Inverse trigonometric functions

Intuition [Beginner]

You know the sine of 30 degrees is one-half. But suppose someone gives you a sine value and asks for the angle. What angle has sine equal to one-half? The answer is 30 degrees — and you just "undid" the sine function.

Every function that pairs inputs with outputs can be run in reverse: given the output, recover the input. Reversing sine, cosine, and tangent produces the inverse trigonometric functions: arcsine, arccosine, and arctangent. They answer the question "what angle produces this ratio?"

There is a catch. The sine function repeats: both 30 degrees and 150 degrees give sine equal to one-half. A repeated output means the reverse direction is ambiguous. The fix is to agree on a standard range — only return angles between $-90$ degrees and $90$ degrees for arcsine, for instance. With this convention, each output corresponds to exactly one angle.

Visual [Beginner]

A graph of $y=\sin \theta$ for angles from $-90$ degrees to $90$ degrees shows a curve that rises from $-1$ to $1$. If you reflect this curve across the line $y=\theta$, you get the arcsine curve: it takes a value between $-1$ and $1$ on the horizontal axis and returns an angle between $-90$ degrees and $90$ degrees on the vertical axis.

The reflection picture shows why the domain restriction matters: the full sine wave fails the horizontal line test, but the restricted piece passes it and therefore has a clean inverse.

Worked example [Beginner]

Find the angle $\theta$ in a right triangle where the opposite side is $3$ and the hypotenuse is $5$.

Step 1. Compute the sine: $\sin \theta =3/5=0.6$.

Step 2. Apply arcsine: $\theta =\arcsin (0.6)$. From a calculator or table, this is approximately $36.87$ degrees.

Step 3. Check: $\sin (36.87^{\circ })=0.6004\approx 0.6$. The small difference comes from rounding.

What this tells us: arcsine converts a ratio back to the angle that produced it. In any right triangle, knowing the ratio of two sides is enough to recover the angle.

Check your understanding [Beginner]

Formal definition [Intermediate+]

The trigonometric functions $\sin$, $\cos$, and $\tan$ are periodic and therefore not injective on $\mathbb{R}$. To define inverses, each must be restricted to a domain on which it is both injective and surjective onto a convenient codomain.

Definition. The inverse trigonometric functions are defined by restriction as follows:

$$\arcsin :[-1,1]\to \left[-\frac{\pi }{2},\frac{\pi }{2}\right],\arcsin (x)=\theta \text{ where }\sin \theta =x.$$

$$\arccos :[-1,1]\to [0,\pi ],\arccos (x)=\theta \text{ where }\cos \theta =x.$$

$$\arctan :\mathbb{R}\to \left(-\frac{\pi }{2},\frac{\pi }{2}\right),\arctan (x)=\theta \text{ where }\tan \theta =x.$$

The restricted domains $[-\pi /2,\pi /2]$ for sine and $[0,\pi ]$ for cosine are called the principal branches. On these intervals each function is strictly monotone (sine increasing, cosine decreasing), hence injective, and attains every value in $[-1,1]$, hence surjective. The arctangent needs no restriction of its domain (the argument $x$ can be any real number) because tangent is already strictly increasing on $(-\pi /2,\pi /2)$ and its range is all of $\mathbb{R}$.

Counterexamples to common slips

• $\arcsin (\sin \theta )=\theta$ does not hold for all $\theta$. If $\theta =150^{\circ }=5\pi /6$, then $\sin \theta =1/2$ and $\arcsin (1/2)=\pi /6\ne 5\pi /6$. The identity $\arcsin (\sin \theta )=\theta$ holds only for $\theta \in [-\pi /2,\pi /2]$.
• The domain of arcsine is not all real numbers. Values $|x|>1$ are outside the range of sine and have no arcsine value.
• Arctangent has no discontinuities. Unlike arcsine and arccosine, which have finite domains, $\arctan (x)$ is defined for every real $x$ and is continuous everywhere. Its graph has horizontal asymptotes at $y=\pm \pi /2$ but no breaks.

