Inverse Trigonometric Functions

Q: What is the derivative of arcsin x?

The derivative of arcsin x is 1 over the square root of 1 minus x squared, for x between minus 1 and 1. The slope is at least 1 and grows without bound near plus and minus 1, where the graph has vertical tangent lines.

Q: What is the derivative of arctan x?

The derivative of arctan x is 1 over 1 plus x squared, for every real x. It is largest, equal to 1, at the origin and tends to zero for large x, which is why the graph flattens towards its asymptotes at plus and minus pi over 2.

Q: What is the derivative of arcsec x?

It is 1 over the absolute value of x times the square root of x squared minus 1, for absolute x greater than 1. The absolute value is there because the arcsecant is increasing on both pieces of its graph, so its slope is positive also for x below minus 1.

Q: What is the derivative of arccos x?

It is minus 1 over the square root of 1 minus x squared, the negative of the derivative of arcsin x. This follows from arccos x equal to pi over 2 minus arcsin x. The same pattern gives the derivatives of arccot and arccsc.

Q: What is the difference between sin inverse x and 1 over sin x?

The exponent minus 1 in sin inverse x means the inverse function, written arcsin x, which gives an angle. The reciprocal 1 over sin x is the cosecant of x, which is a number computed from an angle. They are unrelated functions.

Q: Why is arcsin of sin x not always x?

Because the sine repeats, many angles share the same sine, and the arcsine returns only the one between minus pi over 2 and pi over 2. So arcsin of sin x equals x only on that interval; for x equal to pi it gives 0.

Q: What is cos of arcsin x?

It is the square root of 1 minus x squared. Draw a right triangle with the opposite side x and the hypotenuse 1; the adjacent side is the square root of 1 minus x squared, and the cosine is adjacent over hypotenuse.

Q: How do you differentiate arctan of a function?

Use the chain rule: the derivative of arctan u is the derivative of u divided by 1 plus u squared. For arctan of 2x that is 2 over 1 plus 4x squared. The same pattern works for arcsin, with the square root of 1 minus u squared.

Q: What are the limits of arctan x at infinity?

As x tends to plus infinity, arctan x tends to pi over 2, and as x tends to minus infinity it tends to minus pi over 2. The graph therefore has the two horizontal asymptotes y equal to plus and minus pi over 2.

Updated 18 October 2026 13 min read Intermediate Baca dalam Bahasa Indonesia

Calculus Proof Computing

Quick answer

The inverse trigonometric functions give the angle that has a given trigonometric value, and their derivatives are algebraic: (d)/(dx)\arcsin x=(1)/(√(1-x^2)), (d)/(dx)\arctan x=(1)/(1+x^2) and (d)/(dx)\arcsec x=(1)/(|x|√(x^2-1)). Each comes from the rule for the derivative of an inverse, read off a reference triangle. The three "co" functions have the negatives of these derivatives, because \arccos x=\fracπ2-\arcsin x and likewise for the others.

Key takeaways

Each trigonometric function is made one-to-one on an interval, and its inverse is the angle on that interval with the given value: \arcsin and \arctan give angles in \left(-\fracπ2,\fracπ2\right), \arccos and \arccot in (0,π).
\sin(\arcsin x)=x for -1\le x\le1, but \arcsin(\sin y)=y only for -\fracπ2\le y\le\fracπ2; for example \arcsin(\sinπ)=0.
A right triangle with the angle θ=\arcsin x has sides x, √(1-x^2) and 1, which gives \cos(\arcsin x)=√(1-x^2) and the derivative (1)/(\cosθ)=(1)/(√(1-x^2)).
\arcsin has vertical tangents at x=±1; \arctan is differentiable everywhere, with a slope at most 1 that tends to 0; \arcsec has a positive slope on both pieces, so its formula needs |x|.
With the chain rule, (d)/(dx)\arctan u=(u')/(1+u^2) and (d)/(dx)\arcsin u=(u')/(√(1-u^2)); for instance (d)/(dx)\arcsin\left(x^2\right)=(2x)/(√(1-x^4)).
A picture hanging p to q above your eye is seen at the angle \arctan\frac qx-\arctan\frac px, which is largest at the distance x=√(pq), where its derivative is 0.

