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	$sin^{- 1} x \in [- \frac{π}{2}, \frac{π}{2}], - 1 \leq x \leq 1$
Inverse cosine	$cos^{- 1} x \in [0, π], - 1 \leq x \leq 1$
Inverse tangent	$tan^{- 1} x \in (- \frac{π}{2}, \frac{π}{2}), x \in R$
Undoing sine	$sin (sin^{- 1} x) = x sin^{- 1} (sin θ) = θ ⟺ - \frac{π}{2} \leq θ \leq \frac{π}{2}$

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.
$sin \frac{π}{6} = sin \frac{5 π}{6} = sin (- \frac{7 π}{6}) = \frac{1}{2}$
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.
$y = sin^{- 1} x ⟺ sin y = x, - \frac{π}{2} \leq y \leq \frac{π}{2}$
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.
$sin (sin^{- 1} x) = x (- 1 \leq x \leq 1), sin^{- 1} (sin \frac{5 π}{6}) = \frac{π}{6}$
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.
$cos (sin^{- 1} x) = 1 - x^{2}$

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:
  
  $- \frac{π}{2} \leq sin^{- 1} x \leq \frac{π}{2}$
2. π/6 has sine 1/2, and sine is odd, so the angle in the range is −π/6:
  
  $sin \frac{π}{6} = \frac{1}{2} \Rightarrow sin (- \frac{π}{6}) = - \frac{1}{2}$
Answer

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

$sin^{- 1} (- \frac{1}{2}) = - \frac{π}{6} \approx - 0.5236$
A negative cosine 2 steps
Find cos⁻¹(−√3/2) exactly.

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Show solution
1. The range runs from 0 to π, and the reference angle is π/6:
  
  $0 \leq cos^{- 1} x \leq π, cos \frac{π}{6} = \frac{3}{2}$
2. Inside the range, cosine is negative only past π/2, so the angle is π − π/6:
  
  $cos (π - \frac{π}{6}) = - cos \frac{π}{6} = - \frac{3}{2}$
Answer

cos⁻¹(−√3/2) = 5π/6.

$cos^{- 1} (- \frac{3}{2}) = \frac{5 π}{6} \approx 2.618$
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:
  
  $- \frac{π}{2} < tan^{- 1} x < \frac{π}{2}, x \in R$
2. tan(π/3) = √3, and tangent is odd:
  
  $tan \frac{π}{3} = 3 \Rightarrow tan (- \frac{π}{3}) = - 3$
Answer

tan⁻¹(−√3) = −π/3.

$tan^{- 1} (- 3) = - \frac{π}{3} \approx - 1.0472$
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:
  
  $sin^{- 1} x, cos^{- 1} x : - 1 \leq x \leq 1; 1.5 \in / [- 1, 1], - 1.2 \in / [- 1, 1]$
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:
  
  $tan^{- 1} 1.5 \approx 0.9828 \approx 56.309 9^{\circ}$
Answer

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

$tan^{- 1} 1.5 \approx 0.9828$
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:
  
  $\frac{5 π}{6} > \frac{π}{2}$
2. π − θ has the same sine as θ, and π − 5π/6 lies in the range:
  
  $sin (π - \frac{5 π}{6}) = sin \frac{5 π}{6}, π - \frac{5 π}{6} = \frac{π}{6} \in [- \frac{π}{2}, \frac{π}{2}]$
Answer

sin⁻¹(sin 5π/6) = π/6.

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

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Show solution
1. 4π/3 is past π, outside the range:
  
  $\frac{4 π}{3} > π$
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:
  
  $cos (2 π - \frac{4 π}{3}) = cos \frac{4 π}{3}, 2 π - \frac{4 π}{3} = \frac{2 π}{3} \in [0, π]$
Answer

cos⁻¹(cos 4π/3) = 2π/3.

$cos^{- 1} (cos \frac{4 π}{3}) = \frac{2 π}{3}$
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:
  
  $y = cos^{- 1} (- \frac{3}{5}) : cos y = - \frac{3}{5}, \frac{π}{2} < y \leq π$
2. Sine is positive there, so from sin²y + cos²y = 1
  
  $sin y = + 1 - \frac{9}{25} = \frac{4}{5}$
3. and tan y = sin y / cos y:
  
  $tan y = \frac{4/5}{- 3/5} = - \frac{4}{3}$
Answer

tan(cos⁻¹(−3/5)) = −4/3.

$tan (cos^{- 1} (- \frac{3}{5})) = - \frac{4}{3} \approx - 1.3333$
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:
  
  $y = sin^{- 1} x : sin y = x, - \frac{π}{2} \leq y \leq \frac{π}{2} \Rightarrow cos y \geq 0$
2. so the root takes its positive sign:
  
  $cos y = + 1 - sin^{2} y = 1 - x^{2}$
3. At x = −3/5:
  
  $cos (sin^{- 1} (- \frac{3}{5})) = 1 - \frac{9}{25} = \frac{4}{5}$
Answer

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

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

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

$x = - \frac{1}{2}; - \frac{π}{3} \in / [0, π] \Rightarrow \emptyset$

Supplementary problems, answers only

Find tan⁻¹(tan 2π/3).

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−π/3, which is 2π/3 − π.

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

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2π/5, since cosine is even.

$cos^{- 1} (cos (- \frac{2 π}{5})) = \frac{2 π}{5}$
Find sin⁻¹(sin 4), with 4 in radians.

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

$sin^{- 1} (sin 4) = π - 4 \approx - 0.8584$
Find sin(cos⁻¹(−12/13)) exactly.

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

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

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x/√(1 − x²), for −1 < x < 1.

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

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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.

$sin^{- 1} (- \frac{1}{2}) + cos^{- 1} (- \frac{1}{2}) = - \frac{π}{6} + \frac{2 π}{3} = \frac{π}{2}$

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 42Math team · Reviewed against published sources. Corrections welcome.