Key theorem with proof [Intermediate+]

Theorem (Composition identities). For all $x\in [-1,1]$:

$$\sin (\arcsin x)=x,\cos (\arccos x)=x.$$

For all $x\in \mathbb{R}$:

$$\tan (\arctan x)=x.$$

For all $\theta \in [-\pi /2,\pi /2]$:

$$\arcsin (\sin \theta )=\theta .$$

For all $\theta \in [0,\pi ]$:

$$\arccos (\cos \theta )=\theta .$$

Proof. For the first identity: let $\theta =\arcsin x$. By definition of arcsine, $\sin \theta =x$ and $\theta \in [-\pi /2,\pi /2]$. Substituting: $\sin (\arcsin x)=\sin \theta =x$. The argument for $\cos (\arccos x)=x$ is identical, using the principal branch $[0,\pi ]$.

For $\tan (\arctan x)=x$: let $\varphi =\arctan x$. By definition, $\tan \varphi =x$ and $\varphi \in (-\pi /2,\pi /2)$. Substituting gives the result.

For the converse direction, $\arcsin (\sin \theta )=\theta$: since $\theta \in [-\pi /2,\pi /2]$ and $\sin \theta$ lies in $[-1,1]$, the definition of arcsine returns the unique angle $\alpha \in [-\pi /2,\pi /2]$ with $\sin \alpha =\sin \theta$. Because $\theta$ itself is in $[-\pi /2,\pi /2]$, uniqueness forces $\alpha =\theta$. $\square$

Bridge. The foundational reason these composition identities hold is that restricting each trigonometric function to its principal branch makes it a genuine bijection onto its codomain, and the inverse of a bijection satisfies $f(f^{-1}(x))=x$ and $f^{-1}(f(x))=x$ on the appropriate domains. This is exactly the content that identifies the inverse trigonometric functions with the general inverse-function construction from 00.02.05, and the bridge is between the periodic, many-to-one behaviour of the unrestricted trigonometric functions and the single-valued, invertible behaviour on the principal branch. The central insight is that domain restriction restores invertibility, and putting these together, the composition identities generalise from the trigonometric context to all periodic functions with suitable monotone restrictions. This pattern builds toward 00.05.03 where the Euler identity connects inverse trigonometric functions to the complex logarithm via $\arctan x=\frac{1}{2i}\ln \frac{1+ix}{1-ix}$, and appears again in 00.07.01 where the unit-circle definition extends these inverses to all quadrants.

Exercises [Intermediate+]

Advanced results [Master]

Theorem 1 (derivatives of the inverse trigonometric functions). The inverse trigonometric functions are differentiable on the interior of their domains:

$$\frac{d}{dx}\arcsin x=\frac{1}{\sqrt{1-x^2}},x\in (-1,1).$$

$$\frac{d}{dx}\arccos x=-\frac{1}{\sqrt{1-x^2}},x\in (-1,1).$$

$$\frac{d}{dx}\arctan x=\frac{1}{1+x^2},x\in \mathbb{R}.$$

These are derived via the inverse function theorem. For arcsine: $\frac{d}{dx}\arcsin x=1/\cos (\arcsin x)=1/\sqrt{1-x^2}$. The negative sign in the arccosine derivative reflects the decreasing nature of cosine on $[0,\pi ]$.

Theorem 2 (logarithmic representations). The inverse trigonometric functions can be expressed using complex logarithms:

$$\arcsin x=-i\ln \left(ix+\sqrt{1-x^2}\right),\arccos x=-i\ln \left(x+i\sqrt{1-x^2}\right),$$

$$\arctan x=\frac{1}{2i}\ln \frac{1+ix}{1-ix}.$$

These identities follow from the Euler formula $e^{i\theta }=\cos \theta +i\sin \theta$ and provide the bridge between inverse trigonometry and the complex logarithm.

