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Inverse Trigonometric Functions



Why Restriction Is Necessary
The Arcsine Function
The Arccosine Function
The Arctangent Function
Inverse Reciprocal Functions
Notation: sin⁡−1(x)\sin^{-1}(x) vs 1sin⁡(x)\frac{1}{\sin(x)}
Evaluating Inverse Trigonometric Functions
Compositions of Trigonometric and Inverse Trigonometric Functions
Graphs of Inverse Trigonometric Functions



Recovering Angles from Known Ratios

The six trigonometric functions take an angle and return a number. The inverse trigonometric functions reverse this: they take a number and return an angle. Given that sin⁡(θ)=12\sin(\theta) = \frac{1}{2}, what is θ\theta? The answer is not unique — infinitely many angles satisfy this equation, as the equations page makes clear. But an inverse function must return exactly one value. This forces a restriction: each trigonometric function must be confined to an interval where it is strictly monotonic (always increasing or always decreasing) before an inverse can be defined.

The resulting functions — arcsin⁡\arcsin, arccos⁡\arccos, arctan⁡\arctan, and their reciprocal counterparts — are not merely notational conveniences. They appear as solutions to equations, as building blocks in compositions that simplify using the Pythagorean identity, and as essential tools in calculus (where they arise as antiderivatives of certain algebraic expressions). Their graphs are reflections of the restricted trigonometric graphs over the line y=xy = x, and their domains and ranges are dictated entirely by the properties — specifically the monotonicity — of the original functions.




Why Restriction Is Necessary

A function can have an inverse only if it is one-to-one: each output corresponds to exactly one input. The trigonometric functions, being periodic, are emphatically not one-to-one on their full domains. The equation sin⁡(x)=12\sin(x) = \frac{1}{2} is satisfied by π6\frac{\pi}{6}, 5π6\frac{5\pi}{6}, π6+2π\frac{\pi}{6} + 2\pi, 5π6+2π\frac{5\pi}{6} + 2\pi, and infinitely many others. Defining "sin⁡−1(12)\sin^{-1}\left(\frac{1}{2}\right)" as all of these would not produce a function — a function must assign a single output to each input.

The solution is to restrict each trigonometric function to an interval where it is strictly monotonic — always increasing or always decreasing — and therefore one-to-one. On such an interval, the horizontal line test is passed, and an inverse function exists. The restricted function must still cover the entire range of the original, so that the inverse is defined for every relevant input.

The choice of restriction interval is a convention, universally agreed upon. Different intervals could work (sine is also one-to-one on [π2,3π2]\left[\frac{\pi}{2}, \frac{3\pi}{2}\right], for instance), but the standard choices have been selected for mathematical convenience — they are centered at or near the origin and produce the most natural behavior for compositions and calculus applications.

The Arcsine Function

    The sine function is restricted to [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right], where it is strictly increasing and maps onto its full range [−1,1][-1, 1]. The inverse of this restricted sine is called arcsine:

    arcsin⁡(x)=θmeanssin⁡(θ)=xwithθ∈[−π2,π2]\arcsin(x) = \theta \quad \text{means} \quad \sin(\theta) = x \quad \text{with} \quad \theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]


    Domain of $\arcsin$: [−1,1][-1, 1] — the range of sine. Inputs outside this interval have no corresponding angle.

    Range of $\arcsin$: [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] — outputs are always in Quadrant I (for positive inputs), Quadrant IV expressed as negative angles (for negative inputs), or zero.

    Exact values at standard inputs:

  • arcsin⁡(0)=0\arcsin(0) = 0
  • arcsin⁡(12)=π6\arcsin\left(\frac{1}{2}\right) = \frac{\pi}{6}
  • arcsin⁡(22)=π4\arcsin\left(\frac{\sqrt{2}}{2}\right) = \frac{\pi}{4}
  • arcsin⁡(32)=π3\arcsin\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3}
  • arcsin⁡(1)=π2\arcsin(1) = \frac{\pi}{2}
  • arcsin⁡(−12)=−π6\arcsin\left(-\frac{1}{2}\right) = -\frac{\pi}{6}
  • arcsin⁡(−1)=−π2\arcsin(-1) = -\frac{\pi}{2}

    • The output is always an angle — a number in radians (or degrees, depending on context). The function answers the question: "What angle between −π2-\frac{\pi}{2} and π2\frac{\pi}{2} has this sine value?"

