Inverse Trigonometric Functions

Type x as a decimal, a fraction or with a root, and the page gives the one angle in the principal range whose sine, cosine or tangent is x, or says why there is none. Or drag θ and watch f⁻¹(f(θ)) fold back into the range.

Inverse sine	$\sin^{-1}x\in\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right],\quad -1\le x\le 1$
Inverse cosine	$\cos^{-1}x\in[0,\pi],\quad -1\le x\le 1$
Inverse tangent	$\tan^{-1}x\in\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right),\quad x\in\mathbb{R}$
Undoing sine	$\begin{gathered}\sin\left(\sin^{-1}x\right)=x\\ \sin^{-1}(\sin\theta)=\theta\iff-\tfrac{\pi}{2}\le\theta\le\tfrac{\pi}{2}\end{gathered}$

Worked example — Inverse Trigonometric Functions

Given value	−0.5
Principal value, in radians	−0.5236
In degrees	−30°
cos(sin⁻¹ x), by the triangle	0.866

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Inverse Trigonometric Functions

The inverse trigonometric functions sin⁻¹ x, cos⁻¹ x and tan⁻¹ x (also written arcsin, arccos and arctan) give the one angle in a fixed range, the principal value, whose sine, cosine or tangent is x. The −1 marks an inverse function, not a power: sin⁻¹ x is not 1/sin x.

Where the formula comes from

Sine is not one-to-one. Infinitely many angles share each sine, so “the angle whose sine is ½” names no angle until something chooses one.
$\sin\tfrac{\pi}{6}=\sin\tfrac{5\pi}{6}=\sin\left(-\tfrac{7\pi}{6}\right)=\tfrac{1}{2}$
Restrict, then invert. From −π/2 to π/2 sine rises from −1 to 1 and takes each value once, so there it has an inverse. Cosine is restricted to 0 ≤ y ≤ π and tangent to −π/2 < y < π/2, for the same reason.
$y=\sin^{-1}x\iff\sin y=x,\quad-\tfrac{\pi}{2}\le y\le\tfrac{\pi}{2}$
It undoes sine in one direction only. sin(sin⁻¹ x) = x for every x it accepts, but sin⁻¹(sin θ) returns θ only when θ is already in the range. Otherwise it returns the angle in the range with the same sine.
$\sin\left(\sin^{-1}x\right)=x\quad(-1\le x\le1),\qquad\sin^{-1}\left(\sin\tfrac{5\pi}{6}\right)=\tfrac{\pi}{6}$
Compositions come off a right triangle. Put y = sin⁻¹ x, so sin y = x. Pythagoras gives the other leg, and the range decides its sign.
$\cos\left(\sin^{-1}x\right)=\sqrt{1-x^{2}}$

Assumes: sin⁻¹ and cos⁻¹ accept only −1 ≤ x ≤ 1; tan⁻¹ accepts every real number.

Problems solved in full

A negative sine 2 steps
Find sin⁻¹(−1/2) exactly.

Open this problem in the tool
1. The answer has to lie in the range of sin⁻¹:
  
  $-\tfrac{\pi}{2}\le\sin^{-1}x\le\tfrac{\pi}{2}$
2. π/6 has sine 1/2, and sine is odd, so the angle in the range is −π/6:
  
  $\sin\tfrac{\pi}{6}=\tfrac12\ \Rightarrow\ \sin\left(-\tfrac{\pi}{6}\right)=-\tfrac12$
Answer

sin⁻¹(−1/2) = −π/6. 7π/6 and 11π/6 have the same sine but lie outside the range.

$\sin^{-1}\left(-\tfrac12\right)=-\tfrac{\pi}{6}\approx-0{.}5236$
A negative cosine 2 steps
Find cos⁻¹(−√3/2) exactly.

Open this problem in the tool
Show solution
1. The range runs from 0 to π, and the reference angle is π/6:
  
  $0\le\cos^{-1}x\le\pi,\qquad\cos\tfrac{\pi}{6}=\tfrac{\sqrt3}{2}$
2. Inside the range, cosine is negative only past π/2, so the angle is π − π/6:
  
  $\cos\left(\pi-\tfrac{\pi}{6}\right)=-\cos\tfrac{\pi}{6}=-\tfrac{\sqrt3}{2}$
Answer

cos⁻¹(−√3/2) = 5π/6.

$\cos^{-1}\left(-\tfrac{\sqrt3}{2}\right)=\tfrac{5\pi}{6}\approx2{.}618$
An inverse tangent 2 steps
Find tan⁻¹(−√3) exactly.

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Show solution
1. tan⁻¹ accepts every real number and answers strictly between −π/2 and π/2:
  
  $-\tfrac{\pi}{2}<\tan^{-1}x<\tfrac{\pi}{2},\qquad x\in\mathbb{R}$
2. tan(π/3) = √3, and tangent is odd:
  
  $\tan\tfrac{\pi}{3}=\sqrt3\ \Rightarrow\ \tan\left(-\tfrac{\pi}{3}\right)=-\sqrt3$
Answer

tan⁻¹(−√3) = −π/3.