Theorem 3 (power series for arctangent). For $|x|\le 1$:

$$\arctan x=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\cdots =\sum _{n=0}^{\infty }\frac{(-1{)}^n{x}^{2n+1}}{2n+1}.$$

This is the Gregory series (1671), obtained by integrating the geometric series $1/(1+x^2)=1-x^2+x^4-x^6+\cdots$ term by term. Setting $x=1$ gives the Leibniz formula $\pi /4=1-1/3+1/5-1/7+\cdots$.

Theorem 4 (power series for arcsine). For $|x|\le 1$:

$$\arcsin x=x+\frac{1}{2}\frac{x^3}{3}+\frac{1\cdot 3}{2\cdot 4}\frac{x^5}{5}+\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\frac{x^7}{7}+\cdots$$

This series is derived from the binomial expansion of $(1-x^2{)}^{-1/2}$ integrated term by term.

Theorem 5 (composition identities: the converse direction). The identities $\arcsin x+\arccos x=\pi /2$ for all $x\in [-1,1]$, and $\arctan x+\mathrm{arccot}x=\pi /2$ for all $x\in \mathbb{R}$, reflect the complementary-angle relations. The first is proved by differentiating: $\frac{d}{dx}(\arcsin x+\arccos x)=\frac{1}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-x^2}}=0$, so the sum is constant; evaluating at $x=0$ gives $\arcsin 0+\arccos 0=0+\pi /2$.

Theorem 6 (addition formula for arctangent). For $xy<1$:

$$\arctan x+\arctan y=\arctan \frac{x+y}{1-xy}.$$

When $xy>1$ and $x,y>0$, the sum equals $\arctan \frac{x+y}{1-xy}+\pi$. This two-case structure is the source of the Machin-type formulas for computing $\pi$.

Synthesis. The foundational reason inverse trigonometric functions connect analysis to geometry is that they convert ratios (outputs of the trigonometric functions) back into angles (inputs), and the central insight is that domain restriction via principal branches is the mechanism that makes this reversal single-valued. This is exactly the content that identifies the inverse functions with the branches of the complex logarithm through the Euler formula, and the bridge is between the real-variable theory of inverse functions from 00.02.05 and the complex-analytic theory of the logarithm in 00.05.03. Putting these together, the derivatives of the inverse trigonometric functions are algebraic expressions (reciprocals of square roots and quadratic polynomials) rather than transcendental ones, the power series for arctangent gives the Leibniz formula for $\pi$, and the logarithmic representations generalise into the inverse-function theorem for complex-analytic maps. The pattern recurs throughout analysis: every inverse trigonometric identity has a complex-exponential counterpart, and the composition identities build toward 00.08.01 where the addition formulas produce further inverse-function relations.

Full proof set [Master]

Proposition 1 (derivative of arcsine). For $x\in (-1,1)$, $\frac{d}{dx}\arcsin x=\frac{1}{\sqrt{1-x^2}}$.

Proof. Let $y=\arcsin x$, so $\sin y=x$ and $y\in (-\pi /2,\pi /2)$. Differentiate both sides with respect to $x$: $\cos y\cdot \frac{dy}{dx}=1$, hence $\frac{dy}{dx}=1/\cos y$. On $(-\pi /2,\pi /2)$, $\cos y>0$, so $\cos y=\sqrt{1-{\sin }^{2}y}=\sqrt{1-x^2}$. Therefore $\frac{d}{dx}\arcsin x=1/\sqrt{1-x^2}$. $\square$

Proposition 2 (derivative of arctangent). For $x\in \mathbb{R}$, $\frac{d}{dx}\arctan x=\frac{1}{1+x^2}$.

Proof. Let $y=\arctan x$, so $\tan y=x$ and $y\in (-\pi /2,\pi /2)$. Differentiate: ${\sec }^{2}y\cdot \frac{dy}{dx}=1$, hence $\frac{dy}{dx}=1/{\sec }^{2}y={\cos }^{2}y$. Using ${\cos }^{2}y=1/(1+{\tan }^{2}y)=1/(1+x^2)$. Therefore $\frac{d}{dx}\arctan x=1/(1+x^2)$. $\square$

Proposition 3 ($\arcsin x+\arccos x=\pi /2$). For all $x\in [-1,1]$, $\arcsin x+\arccos x=\pi /2$.