    The graph of y=arcsin⁡(x)y = \arcsin(x) is obtained by reflecting the restricted sine graph over the line y=xy = x. It is an increasing S-shaped curve, starting at (−1,−π2)(-1, -\frac{\pi}{2}), passing through the origin, and ending at (1,π2)(1, \frac{\pi}{2}).

The Arccosine Function

    The cosine function is restricted to [0,π][0, \pi], where it is strictly decreasing and maps onto [−1,1][-1, 1]. The inverse is arccosine:

    arccos⁡(x)=θmeanscos⁡(θ)=xwithθ∈[0,π]\arccos(x) = \theta \quad \text{means} \quad \cos(\theta) = x \quad \text{with} \quad \theta \in [0, \pi]


    Domain of $\arccos$: [−1,1][-1, 1].

    Range of $\arccos$: [0,π][0, \pi] — outputs are always in Quadrant I (for positive inputs) or Quadrant II (for negative inputs).

    Exact values:

  • arccos⁡(1)=0\arccos(1) = 0
  • arccos⁡(32)=π6\arccos\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6}
  • arccos⁡(22)=π4\arccos\left(\frac{\sqrt{2}}{2}\right) = \frac{\pi}{4}
  • arccos⁡(12)=π3\arccos\left(\frac{1}{2}\right) = \frac{\pi}{3}
  • arccos⁡(0)=π2\arccos(0) = \frac{\pi}{2}
  • arccos⁡(−12)=2π3\arccos\left(-\frac{1}{2}\right) = \frac{2\pi}{3}
  • arccos⁡(−1)=π\arccos(-1) = \pi

    • A fundamental relationship connects arcsine and arccosine:

    arcsin⁡(x)+arccos⁡(x)=π2for all x∈[−1,1]\arcsin(x) + \arccos(x) = \frac{\pi}{2} \quad \text{for all } x \in [-1, 1]


    This is the inverse-function version of the cofunction identity sin⁡θ=cos⁡(π2−θ)\sin\theta = \cos\left(\frac{\pi}{2} - \theta\right). It means knowing one of arcsin⁡(x)\arcsin(x) or arccos⁡(x)\arccos(x) immediately gives the other.

    The graph of y=arccos⁡(x)y = \arccos(x) is a decreasing curve from (−1,π)(−1, \pi) to (1,0)(1, 0), passing through (0,π2)(0, \frac{\pi}{2}). It is the reflection of the restricted cosine graph over y=xy = x.

The Arctangent Function

    The tangent function is restricted to (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right), where it is strictly increasing and maps onto (−∞,∞)(-\infty, \infty). The inverse is arctangent:

    arctan⁡(x)=θmeanstan⁡(θ)=xwithθ∈(−π2,π2)\arctan(x) = \theta \quad \text{means} \quad \tan(\theta) = x \quad \text{with} \quad \theta \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)


    Domain of $\arctan$: (−∞,∞)(-\infty, \infty) — all real numbers, since tangent's range is unbounded.

    Range of $\arctan$: (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) — an open interval, since tangent approaches but never reaches ±π2\pm\frac{\pi}{2} on its restricted domain.

    Exact values:

  • arctan⁡(0)=0\arctan(0) = 0
  • arctan⁡(33)=π6\arctan\left(\frac{\sqrt{3}}{3}\right) = \frac{\pi}{6}
  • arctan⁡(1)=π4\arctan(1) = \frac{\pi}{4}
  • arctan⁡(3)=π3\arctan(\sqrt{3}) = \frac{\pi}{3}
  • arctan⁡(−1)=−π4\arctan(-1) = -\frac{\pi}{4}
  • arctan⁡(−3)=−π3\arctan(-\sqrt{3}) = -\frac{\pi}{3}

    • As x→∞x \to \infty, arctan⁡(x)→π2\arctan(x) \to \frac{\pi}{2}. As x→−∞x \to -\infty, arctan⁡(x)→−π2\arctan(x) \to -\frac{\pi}{2}. These are horizontal asymptotes of the arctangent graph — a feature unique among the three primary inverse functions.