What are the inverse trigonometric functions?

An inverse trigonometric function gives the angle, in radians, whose sine, cosine, tangent, cotangent, secant or cosecant is a given number, chosen from one fixed interval so that it is a function. Each trigonometric function repeats, so it is not one-to-one and has no inverse; restricting it to an interval where it is one-to-one fixes that, and reflecting the restricted graph in the line y=x draws the inverse.

Function	Domain	Range
y=\arcsin x	-1\le x\le1	-\fracπ2\le y\le\fracπ2
y=\arccos x	-1\le x\le1	0\le y\leπ
y=\arctan x	all real x	-\fracπ2<y<\fracπ2
y=\arccot x	all real x	0<y<π
y=\arcsec x	x\le-1 or x\ge1	0\le y\leπ, y\ne\fracπ2
y=\arccsc x	x\le-1 or x\ge1	-\fracπ2\le y\le\fracπ2, y\ne0

So \arctan x is the angle between -\fracπ2 and \fracπ2 whose tangent is x. The open intervals avoid the values where the tangent, cotangent, secant or cosecant is not defined.

Notation. The inverse sine is also written \sin^-1x, and likewise for the others. The exponent -1 here means inverse function, and not reciprocal: \sin^-1x=\arcsin x is not (1)/(\sin x), which is \csc x.

Symmetry and asymptotes. The graph of y=\arctan x is symmetric about the origin, so \arctan(-x)=-\arctan x: the arctangent is an odd function. It has two horizontal asymptotes, y=\fracπ2 and y=-\fracπ2, because \arctan x→\fracπ2 as x→∞ and →-\fracπ2 as x→-∞.

The branch of the arcsecant. There is no general agreement about \arcsec x for negative x. Here it is taken in the second quadrant, between \fracπ2 and π. This choice makes \arcsec x=\arccos\frac1x, the identity that calculators use, and it makes \arcsec increasing on each piece. Some texts choose the third quadrant instead, which simplifies the derivative but breaks the identity.

The cofunction identities. The three "co" functions are the other three reflected and shifted: \arccos x=\fracπ2-\arcsin x \arccot x=\fracπ2-\arctan x \arccsc x=\fracπ2-\arcsec x and \arcsec x=\arccos\frac1x, \arccsc x=\arcsin\frac1x.

Example 1. The values of \arctan come from the first and fourth quadrants. \arctan1=\fracπ4, because \tan\fracπ4=1. \arctan\left(-√3\right)=-\fracπ3, because \tan\left(-\fracπ3\right)=-√3 and -\fracπ3 lies in \left(-\fracπ2,\fracπ2\right). Others: \arcsin\frac12=\fracπ6, \arccos\left(-\frac12\right)=(2π)/(3) (in [0,π], not -\fracπ3), \arccot(-1)=(3π)/(4).

Choose a function below and slide x: its graph, the tangent line there and the graph of the slope are drawn.

Interactive: an inverse trigonometric function and its slope

Slide a point along the graph of arcsin, arccos, arctan, arccot, arcsec or arccsc and watch the slope of its tangent line trace out the graph of the derivative, with the domain and range of each.

What are sin(arcsin x), arcsin(sin y) and cos(arcsin x)?

A trigonometric function undoes its inverse everywhere on the domain of the inverse, \sin(\arcsin x)=x for -1\le x\le1, but the inverse undoes the trigonometric function only on the restricted interval: \arcsin(\sin y)=y only when -\fracπ2\le y\le\fracπ2. The sine repeats, so many angles share a sine, and the arcsine returns the one in its range. For example \arcsin(\sinπ)=\arcsin0=0 \arcsin(\sin2)=π-2≈1.14, because \sin2=\sin(π-2) and π-2 lies in \left[-\fracπ2,\fracπ2\right]. Likewise \arccos(\cos(-1))=1.

A trigonometric function of an inverse is algebraic. Draw the reference triangle: if θ=\arcsin x with 0<x<1, then \sinθ=x, so a right triangle with the opposite side x and the hypotenuse 1 has the angle θ. By Pythagoras the adjacent side is √(1-x^2). Reading the triangle, \cos(\arcsin x)=√(1-x^2) \tan(\arcsin x)=(x)/(√(1-x^2)). For θ=\arctan x with x>0 take the opposite side x and the adjacent side 1, so the hypotenuse is √(1+x^2) and \sec(\arctan x)=√(1+x^2), \sin(\arctan x)=(x)/(√(1+x^2)). For θ=\arcsec x with x>1 take the hypotenuse x and the adjacent side 1, so the opposite side is √(x^2-1) and \tan(\arcsec x)=√(x^2-1).

For example \cos\left(\arcsin\frac35\right)=√(1-(9)/(25))=\frac45 and \tan\left(\arcsin\frac35\right)=\frac34.

These are the very expressions that appear in the derivatives, which is no accident: the derivative of an inverse is the reciprocal of the derivative of the trigonometric function, written in terms of the triangle.

Move x below and watch the triangle change: its third side comes from Pythagoras, and the derivative is read straight from it.

Interactive: the reference triangle of an inverse trigonometric function

Move x and see the right triangle whose angle is arcsin x, arctan x or arcsec x, its third side from Pythagoras, and the derivative read straight off the triangle.

What are the derivatives of arcsin and arctan?

(d)/(dx)\arcsin x=(1)/(√(1-x^2)) for -1<x<1 and (d)/(dx)\arctan x=(1)/(1+x^2) for every real x. Both come from the derivative rule for inverses: \left(f^-1\right)'(x)=(1)/(f'\left(f^-1(x)\right)).

Arcsine. The sine is differentiable with derivative \cos y>0 on -\fracπ2<y<\fracπ2, so the arcsine is differentiable on -1<x<1. With f=\sin: \left(\arcsin\right)'(x)=(1)/(\cos(\arcsin x)) =(1)/(√(1-\sin^2(\arcsin x))) =(1)/(√(1-x^2)). We cannot expect a derivative at x=±1: the graph has vertical tangent lines there, and the slope (1)/(√(1-x^2)) grows without bound as x→±1. In between the slope is at least 1.

Arctangent. The tangent has derivative \sec^2y>0 on -\fracπ2<y<\fracπ2: \left(\arctan\right)'(x)=(1)/(\sec^2(\arctan x)) =(1)/(1+\tan^2(\arctan x)) =(1)/(1+x^2). This is defined for all real x, is largest, 1, at x=0, and tends to 0 as x→±∞: the graph flattens towards its horizontal asymptotes.

With the chain rule. If u is a differentiable function of x, (d)/(dx)\arcsin u=(1)/(√(1-u^2))·(du)/(dx) (|u|<1) (d)/(dx)\arctan u=(1)/(1+u^2)·(du)/(dx).

Example 2. (d)/(dx)\arcsin\left(x^2\right)=(1)/(√(1-\left(x^2\right)^2))·2x=(2x)/(√(1-x^4)) for |x|<1.

Example 3. (d)/(dx)(1)/(\arctan x)=-(1)/(\arctan^2x)·(1)/(1+x^2)=-(1)/(\left(1+x^2\right)\arctan^2x) for x\ne0. This is the derivative of the reciprocal of the arctangent, which is not its inverse: the chain rule with the power -1.

A surprising simplification. (d)/(dx)\arctan\frac1x=(1)/(1+(1)/(x^2))·\left(-(1)/(x^2)\right)=-(1)/(x^2+1), the derivative of -\arctan x. Indeed \arctan\frac1x=\fracπ2-\arctan x for x>0.

An antiderivative in disguise. The two formulas say that \arcsin x and \arctan x are functions whose slopes are the algebraic expressions (1)/(√(1-x^2)) and (1)/(1+x^2). Reading them backwards is what a later chapter of integration does.

What are the derivatives of arcsec, arccos, arccot and arccsc?

(d)/(dx)\arcsec x=(1)/(|x|√(x^2-1)) for |x|>1, and the derivatives of \arccos, \arccot and \arccsc are the negatives of those of \arcsin, \arctan and \arcsec.

Arcsecant, by implicit differentiation. The rule for inverses does not apply to \sec directly, because its domain is not connected, but it applies on each piece. Let y=\arcsec x, so \sec y=x. Differentiate both sides with respect to x, using the chain rule: \sec y\tan y·(dy)/(dx)=1, (dy)/(dx)=(1)/(\sec y\tan y). For |x|>1 the angle y lies in \left(0,\fracπ2\right)∪\left(\fracπ2,π\right), where \sec y\tan y>0. Since \sec y=x and \tan y=±√(\sec^2y-1)=±√(x^2-1), we get (dy)/(dx)=(1)/(± x√(x^2-1)). The slope of the graph is positive everywhere, so the sign is the one that makes the denominator positive: x√(x^2-1) for x>1 and -x√(x^2-1) for x<-1. One expression covers both: (d)/(dx)\arcsec x=(1)/(|x|√(x^2-1)).

The co-functions. Because \arccos x=\fracπ2-\arcsin x, differentiating gives (d)/(dx)\arccos x=-(d)/(dx)\arcsin x =-(1)/(√(1-x^2)). The same works for the others: \arccot x=\fracπ2-\arctan x and \arccsc x=\fracπ2-\arcsec x.

Function	Derivative
\arcsin x	(1)/(√(1-x^2))
\arccos x	-(1)/(√(1-x^2))
\arctan x	(1)/(1+x^2)
\arccot x	-(1)/(1+x^2)
\arcsec x	(1)/(\left\|x\right\|√(x^2-1))
\arccsc x	-(1)/(\left\|x\right\|√(x^2-1))

With a function u(x) inside, each formula gets the factor (du)/(dx), and the one for arcsec needs |u|>1: (d)/(dx)\arcsec u=(1)/(|u|√(u^2-1))·(du)/(dx).

Example 4. (d)/(dx)\arcsec\left(5x^4\right)=(1)/(5x^4√(25x^8-1))·20x^3=(4)/(x√(25x^8-1)) for 5x^4>1. The absolute value \left|5x^4\right|=5x^4 is just 5x^4 because it is never negative, and the x left in the denominator keeps its sign.

Type any function below, built from these and the earlier functions, and see each rule applied.

Interactive: differentiate with inverse trigonometric functions

Type a function with arcsin, arccos, arctan, arccot, arcsec, arccsc or abs of other functions, and see each rule applied, with the outside and inside function at every chain step, the derivative checked against a difference quotient.

Where should you stand to see a picture best?

A picture whose lower edge hangs p and whose upper edge hangs q above your eye is seen at the angle θ(x)=\arctan\frac qx-\arctan\frac px when you stand x away from the wall, and the angle is largest at x=√(pq), the geometric mean of the two heights. The angle between the line of sight to the top and the line of sight to the bottom is the difference of two angles of elevation, each the arctangent of a height over a distance.

The derivative. By the chain rule, (d)/(dx)\arctan\frac qx=(1)/(1+(q^2)/(x^2))·\left(-(q)/(x^2)\right)=-(q)/(x^2+q^2), and the same with p, so θ'(x)=-(q)/(x^2+q^2)+(p)/(x^2+p^2). This is zero when q\left(x^2+p^2\right)=p\left(x^2+q^2\right), that is (q-p)x^2=pq(q-p), so x^2=pq and x=√(pq). For x a little smaller θ'>0 and for x a little larger θ'<0: the angle grows as you step back from very close, and shrinks as you move far away, so x=√(pq) is the top of the curve.

Example. A picture hangs from 1 m to 3 m above your eye. The best distance is x=√(1·3)=√3≈1.73 m, and the angle there is θ=\arctan(3)/(√3)-\arctan(1)/(√3) =\arctan√3-\arctan(1)/(√3) =\fracπ3-\fracπ6=\fracπ6, which is 30^\circ. In general the largest angle is \arctan(q-p)/(2√(pq)). Standing very close, the picture is above you and squeezed into a narrow angle; far away it is small. The best view is in between.

Choose the two heights and walk back and forth below.

Interactive: the best place to look at a picture

Choose how high the lower and upper edges of a picture hang above your eye and move away from the wall: the viewing angle is arctan(q/x) − arctan(p/x), and it is largest where its derivative is zero, at the geometric mean of the two heights.

How do you compute inverse trigonometric functions and their derivatives in code?

Programming languages name them asin, acos, atan and atan2: math.asin(x) returns an angle in radians in \left[-\fracπ2,\fracπ2\right], and SymPy's sp.asin, sp.atan and sp.asec differentiate to the formulas of this article. There is no asec in Python's math; use math.acos(1/x), which is the branch chosen here. Remember that math.asin(math.sin(x)) is x only for x in \left[-\fracπ2,\fracπ2\right]. The last lines scan the viewing angle of the picture from 1 m to 3 m and find the best distance and the largest angle without using the derivative: √3≈1.732 and 30^\circ.

Python and SymPy: inverse trigonometric derivatives, compositions and the viewing angle

import math
import sympy as sp

x = sp.symbols('x', positive=True)
for f in (sp.asin(x), sp.acos(x), sp.atan(x), sp.acot(x), sp.asec(x), sp.acsc(x), sp.asin(x**2), sp.atan(1 / x), sp.asec(5 * x**4)):
    print(f, '->', sp.simplify(sp.diff(f, x)))
# asin(x) -> 1/sqrt(1 - x**2)
# acos(x) -> -1/sqrt(1 - x**2)
# atan(x) -> 1/(x**2 + 1)
# acot(x) -> -1/(x**2 + 1)
# asec(x) -> 1/(x*sqrt(x**2 - 1))
# acsc(x) -> -1/(x*sqrt(x**2 - 1))
# asin(x**2) -> 2*x/sqrt(1 - x**4)
# atan(1/x) -> -1/(x**2 + 1)
# asec(5*x**4) -> 4/(x*sqrt(25*x**8 - 1))

print(round(math.asin(math.sin(math.pi)), 6), round(math.asin(math.sin(2)), 6), round(math.pi - 2, 6))   # 0.0 1.141593 1.141593
print(round(math.cos(math.asin(0.6)), 6), round(math.tan(math.asin(0.6)), 6))                             # 0.8 0.75
print(round(math.degrees(math.atan(1)), 6), round(math.degrees(math.acos(-0.5)), 6))                     # 45.0 120.0

angle = lambda t, p=1, q=3: math.atan(q / t) - math.atan(p / t)      # the viewing angle
best = max((k / 1000 for k in range(1, 20001)), key=angle)           # a scan, no derivative
print(best, round(math.degrees(angle(best)), 4))                     # 1.732 30.0
print(round(math.sqrt(1 * 3), 4), round(math.degrees(math.atan((3 - 1) / (2 * math.sqrt(1 * 3)))), 4))   # 1.7321 30.0

C++

#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;

const double PI = acos(-1.0);

double angle(double t, double p, double q) {           // the viewing angle of a picture from p to q above the eye
  return atan(q / t) - atan(p / t);
}

int main() {
  printf("%.6f %.6f %.6f\n", asin(sin(PI)), asin(sin(2.0)), PI - 2);        // 0 1.141593 1.141593
  printf("%.6f %.6f\n", cos(asin(0.6)), tan(asin(0.6)));                    // 0.8 0.75
  printf("%.6f %.6f\n", atan(1.0) * 180 / PI, acos(-0.5) * 180 / PI);      // 45 120

  double best = 1, top = 0;                            // a scan, no derivative
  for (int k = 1; k <= 20000; k++) {
    double t = k / 1000.0;
    if (angle(t, 1, 3) > top) { top = angle(t, 1, 3); best = t; }
  }
  printf("%.3f %.4f\n", best, top * 180 / PI);                              // 1.732 30
  printf("%.4f %.4f\n", sqrt(1.0 * 3), atan((3 - 1) / (2 * sqrt(1.0 * 3))) * 180 / PI);   // 1.7321 30
  return 0;
}

/* output:
0.000000 1.141593 1.141593
0.800000 0.750000
45.000000 120.000000
1.732 30.0000
1.7321 30.0000
*/

Who named the inverse trigonometric functions, and who found their series?

The arctangent had its series long before it had its derivative, and its notation settled only in the 19th century.

Series. The series \arctan x=x-(x^3)/(3)+(x^5)/(5)-… was found by Madhava of Sangamagrama in India around the 14th century, and again by James Gregory in 1671, and by Leibniz soon after. Putting x=1 gives \fracπ4=1-\frac13+\frac15-…, the Leibniz series for π.
Derivative. With the calculus of Newton and Leibniz, the slope (1)/(1+x^2) of the arctangent explains that series: the series of (1)/(1+x^2) is a geometric series, and integrating it term by term gives the series of \arctan x.
Names. The prefix arc for the inverse trigonometric functions, because the angle measures an arc of the unit circle, is credited to Karl Scherffer in the 18th century. The exponent notation \sin^-1x was introduced by John Herschel in 1813, which is why \sin^-1 and (1)/(\sin) are so often confused.
Euler. Leonhard Euler treated the inverse trigonometric functions together with the exponential and the logarithm, and found their remarkable connection through complex numbers, \arctan x=(1)/(2i)\ln(1+ix)/(1-ix).

What mistakes do people make with inverse trigonometric functions?

The most common mistakes with inverse trigonometric functions are the nine below, each with the correct statement.

Mistake	Correct
❌ \sin^-1x=(1)/(\sin x)	The -1 means the inverse function, \arcsin x; the reciprocal (1)/(\sin x) is \csc x.
❌ \arcsin(\sin y)=y for every y	Only for -\fracπ2\le y\le\fracπ2: \arcsin(\sinπ)=0.
❌ \arccos\left(-\frac12\right)=-\fracπ3	The range of \arccos is [0,π], so the answer is (2π)/(3).
❌ (d)/(dx)\arcsin x=(1)/(1+x^2)	That is the arctangent. The arcsine has (1)/(√(1-x^2)).
❌ (d)/(dx)\arctan x=(1)/(1-x^2)	The sign is plus: (1)/(1+x^2), defined for every x.
❌ (d)/(dx)\arccos x=(1)/(√(1-x^2))	The arccosine falls: -(1)/(√(1-x^2)).
❌ (d)/(dx)\arcsin(2x)=(2)/(√(1-x^2))	The inside is 2x, so (2)/(√(1-4x^2)) by the chain rule.
❌ (d)/(dx)\arcsec x=(1)/(x√(x^2-1)) for all x	The slope is positive on both pieces: use \|x\|, as (1)/(\|x\|√(x^2-1)).
❌ (d)/(dx)(1)/(\arctan x)=(1)/(1+x^2)	The reciprocal is a power -1: -(1)/(\left(1+x^2\right)\arctan^2x).

Practice: test your understanding

These questions mix everything above. A wrong answer costs nothing here: read the hint and try again.

True or false?

Decide whether each statement is True or False.

(d)/(dx)\arctan x=(1)/(1+x^2) for every real x.
(d)/(dx)\arccos x=(1)/(√(1-x^2)).
\sin^-1x means (1)/(\sin x).
\arcsin(\sinπ)=π.
The slope of \arcsec is positive on both pieces of its graph.
\arcsin x has vertical tangent lines at x=±1.

Select all that apply

Choose all the correct derivatives.

(d)/(dx)\arcsin\left(x^2\right)=(2x)/(√(1-x^4))
(d)/(dx)\arctan(2x)=(2)/(1+4x^2)
(d)/(dx)\arccot x=-(1)/(1+x^2)
(d)/(dx)\arctan\frac1x=-(1)/(1+x^2)
(d)/(dx)\arcsin(2x)=(2)/(√(1-x^2))

The slope of the arctangent

Find the slope of y=\arctan x at x=1 (as a decimal).

The slope of the arcsine

Find the slope of y=\arcsin x at x=0.6.

An angle in the second quadrant

What is \arccos\left(-\frac12\right)?

(2π)/(3)
-\fracπ3
\fracπ3
(4π)/(3)

The arcsine of a sine

What is \arcsin(\sinπ)?

0
π
-π
\fracπ2

A chain rule at the origin

Find the slope of y=\arcsin(2x) at x=0.

The slope of the arcsecant

Find the slope of y=\arcsec x at x=\frac53 (as a decimal).

A cosine of an arcsine

Find \cos\left(\arcsin\frac35\right) (as a decimal).

The arctangent of a reciprocal

What is (d)/(dx)\arctan\frac1x for x\ne0?

-(1)/(1+x^2)
(1)/(1+x^2)
-(1)/(x^2\left(1+x^2\right))
(1)/(x^2)

The best place to stand

A picture hangs from 1 m to 4 m above your eye. At what distance from the wall (in m) is the viewing angle largest?

Why the absolute value

Why does the derivative of \arcsec x have |x| in it?

its slope is positive on both pieces of the graph, also for x<-1
because x can be zero
to make the square root defined
because the function is odd

Summary: inverse trigonometric functions at a glance

Definition: the angle with a given trigonometric value, in one fixed range: \arcsin and \arctan in \left(-\fracπ2,\fracπ2\right) (closed for arcsin), \arccos in [0,π], \arccot in (0,π).
Compositions: \sin(\arcsin x)=x but \arcsin(\sin y)=y only on \left[-\fracπ2,\fracπ2\right]; \cos(\arcsin x)=√(1-x^2) from the reference triangle.
Derivatives: (1)/(√(1-x^2)) for arcsin, (1)/(1+x^2) for arctan, (1)/(|x|√(x^2-1)) for arcsec; the co-functions have the negatives.
Chain rule: put u for x and multiply by u'; (d)/(dx)\arcsin\left(x^2\right)=(2x)/(√(1-x^4)).
Identities: \arccos x=\fracπ2-\arcsin x, \arcsec x=\arccos\frac1x; the arctangent is odd with asymptotes ±\fracπ2.
Viewing angle: \arctan\frac qx-\arctan\frac px is largest at x=√(pq).
Code: math.asin, math.acos(1/x) for arcsec, sp.diff(sp.asin(x), x).

Glossary: key terms

Inverse trigonometric function: A function that gives the angle with a given sine, cosine, tangent, cotangent, secant or cosecant, chosen from one fixed interval so that it is single-valued.
Arcsine: The angle in the interval from minus pi over 2 to pi over 2 whose sine is x, defined for x from minus 1 to 1; it is written arcsin x or sin to the minus 1 of x.
Arctangent: The angle strictly between minus pi over 2 and pi over 2 whose tangent is x, defined for every real x, with horizontal asymptotes at plus and minus pi over 2.
Reference triangle: The right triangle drawn for an angle such as arcsin x, with sides chosen so that the right ratio equals x, used to read off the other trigonometric values.
Cofunction identity: The relation that an inverse cofunction is pi over 2 minus the inverse function, such as arccos x equals pi over 2 minus arcsin x, so their derivatives are negatives of each other.
Angle of elevation: The angle between the horizontal and the line of sight to a point above it; for a height h at distance x it is arctan of h over x.

Frequently asked questions

What is the derivative of arcsin x?