Proof. Differentiate the left side: $\frac{d}{dx}(\arcsin x+\arccos x)=\frac{1}{\sqrt{1-x^2}}+\left(-\frac{1}{\sqrt{1-x^2}}\right)=0$ for $x\in (-1,1)$. Therefore $\arcsin x+\arccos x$ is constant on $(-1,1)$, and by continuity on $[-1,1]$. Evaluate at $x=0$: $\arcsin 0+\arccos 0=0+\pi /2=\pi /2$. $\square$

Connections [Master]

• Complex numbers and Euler's identity 00.05.03. The inverse trigonometric functions acquire their deepest interpretation through the complex exponential. The identity $\arctan x=\frac{1}{2i}\ln \frac{1+ix}{1-ix}$ identifies each inverse trigonometric value with a branch of the complex logarithm, and the Euler identity $e^{i\theta }=\cos \theta +i\sin \theta$ is the foundational reason the logarithmic representations work. The inverse functions are the analytic continuation of the angle-extraction problem from real ratios to complex arguments.

• Functions, domain, and range 00.02.05. The inverse trigonometric functions are the precalc strand's most instructive example of restricted-domain inversion. A function that fails the horizontal line test on its natural domain becomes invertible after restriction to a monotone piece, and this procedure — motivated and carried out for sine, cosine, and tangent — is the template for defining inverse functions in every context where periodicity or non-monotonicity obstructs invertibility.

• Right-triangle trigonometry 00.06.01. The six trigonometric ratios defined in 00.06.01 acquire their computational power from the ability to run in reverse: given a ratio, the inverse function returns the angle. Every application of trigonometry to measurement (surveying, navigation, construction) relies on this reversal, and the principal-branch convention is the mathematical formalisation of the physical constraint that angles in a triangle are between $0$ and $\pi$.

Historical & philosophical context [Master]

Euler 1748 Introductio in analysin infinitorum ^[Euler1748] introduced the inverse trigonometric functions as analytic objects, giving them the notation $\arcsin$ and deriving the power series for arctangent via term-by-term integration of the geometric series. The concept of restricting the domain to obtain a single-valued inverse was implicit in the chord and sine tables of Ptolemy ~150 ^[Ptolemy150] and the Indian mathematicians (Aryabhata 476, who tabulated half-chords for specific arcs), but the formal machinery of inverse functions and principal branches is an Eulerian contribution. The Gregory series for arctangent was discovered independently by Gregory 1671 and Leibniz 1674; the Leibniz formula $\pi /4=1-1/3+1/5-\cdots$ was the first infinite-series representation of $\pi$. The Machin formula $\pi /4=4\arctan (1/5)-\arctan (1/239)$ (Machin 1706) and its descendants remained the primary computational tool for $\pi$ until the 20th century.

Bibliography [Master]

@book{Euler1748,
  author = {Euler, Leonhard},
  title = {Introductio in analysin infinitorum},
  publisher = {Marcus-Michaelis Bousquet},
  address = {Lausanne},
  year = {1748}
}

@book{Ptolemy150,
  author = {Ptolemy, Claudius},
  title = {Mathematike Syntaxis (Almagest)},
  year = {~150 CE}
}

@book{Granville1911,
  author = {Granville, William Anthony},
  title = {Elements of the Differential and Integral Calculus},
  publisher = {Ginn and Company},
  year = {1911}
}

@article{Machin1706,
  author = {Machin, John},
  title = {The true value of pi},
  journal = {Philosophical Transactions of the Royal Society},
  year = {1706}
}

@book{Lang1988,
  author = {Lang, Serge},
  title = {Basic Mathematics},
  publisher = {Springer},
  year = {1988}
}