \cos^{-1} x \quad \Longleftrightarrow \quad \cos y = x \;\text{ and }\; y \in [0,\, \pi]

Domain: [-1, 1]. Range: [0, \pi].

Inverse tangent ($\tan^{-1}$ or $\arctan$)

y = \tan^{-1} x \quad \Longleftrightarrow \quad \tan y = x \;\text{ and }\; y \in \left(-\frac{\pi}{2},\, \frac{\pi}{2}\right)

Domain: (-\infty, \infty). Range: \left(-\frac{\pi}{2}, \frac{\pi}{2}\right).

Notice the patterns. For each function, the chosen range is an interval where the original trig function is one-to-one and covers its full range of values:

Why [0, \pi] for cosine instead of [-\frac{\pi}{2}, \frac{\pi}{2}]? Because cosine is not one-to-one on [-\frac{\pi}{2}, \frac{\pi}{2}] — it takes the value \frac{\sqrt{3}}{2} at both \frac{\pi}{6} and -\frac{\pi}{6}. On [0, \pi], cosine is strictly decreasing, so each value appears exactly once.

Reading the notation

The notation \sin^{-1} x is standard in Indian textbooks (NCERT, RD Sharma, and most JEE materials). The notation \arcsin x means exactly the same thing and is more common in some international contexts. Both are used interchangeably.

One critical warning: \sin^{-1} x does not mean \frac{1}{\sin x}. The superscript -1 here denotes the inverse function, not a reciprocal. The reciprocal of \sin x is \csc x (or (\sin x)^{-1} if you must use a power notation). This is an unfortunate ambiguity in mathematical notation, and the only way to handle it is to memorise the convention.

The graphs

The graph of an inverse function is obtained by reflecting the graph of the original function across the line y = x. What was horizontal becomes vertical; what was the x-axis becomes the y-axis.

Graph of \sin^{-1} x

Take the principal branch of sine (from -\frac{\pi}{2} to \frac{\pi}{2}) and reflect it across y = x. The result is an S-shaped curve that starts at (-1, -\frac{\pi}{2}) and ends at (1, \frac{\pi}{2}).

The graph of $y = \sin^{-1} x$. It exists only for $x \in [-1, 1]$, and its output is always between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$. The curve passes through the origin and is symmetric about it (odd function). It gets steeper near the endpoints — that is because the sine curve is nearly flat at its peaks, so the inverse must be nearly vertical there.

Key features:

Graph of \cos^{-1} x

The graph of $y = \cos^{-1} x$. It goes from $(−1, \pi)$ down to $(1, 0)$ — a decreasing curve. The output is always between $0$ and $\pi$. Unlike $\sin^{-1}$, this function is neither odd nor even.

Key features:

Graph of \tan^{-1} x

The graph of $y = \tan^{-1} x$. Unlike the other two, this function accepts *any* real number as input. As $x \to \infty$, the output approaches $\frac{\pi}{2}$ but never reaches it; as $x \to -\infty$, it approaches $-\frac{\pi}{2}$. These are horizontal asymptotes. The curve passes through the origin and is an odd function.

Key features:

Basic properties

These identities follow directly from the definitions and are used constantly.

Cancellation identities

For \sin^{-1}:

The first identity says: take a number, find the angle whose sine is that number, then take the sine of that angle — you get back where you started. This always works within the domain.

The second identity is the one that trips people up. If x = \frac{5\pi}{6}, then \sin \frac{5\pi}{6} = \frac{1}{2}, and \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}, not \frac{5\pi}{6}. You do not get back to where you started because \frac{5\pi}{6} is outside the range \left[-\frac{\pi}{2}, \frac{\pi}{2}\right].

The general rule: \sin^{-1}(\sin x) returns the angle in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] that has the same sine as x. If x is already in that interval, the answer is x. If not, you have to find the equivalent angle in the principal range.

The same logic applies to cosine and tangent:

The complementary identity

\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \quad \text{for all } x \in [-1, 1]

Here is why. Let \sin^{-1} x = \alpha. Then \sin \alpha = x and \alpha \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right].

Now \cos\left(\frac{\pi}{2} - \alpha\right) = \sin \alpha = x. The angle \frac{\pi}{2} - \alpha lies in [0, \pi] (check: when \alpha = -\frac{\pi}{2}, you get \pi; when \alpha = \frac{\pi}{2}, you get 0). So \frac{\pi}{2} - \alpha is the angle in [0, \pi] whose cosine is x — which is exactly the definition of \cos^{-1} x.

Therefore \cos^{-1} x = \frac{\pi}{2} - \alpha = \frac{\pi}{2} - \sin^{-1} x, and rearranging gives \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}.

This identity says that the "inverse sine angle" and the "inverse cosine angle" for the same input always add up to a right angle. Geometrically, this is just the fact that the two acute angles in a right triangle sum to \frac{\pi}{2}.