    The graph of y=arctan⁡(x)y = \arctan(x) is an increasing S-shaped curve spanning the entire horizontal axis, bounded vertically between −π2-\frac{\pi}{2} and π2\frac{\pi}{2}. It passes through the origin with slope 1 (since ddxarctan⁡(x)∣x=0=1\frac{d}{dx}\arctan(x)\big|_{x=0} = 1) and flattens toward the asymptotes.

    Arctangent is particularly well-behaved: it is defined for all real numbers, it is continuous and differentiable everywhere, and its output is always finite. These properties make it a natural tool in calculus and applied mathematics.

Inverse Reciprocal Functions

    The reciprocal trigonometric functions — cosecant, secant, and cotangent — also have inverses, though they are used less frequently and their conventions vary more across textbooks.

    Arccosecant (arccsc\text{arccsc}): the inverse of cosecant restricted to [−π2,π2]∖{0}\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \setminus \{0\}.

  • (−∞,−1]∪[1,∞)(-\infty, -1] \cup [1, \infty)
  • [−π2,0)∪(0,π2]\left[-\frac{\pi}{2}, 0\right) \cup \left(0, \frac{\pi}{2}\right]
  • arccsc(x)=arcsin⁡(1x)\text{arccsc}(x) = \arcsin\left(\frac{1}{x}\right)

    • Arcsecant (arcsec\text{arcsec}): the inverse of secant restricted to [0,π]∖{π2}[0, \pi] \setminus \left\{\frac{\pi}{2}\right\}.

  • (−∞,−1]∪[1,∞)(-\infty, -1] \cup [1, \infty)
  • [0,π2)∪(π2,π]\left[0, \frac{\pi}{2}\right) \cup \left(\frac{\pi}{2}, \pi\right]
  • arcsec(x)=arccos⁡(1x)\text{arcsec}(x) = \arccos\left(\frac{1}{x}\right)

    • Arccotangent (arccot\text{arccot}): the inverse of cotangent restricted to (0,π)(0, \pi).

  • (−∞,∞)(-\infty, \infty)
  • (0,π)(0, \pi)
  • arccot(x)=arctan⁡(1x)\text{arccot}(x) = \arctan\left(\frac{1}{x}\right) for x>0x > 0; requires adjustment for x<0x < 0

    • In practice, these are rarely evaluated directly. When an expression involves arcsec(x)\text{arcsec}(x), it is usually converted to arccos⁡(1x)\arccos\left(\frac{1}{x}\right) for computation. The same applies to arccosecant via arcsine and arccotangent via arctangent. Their primary role is theoretical — they appear in integral formulas in calculus (for example, ∫1xx2−1 dx=arcsec∣x∣+C\int \frac{1}{x\sqrt{x^2 - 1}}\,dx = \text{arcsec}|x| + C) and in certain identity derivations.

Notation: sin⁡−1(x)\sin^{-1}(x) vs 1sin⁡(x)\frac{1}{\sin(x)}

    The notation sin⁡−1(x)\sin^{-1}(x) is standard for the inverse sine function — it means the same thing as arcsin⁡(x)\arcsin(x). This is a potential source of serious confusion because of how the superscript −1-1 is used elsewhere in trigonometry.

    When we write sin⁡2(x)\sin^2(x), the exponent 2 means squaring: sin⁡2(x)=(sin⁡(x))2\sin^2(x) = (\sin(x))^2. By this pattern, sin⁡−1(x)\sin^{-1}(x) should mean (sin⁡(x))−1=1sin⁡(x)=csc⁡(x)(\sin(x))^{-1} = \frac{1}{\sin(x)} = \csc(x). But it does not. In the specific case of the exponent −1-1, the notation is hijacked to mean the inverse function rather than the reciprocal.

    This is an inconsistency in mathematical notation, not a logical rule. The exponent −1-1 on a function name means "inverse function" by convention, overriding the algebraic meaning of raising to the power −1-1. It applies to all six trigonometric functions: cos⁡−1(x)=arccos⁡(x)\cos^{-1}(x) = \arccos(x), tan⁡−1(x)=arctan⁡(x)\tan^{-1}(x) = \arctan(x), and so on.

    To avoid ambiguity:

  • sin⁡−1(x)\sin^{-1}(x) or arcsin⁡(x)\arcsin(x) = the inverse function (returns an angle)
  • (sin⁡(x))−1(\sin(x))^{-1} or csc⁡(x)\csc(x) = the reciprocal (returns a number)
  • 1sin⁡(x)\frac{1}{\sin(x)} = the reciprocal, written unambiguously

    • The arcsin⁡\arcsin / arccos⁡\arccos / arctan⁡\arctan notation eliminates the confusion entirely and is preferred in any context where the −1-1 superscript might be misread. Many textbooks and scientific publications use the "arc" notation exclusively for this reason.

Evaluating Inverse Trigonometric Functions

Evaluating an inverse trigonometric function means answering: "What angle in the restricted range has this function value?"

For standard inputs — the values 0,±12,±22,±32,±10, \pm\frac{1}{2}, \pm\frac{\sqrt{2}}{2}, \pm\frac{\sqrt{3}}{2}, \pm 1 — the answer comes from the unit circle values, filtered through the range restriction.

arcsin⁡(32)=π3\arcsin\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3} because sin⁡(π3)=32\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} and π3∈[−π2,π2]\frac{\pi}{3} \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right].

arccos⁡(−22)=3π4\arccos\left(-\frac{\sqrt{2}}{2}\right) = \frac{3\pi}{4} because cos⁡(3π4)=−22\cos\left(\frac{3\pi}{4}\right) = -\frac{\sqrt{2}}{2} and 3π4∈[0,π]\frac{3\pi}{4} \in [0, \pi].

arctan⁡(−1)=−π4\arctan(-1) = -\frac{\pi}{4} because tan⁡(−π4)=−1\tan\left(-\frac{\pi}{4}\right) = -1 and −π4∈(−π2,π2)-\frac{\pi}{4} \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right).

A common trap: arcsin⁡(12)≠5π6\arcsin\left(\frac{1}{2}\right) \neq \frac{5\pi}{6}, even though sin⁡(5π6)=12\sin\left(\frac{5\pi}{6}\right) = \frac{1}{2}. The angle 5π6\frac{5\pi}{6} is outside the range [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right], so it is not the arcsine output. The correct answer is π6\frac{\pi}{6}.

For non-standard inputs — values like arcsin⁡(0.7)\arcsin(0.7) or arctan⁡(3.5)\arctan(3.5) — a calculator is required. Ensure the calculator is in the correct angle mode (degrees or radians) for the desired output format. Most calculators return radians by default for inverse trigonometric functions.

Compositions of Trigonometric and Inverse Trigonometric Functions

Compositions like sin⁡(arccos⁡(x))\sin(\arccos(x)) or arcsin⁡(sin⁡(x))\arcsin(\sin(x)) combine a trigonometric function with an inverse. The behavior of these compositions depends on the direction and on whether the input falls within the restricted range.

Direct compositions (function applied to its own inverse):

sin⁡(arcsin⁡(x))=xfor all x∈[−1,1]\sin(\arcsin(x)) = x \quad \text{for all } x \in [-1, 1]

cos⁡(arccos⁡(x))=xfor all x∈[−1,1]\cos(\arccos(x)) = x \quad \text{for all } x \in [-1, 1]

tan⁡(arctan⁡(x))=xfor all x∈(−∞,∞)\tan(\arctan(x)) = x \quad \text{for all } x \in (-\infty, \infty)


These hold universally within the domain — applying a function to its inverse always recovers the input.

Reverse compositions (inverse applied to its own function):

arcsin⁡(sin⁡(x))=xonly if x∈[−π2,π2]\arcsin(\sin(x)) = x \quad \text{only if } x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]


If xx is outside this range, the arcsine "folds" the result back into the restricted range. For example, arcsin⁡(sin⁡(5π6))=arcsin⁡(12)=π6\arcsin\left(\sin\left(\frac{5\pi}{6}\right)\right) = \arcsin\left(\frac{1}{2}\right) = \frac{\pi}{6}, not 5π6\frac{5\pi}{6}. The same caution applies to arccos⁡(cos⁡(x))\arccos(\cos(x)) (valid only on [0,π][0, \pi]) and arctan⁡(tan⁡(x))\arctan(\tan(x)) (valid only on (−π2,π2)(-\frac{\pi}{2}, \frac{\pi}{2})).

Mixed compositions (different functions composed):

sin⁡(arccos⁡(x))\sin(\arccos(x)), cos⁡(arctan⁡(x))\cos(\arctan(x)), tan⁡(arcsin⁡(x))\tan(\arcsin(x)), etc. These are simplified using a right triangle construction:

To evaluate sin⁡(arccos⁡(x))\sin(\arccos(x)): let θ=arccos⁡(x)\theta = \arccos(x), so cos⁡θ=x=x1\cos\theta = x = \frac{x}{1}. Construct a right triangle with adjacent side xx and hypotenuse 11. The opposite side is 1−x2\sqrt{1 - x^2} (by the Pythagorean theorem). Therefore:

sin⁡(arccos⁡(x))=oppositehypotenuse=1−x2\sin(\arccos(x)) = \frac{\text{opposite}}{\text{hypotenuse}} = \sqrt{1 - x^2}


This is valid for x∈[−1,1]x \in [-1, 1], and the result is always non-negative because arccos⁡(x)∈[0,π]\arccos(x) \in [0, \pi], where sine is non-negative.

To evaluate cos⁡(arctan⁡(x))\cos(\arctan(x)): let θ=arctan⁡(x)\theta = \arctan(x), so tan⁡θ=x=x1\tan\theta = x = \frac{x}{1}. Opposite =x= x, adjacent =1= 1, hypotenuse =1+x2= \sqrt{1 + x^2}. Therefore:

cos⁡(arctan⁡(x))=11+x2\cos(\arctan(x)) = \frac{1}{\sqrt{1 + x^2}}


The triangle method works for every mixed composition. It converts the problem from inverse trigonometric territory back to right triangle ratios, using the Pythagorean identity implicitly to find the missing side.

Graphs of Inverse Trigonometric Functions

    The graph of each inverse trigonometric function is the reflection of the corresponding restricted trigonometric graph over the line y=xy = x. This reflection swaps the roles of input and output — the domain of the original becomes the range of the inverse, and vice versa.

    $y = \arcsin(x)$: An increasing S-shaped curve.

  • [−1,1][-1, 1] (horizontal extent)
  • [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] (vertical extent)
  • (−1,−π2)(-1, -\frac{\pi}{2}), (0,0)(0, 0), (1,π2)(1, \frac{\pi}{2})
  • [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}]

    • $y = \arccos(x)$: A decreasing curve.

  • [−1,1][-1, 1]
  • [0,π][0, \pi]
  • (−1,π)(-1, \pi), (0,π2)(0, \frac{\pi}{2}), (1,0)(1, 0)
  • [0,π][0, \pi]

    • $y = \arctan(x)$: An increasing S-shaped curve with horizontal asymptotes.

  • (−∞,∞)(-\infty, \infty)
  • (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)
  • (0,0)(0, 0)
  • y=π2y = \frac{\pi}{2} as x→∞x \to \infty
  • y=−π2y = -\frac{\pi}{2} as x→−∞x \to -\infty
  • R\mathbb{R}

    • All three primary inverse functions are continuous on their domains. Arcsine and arccosine have bounded domains (closed intervals), so their graphs are finite curves with endpoints. Arctangent, with its infinite domain, extends without bound horizontally but is squeezed vertically between the asymptotes — a distinctive shape that appears in probability (the Cauchy distribution), physics (the arctangent potential), and many other contexts.