$\tan^{-1}\left(-\sqrt3\right)=-\tfrac{\pi}{3}\approx-1{.}0472$
Which inverses exist 2 steps
Which of sin⁻¹ 1.5, cos⁻¹(−1.2) and tan⁻¹ 1.5 exist? Evaluate those that do.

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Show solution
1. Sine and cosine are coordinates of a point on the unit circle, so sin⁻¹ and cos⁻¹ accept only −1 to 1:
  
  $\sin^{-1}x,\ \cos^{-1}x:\ -1\le x\le1;\qquad1{.}5\notin[-1,1],\quad-1{.}2\notin[-1,1]$
2. Tangent takes every real value, so tan⁻¹ 1.5 exists. It is not one of the standard angles, so take it from a calculator in radians:
  
  $\tan^{-1}1{.}5\approx0{.}9828\approx56{.}3099^{\circ}$
Answer

Only tan⁻¹ 1.5 ≈ 0.9828, about 56.3099°. sin⁻¹ 1.5 and cos⁻¹(−1.2) are undefined.

$\tan^{-1}1{.}5\approx0{.}9828$
Undoing sine outside the range 2 steps
Find sin⁻¹(sin 5π/6).

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Show solution
1. 5π/6 lies outside the range, so the answer is not 5π/6:
  
  $\tfrac{5\pi}{6}>\tfrac{\pi}{2}$
2. π − θ has the same sine as θ, and π − 5π/6 lies in the range:
  
  $\sin\left(\pi-\tfrac{5\pi}{6}\right)=\sin\tfrac{5\pi}{6},\qquad\pi-\tfrac{5\pi}{6}=\tfrac{\pi}{6}\in\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$
Answer

sin⁻¹(sin 5π/6) = π/6.

$\sin^{-1}\left(\sin\tfrac{5\pi}{6}\right)=\tfrac{\pi}{6}$
Undoing cosine outside the range 2 steps
Find cos⁻¹(cos 4π/3).

Open this problem in the tool
Show solution
1. 4π/3 is past π, outside the range:
  
  $\tfrac{4\pi}{3}>\pi$
2. 2π − θ has the same cosine as θ, because cosine is even and repeats every 2π, and 2π − 4π/3 lies in the range:
  
  $\cos\left(2\pi-\tfrac{4\pi}{3}\right)=\cos\tfrac{4\pi}{3},\qquad2\pi-\tfrac{4\pi}{3}=\tfrac{2\pi}{3}\in[0,\pi]$
Answer

cos⁻¹(cos 4π/3) = 2π/3.

$\cos^{-1}\left(\cos\tfrac{4\pi}{3}\right)=\tfrac{2\pi}{3}$
A composition with a negative input 3 steps
Find tan(cos⁻¹(−3/5)) exactly.

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Show solution
1. Call the angle y. Its cosine is −3/5, so the range puts it past π/2:
  
  $y=\cos^{-1}\left(-\tfrac35\right):\quad\cos y=-\tfrac35,\quad\tfrac{\pi}{2}<y\le\pi$
2. Sine is positive there, so from sin²y + cos²y = 1
  
  $\sin y=+\sqrt{1-\tfrac{9}{25}}=\tfrac45$
3. and tan y = sin y / cos y:
  
  $\tan y=\frac{4/5}{-3/5}=-\tfrac43$
Answer

tan(cos⁻¹(−3/5)) = −4/3.

$\tan\left(\cos^{-1}\left(-\tfrac35\right)\right)=-\tfrac43\approx-1{.}3333$
A composition in x 3 steps
Express cos(sin⁻¹ x) in terms of x, and check it at x = −3/5.

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Show solution
1. Call the angle y. The range of sin⁻¹ keeps y between −π/2 and π/2, where cosine is never negative:
  
  $y=\sin^{-1}x:\quad\sin y=x,\quad-\tfrac{\pi}{2}\le y\le\tfrac{\pi}{2}\ \Rightarrow\ \cos y\ge0$
2. so the root takes its positive sign:
  
  $\cos y=+\sqrt{1-\sin^{2}y}=\sqrt{1-x^{2}}$
3. At x = −3/5:
  
  $\cos\left(\sin^{-1}\left(-\tfrac35\right)\right)=\sqrt{1-\tfrac{9}{25}}=\tfrac45$
Answer

cos(sin⁻¹ x) = √(1 − x²), with no ±, for every x from −1 to 1.

$\cos\left(\sin^{-1}x\right)=\sqrt{1-x^{2}},\qquad-1\le x\le1$
An equation in an inverse function 2 steps
Solve cos⁻¹ x = 2π/3, and then cos⁻¹ x = −π/3.

Open this problem in the tool
Show solution
1. Take the cosine of both sides. 2π/3 lies in the range, so this is the one solution:
  
  $\cos^{-1}x=\tfrac{2\pi}{3}\ \Rightarrow\ x=\cos\tfrac{2\pi}{3}=-\tfrac12$
2. cos⁻¹ never returns −π/3, which lies outside its range:
  
  $-\tfrac{\pi}{3}\notin[0,\pi]$
Answer

x = −1/2 for the first; the second has no solution. cos(−π/3) = 1/2, but cos⁻¹(1/2) is π/3.

$x=-\tfrac12;\qquad-\tfrac{\pi}{3}\notin[0,\pi]\Rightarrow\varnothing$

Supplementary problems, answers only

Find tan⁻¹(tan 2π/3).

Show answer

−π/3, which is 2π/3 − π.

$\tan^{-1}\left(\tan\tfrac{2\pi}{3}\right)=-\tfrac{\pi}{3}$
Find cos⁻¹(cos(−2π/5)).

Show answer

2π/5, since cosine is even.

$\cos^{-1}\left(\cos\left(-\tfrac{2\pi}{5}\right)\right)=\tfrac{2\pi}{5}$
Find sin⁻¹(sin 4), with 4 in radians.

Show answer

π − 4 ≈ −0.8584, because 4 lies between π/2 and 3π/2.

$\sin^{-1}(\sin4)=\pi-4\approx-0{.}8584$
Find sin(cos⁻¹(−12/13)) exactly.

Show answer

5/13: the angle lies past π/2, where sine is positive.

$\sin\left(\cos^{-1}\left(-\tfrac{12}{13}\right)\right)=\tfrac{5}{13}$
Express tan(sin⁻¹ x) in terms of x.

Show answer

x/√(1 − x²), for −1 < x < 1.

$\tan\left(\sin^{-1}x\right)=\frac{x}{\sqrt{1-x^{2}}},\qquad-1<x<1$
Show that sin⁻¹ x + cos⁻¹ x = π/2 for −1 ≤ x ≤ 1, and check it at x = −1/2.

Show answer

If y = sin⁻¹ x, then cos(π/2 − y) = sin y = x, and π/2 − y lies between 0 and π, so π/2 − y = cos⁻¹ x.

$\sin^{-1}\left(-\tfrac12\right)+\cos^{-1}\left(-\tfrac12\right)=-\tfrac{\pi}{6}+\tfrac{2\pi}{3}=\tfrac{\pi}{2}$

sin⁻¹(sin θ) runs backwards on half the circle

Press sin⁻¹(sin 150°): the answer is 30°. Now drag θ from 90° to 270°. While θ rises, sin⁻¹(sin θ) falls from 90° to −90° at a slope of exactly −1, because on that stretch it equals 180° − θ. Sine climbs to 1 at 90° and turns back, so whatever undoes it has to turn back with it. Cosine turns at 0° and at 180°, so between −180° and 180° cos⁻¹(cos θ) is |θ|, a V. Tangent never turns. It rises on every stretch between its asymptotes, so tan⁻¹(tan θ) never runs backwards; it drops by 180° each time θ passes an odd multiple of 90°, and that is all. tan⁻¹(tan 135°) is 135° − 180° = −45°.

cos⁻¹ cannot borrow the range of sin⁻¹

Between −90° and 90° cosine is never negative, and it takes every value strictly between 0 and 1 twice, at θ and at −θ. An inverse needs the opposite on both counts: every value from −1 to 1, each exactly once. So cos⁻¹ answers from 0° to 180°, the top half of the circle, where cosine falls steadily from 1 to −1. You pay for it at negative inputs. sin⁻¹(−1/2) is −30°, below the x-axis, but cos⁻¹(−√3/2) is 150°: a negative cosine never gives a negative angle, only one past 90°.

cos(sin⁻¹ x) never needs a ±

With the Pythagorean identities, √(1 − sin²θ) gave only the size of cos θ and the quadrant had to supply the sign. Here the range supplies it before anyone asks. sin⁻¹ x lies between −90° and 90°, where cosine is never negative, so cos(sin⁻¹ x) = √(1 − x²) for every x from −1 to 1. At x = −3/5 it is +4/5, although the angle itself is negative. The same goes for sin(cos⁻¹ x), since sine is never negative from 0° to 180°. Tangent can still come out negative, because the range of cos⁻¹ includes quadrant II, but tan(cos⁻¹ x) = √(1 − x²)/x carries that sign in x and needs no ± either. Press tan(cos⁻¹(−3/5)) and the angle is 126.8699°, so the tangent is −4/3.

Learning path

Trigonometric identities and equations

References (1)

The domains and principal ranges of sin⁻¹, cos⁻¹ and tan⁻¹, sin⁻¹(sin x) = x only for −π/2 ≤ x ≤ π/2, the warning that sin⁻¹ x is not 1/sin x, and compositions such as cos(sin⁻¹ x) by a right triangle: J. Abramson, Precalculus 2e, OpenStax (2021), section 6.3, “Inverse Trigonometric Functions”.

By the 7 Raven Room team · Reviewed against published sources. Corrections welcome.

Inverse Trigonometric Functions

Where the formula comes from

Supplementary problems, answers only

sin⁻¹(sin θ) runs backwards on half the circle 🖖

cos⁻¹ cannot borrow the range of sin⁻¹ 🖖

cos(sin⁻¹ x) never needs a ± 🖖

sin⁻¹(sin θ) runs backwards on half the circle

cos⁻¹ cannot borrow the range of sin⁻¹

cos(sin⁻¹ x) never needs a ±