Negative argument identities

\sin^{-1}(-x) = -\sin^{-1}(x), \qquad \cos^{-1}(-x) = \pi - \cos^{-1}(x), \qquad \tan^{-1}(-x) = -\tan^{-1}(x)

For \sin^{-1}: Let \sin^{-1} x = \alpha. Then \sin \alpha = x, so \sin(-\alpha) = -x, and since -\alpha \in [-\frac{\pi}{2}, \frac{\pi}{2}], we get \sin^{-1}(-x) = -\alpha = -\sin^{-1} x. This is exactly the statement that \sin^{-1} is an odd function.

For \cos^{-1}: Let \cos^{-1} x = \alpha, so \cos \alpha = x and \alpha \in [0, \pi]. Then \cos(\pi - \alpha) = -\cos \alpha = -x, and \pi - \alpha \in [0, \pi]. So \cos^{-1}(-x) = \pi - \alpha = \pi - \cos^{-1} x.

For \tan^{-1}: the argument is the same as for \sin^{-1}, using the fact that tangent is an odd function.

Computing specific values

Here is a table of the most commonly needed values. Every entry follows from the standard trig table and the principal value ranges.

x \sin^{-1} x \cos^{-1} x
0 0 \frac{\pi}{2}
\frac{1}{2} \frac{\pi}{6} \frac{\pi}{3}
\frac{1}{\sqrt{2}} \frac{\pi}{4} \frac{\pi}{4}
\frac{\sqrt{3}}{2} \frac{\pi}{3} \frac{\pi}{6}
1 \frac{\pi}{2} 0
-\frac{1}{2} -\frac{\pi}{6} \frac{2\pi}{3}
-1 -\frac{\pi}{2} \pi

Notice how each row confirms \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}.

x \tan^{-1} x
0 0
\frac{1}{\sqrt{3}} \frac{\pi}{6}
1 \frac{\pi}{4}
\sqrt{3} \frac{\pi}{3}
-1 -\frac{\pi}{4}

Example 1: Find $\sin^{-1}\left(\sin \frac{7\pi}{6}\right)$

This is the classic "trap" problem. The angle \frac{7\pi}{6} is not in the range of \sin^{-1}, so the answer is not \frac{7\pi}{6}.

Step 1. Compute \sin \frac{7\pi}{6}.

\sin \frac{7\pi}{6} = \sin\left(\pi + \frac{\pi}{6}\right) = -\sin \frac{\pi}{6} = -\frac{1}{2}

Why: \frac{7\pi}{6} is in the third quadrant, where sine is negative. The reference angle is \frac{7\pi}{6} - \pi = \frac{\pi}{6}.

Step 2. Now compute \sin^{-1}\left(-\frac{1}{2}\right).

You need the angle in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] whose sine is -\frac{1}{2}. That angle is -\frac{\pi}{6}.

Why: \sin\left(-\frac{\pi}{6}\right) = -\frac{1}{2}, and -\frac{\pi}{6} is in the principal range \left[-\frac{\pi}{2}, \frac{\pi}{2}\right].

Step 3. Verify using the negative argument identity.

\sin^{-1}\left(-\frac{1}{2}\right) = -\sin^{-1}\left(\frac{1}{2}\right) = -\frac{\pi}{6}.

Why: \sin^{-1} is odd, so \sin^{-1}(-x) = -\sin^{-1}(x). This confirms the answer from Step 2.

Result: \sin^{-1}\left(\sin \frac{7\pi}{6}\right) = -\frac{\pi}{6}.

Unit circle showing why sin inverse of sin 7 pi over 6 is negative pi over 6A unit circle with two radii drawn. One is at angle 7 pi over 6 in the third quadrant, and the other is at negative pi over 6 in the fourth quadrant. Both points have the same y-coordinate of negative one-half. The inverse sine function returns the angle in the principal range, which is negative pi over 6. 7π/6 −π/6 y = −1/2 x y
Two angles with the same sine: $\frac{7\pi}{6}$ (third quadrant, dashed) and $-\frac{\pi}{6}$ (fourth quadrant, solid red). Both have $y$-coordinate $-\frac{1}{2}$. The $\sin^{-1}$ function always returns the angle in its principal range $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, which is $-\frac{\pi}{6}$.

The unit circle makes the answer visible. The angle \frac{7\pi}{6} and the angle -\frac{\pi}{6} land on the same horizontal line (same y-coordinate). The inverse sine function always picks the one on the right half of the circle — the one in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right].

Example 2: Simplify $\tan^{-1} 1 + \cos^{-1}\left(-\frac{1}{2}\right) + \sin^{-1}\left(-\frac{1}{2}\right)$

Step 1. Evaluate each term separately.

\tan^{-1} 1: The angle in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) whose tangent is 1 is \frac{\pi}{4}.

Why: \tan \frac{\pi}{4} = 1, and \frac{\pi}{4} is in the principal range of \tan^{-1}.

Step 2. Evaluate \cos^{-1}\left(-\frac{1}{2}\right).

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

Why: the identity \cos^{-1}(-x) = \pi - \cos^{-1}(x) converts a negative argument into a positive one. And \cos^{-1}\frac{1}{2} = \frac{\pi}{3} from the standard table.

Step 3. Evaluate \sin^{-1}\left(-\frac{1}{2}\right).

\sin^{-1}\left(-\frac{1}{2}\right) = -\sin^{-1}\left(\frac{1}{2}\right) = -\frac{\pi}{6}.

Why: \sin^{-1} is odd. And \sin^{-1}\frac{1}{2} = \frac{\pi}{6}.

Step 4. Add.

\frac{\pi}{4} + \frac{2\pi}{3} + \left(-\frac{\pi}{6}\right) = \frac{\pi}{4} + \frac{2\pi}{3} - \frac{\pi}{6}

Common denominator is 12:

= \frac{3\pi}{12} + \frac{8\pi}{12} - \frac{2\pi}{12} = \frac{9\pi}{12} = \frac{3\pi}{4}

Why: pure arithmetic with fractions. The key was knowing each inverse trig value first, then combining.

Result: \tan^{-1} 1 + \cos^{-1}\left(-\frac{1}{2}\right) + \sin^{-1}\left(-\frac{1}{2}\right) = \frac{3\pi}{4}.

Number line showing the three inverse trig values adding up to three pi over fourA number line from negative pi over 2 to pi, with three intervals marked: negative pi over 6 in blue-grey, pi over 4 in red, and 2 pi over 3 in red, stacking up to 3 pi over 4 total. 0 −π/2 π π/2 −π/6 π/4 2π/3 3π/4
A visual sum on the number line. Start at $-\frac{\pi}{6}$, add $\frac{\pi}{4}$, add $\frac{2\pi}{3}$: you land at $\frac{3\pi}{4}$. Each segment corresponds to one of the three inverse trig values.

This kind of problem tests whether you know the principal value ranges cold. There is no trick beyond knowing that \cos^{-1}(-\frac{1}{2}) = \frac{2\pi}{3} (not -\frac{\pi}{3} or \frac{4\pi}{3}) and \sin^{-1}(-\frac{1}{2}) = -\frac{\pi}{6} (not \frac{11\pi}{6}).

Common confusions

Going deeper

If you came here to understand what inverse trig functions are, how to read their graphs, and how to evaluate them at standard values, you have it. The rest of this section covers the remaining three inverse trig functions and the connection to trigonometric equations.

The other three: \cot^{-1}, \sec^{-1}, \csc^{-1}

The three reciprocal trig functions also have inverse functions, though they appear less frequently.

\cot^{-1} x: Domain: (-\infty, \infty). Range: (0, \pi). The principal branch of cotangent is taken on (0, \pi), where it is strictly decreasing from +\infty to -\infty.

\sec^{-1} x: Domain: (-\infty, -1] \cup [1, \infty). Range: [0, \pi] \setminus \left\{\frac{\pi}{2}\right\} — i.e., \left[0, \frac{\pi}{2}\right) \cup \left(\frac{\pi}{2}, \pi\right]. The gap at \frac{\pi}{2} exists because \sec \frac{\pi}{2} is undefined.

\csc^{-1} x: Domain: (-\infty, -1] \cup [1, \infty). Range: \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \setminus \{0\}. Similar reasoning: \csc 0 is undefined.

These three can always be converted to the main three:

\cot^{-1} x = \frac{\pi}{2} - \tan^{-1} x \quad (x > 0)
\sec^{-1} x = \cos^{-1}\frac{1}{x} \quad (|x| \geq 1)
\csc^{-1} x = \sin^{-1}\frac{1}{x} \quad (|x| \geq 1)

So in practice, you can work entirely with \sin^{-1}, \cos^{-1}, and \tan^{-1} and convert when needed.

Connection to trigonometric equations

Inverse trig functions are the formal machinery behind the "principal solution" concept from Trigonometric Equations. When you solve \sin x = \frac{1}{2} and write the general solution as x = n\pi + (-1)^n \frac{\pi}{6}, the value \frac{\pi}{6} is precisely \sin^{-1}\frac{1}{2} — the principal value.

More generally, the general solution of \sin x = a (for |a| \leq 1) can be written as:

x = n\pi + (-1)^n \sin^{-1} a

The inverse function gives you the starting angle; the general solution formula generates the full infinite family from it. Without the inverse function, you would have to look up the principal value from a table or a calculator every time. With it, the answer is a self-contained algebraic expression.

Madhava and the infinite series for \arctan

Long before the notation \tan^{-1} existed, the Kerala mathematician Madhava of Sangamagrama (14th century) discovered the infinite series

\tan^{-1} x = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots

valid for |x| \leq 1. Setting x = 1 gives

\frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \cdots

This is one of the earliest known formulas for computing \pi as an infinite series, predating European discovery of the same result by roughly 250 years. The series converges slowly — you need thousands of terms for even a few decimal places of \pi — but its mathematical beauty is undeniable. It says that \pi is hiding inside the inverse tangent function, which is hiding inside the geometry of the unit circle.

Where this leads next

You now know what inverse trig functions are, why their domains and ranges are what they are, and how to evaluate them. The natural next steps: