Inverse Trigonometry Calculator

Evaluate arcsin, arccos, and arctan with step-by-step solutions

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∑Math Input
arcsin(1/2)
arccos(-sqrt(2)/2)
arctan(sqrt(3))
sin(arccos(3/5))

What are Inverse Trigonometric Functions?

Inverse trigonometric functions reverse the standard trig functions. Given a ratio, they return the angle:

arcsin⁡(x)=θ  ⟺  sin⁡(θ)=x\arcsin(x) = \theta \iff \sin(\theta) = x
arccos⁡(x)=θ  ⟺  cos⁡(θ)=x\arccos(x) = \theta \iff \cos(\theta) = x
arctan⁡(x)=θ  ⟺  tan⁡(θ)=x\arctan(x) = \theta \iff \tan(\theta) = x

Since trig functions are not one-to-one, we restrict their domains to define proper inverses:

FunctionDomainRange (Principal Values)
arcsin⁡(x)\arcsin(x)[−1,1][-1, 1][−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]
arccos⁡(x)\arccos(x)[−1,1][-1, 1][0,π][0, \pi]
arctan⁡(x)\arctan(x)(−∞,∞)(-\infty, \infty)(−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)

Alternate notations: sin⁡−1(x)\sin^{-1}(x), cos⁡−1(x)\cos^{-1}(x), tan⁡−1(x)\tan^{-1}(x) (note: sin⁡−1(x)≠1sin⁡x\sin^{-1}(x) \neq \frac{1}{\sin x}).

Key relationships:

  • arcsin⁡(x)+arccos⁡(x)=π2\arcsin(x) + \arccos(x) = \frac{\pi}{2} for all x∈[−1,1]x \in [-1, 1]
  • arctan⁡(x)+arccot(x)=π2\arctan(x) + \text{arccot}(x) = \frac{\pi}{2} for all xx

Inverse trig functions appear in integration (∫11+x2 dx=arctan⁡x+C\int \frac{1}{1+x^2}\,dx = \arctan x + C), geometry, navigation, and physics.

How to Evaluate Inverse Trig Functions

Method 1: Using Known Values

For standard values, use the unit circle in reverse:

arcsin⁡(12)=π6because sin⁡π6=12\arcsin\left(\frac{1}{2}\right) = \frac{\pi}{6} \quad \text{because } \sin\frac{\pi}{6} = \frac{1}{2}

Common exact values:

Inputarcsin⁡\arcsinarccos⁡\arccosarctan⁡\arctan
0000π2\frac{\pi}{2}00
12\frac{1}{2}π6\frac{\pi}{6}π3\frac{\pi}{3}—
22\frac{\sqrt{2}}{2}π4\frac{\pi}{4}π4\frac{\pi}{4}—
32\frac{\sqrt{3}}{2}π3\frac{\pi}{3}π6\frac{\pi}{6}—
11π2\frac{\pi}{2}00π4\frac{\pi}{4}
3\sqrt{3}——π3\frac{\pi}{3}

Method 2: Right Triangle Method

To evaluate compositions like cos⁡(arcsin⁡(35))\cos(\arcsin(\frac{3}{5})):

  1. Let θ=arcsin⁡(35)\theta = \arcsin(\frac{3}{5}), so sin⁡θ=35\sin\theta = \frac{3}{5}
  2. Draw a right triangle: opposite =3= 3, hypotenuse =5= 5
  3. Find adjacent =25−9=4= \sqrt{25 - 9} = 4 (Pythagorean theorem)
  4. Therefore cos⁡θ=45\cos\theta = \frac{4}{5}

Method 3: Algebraic Identities

Useful identities for simplification:

sin⁡(arccos⁡x)=1−x2\sin(\arccos x) = \sqrt{1 - x^2}
cos⁡(arcsin⁡x)=1−x2\cos(\arcsin x) = \sqrt{1 - x^2}
tan⁡(arcsin⁡x)=x1−x2\tan(\arcsin x) = \frac{x}{\sqrt{1 - x^2}}
sin⁡(arctan⁡x)=x1+x2\sin(\arctan x) = \frac{x}{\sqrt{1 + x^2}}
cos⁡(arctan⁡x)=11+x2\cos(\arctan x) = \frac{1}{\sqrt{1 + x^2}}

Method 4: Derivatives of Inverse Trig Functions

These are essential for calculus:

ddxarcsin⁡x=11−x2\frac{d}{dx}\arcsin x = \frac{1}{\sqrt{1-x^2}}
ddxarccos⁡x=−11−x2\frac{d}{dx}\arccos x = -\frac{1}{\sqrt{1-x^2}}
ddxarctan⁡x=11+x2\frac{d}{dx}\arctan x = \frac{1}{1+x^2}

Comparison of Approaches

MethodBest ForKey Indicator
Known ValuesStandard ratiosInput is 0,12,22,32,10, \frac{1}{2}, \frac{\sqrt{2}}{2}, \frac{\sqrt{3}}{2}, 1
Right TriangleCompositionscos⁡(arcsin⁡(⋅))\cos(\arcsin(\cdot)) type expressions
IdentitiesAlgebraic simplificationNeed to eliminate inverse trig
CalculatorNon-standard decimalsNo exact form expected

Common Mistakes to Avoid

  • Confusing sin⁡−1(x)\sin^{-1}(x) with 1sin⁡x\frac{1}{\sin x}: The notation sin⁡−1(x)\sin^{-1}(x) means arcsin, not cosecant. Use context or prefer the "arc" notation to avoid confusion.
  • Ignoring principal value ranges: arcsin⁡(−12)=−π6\arcsin(-\frac{1}{2}) = -\frac{\pi}{6}, not 11π6\frac{11\pi}{6}. The answer must be in the defined range [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}].
  • Applying cancellation incorrectly: sin⁡(arcsin⁡x)=x\sin(\arcsin x) = x for x∈[−1,1]x \in [-1,1], but arcsin⁡(sin⁡x)=x\arcsin(\sin x) = x only when x∈[−π2,π2]x \in [-\frac{\pi}{2}, \frac{\pi}{2}]. Outside this range, you get the reference angle with appropriate sign.
  • Domain errors: arcsin⁡(2)\arcsin(2) and arccos⁡(−3)\arccos(-3) are undefined in real numbers since their domains are [−1,1][-1, 1].
  • Wrong sign in Pythagorean step: When using the right triangle method, ensure you take the correct sign based on the quadrant implied by the principal value range.

Examples

Step 1: We need θ∈[0,π]\theta \in [0, \pi] such that cos⁡θ=−32\cos\theta = -\frac{\sqrt{3}}{2}
Step 2: We know cos⁡π6=32\cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}. Since cosine is negative, θ\theta is in Quadrant II
Step 3: θ=π−π6=5π6\theta = \pi - \frac{\pi}{6} = \frac{5\pi}{6}
Answer: 5π6\frac{5\pi}{6}

Step 1: Let θ=arctan⁡43\theta = \arctan\frac{4}{3}, so tan⁡θ=43\tan\theta = \frac{4}{3} with θ∈(−π2,π2)\theta \in (-\frac{\pi}{2}, \frac{\pi}{2})
Step 2: Draw right triangle: opposite =4= 4, adjacent =3= 3, hypotenuse =16+9=5= \sqrt{16 + 9} = 5
Step 3: sin⁡θ=oppositehypotenuse=45\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{4}{5}
Answer: 45\frac{4}{5}

Step 1: First compute sin⁡5π4\sin\frac{5\pi}{4}. This angle is in Quadrant III with reference angle π4\frac{\pi}{4}: sin⁡5π4=−22\sin\frac{5\pi}{4} = -\frac{\sqrt{2}}{2}
Step 2: Now find arcsin⁡(−22)\arcsin(-\frac{\sqrt{2}}{2}): we need θ∈[−π2,π2]\theta \in [-\frac{\pi}{2}, \frac{\pi}{2}] with sin⁡θ=−22\sin\theta = -\frac{\sqrt{2}}{2}
Step 3: θ=−π4\theta = -\frac{\pi}{4} (in Quadrant IV of the restricted range)
Answer: −π4-\frac{\pi}{4}

Frequently Asked Questions

Arcsin(x) answers 'what angle has a sine of x?' Similarly for arccos and arctan. They are the inverse operations of sin, cos, and tan. For example, arcsin(1/2) = 30 degrees (or pi/6 radians) because sin(30 degrees) = 1/2.

Because sine, cosine, and tangent are periodic, each output value corresponds to infinitely many angles. To make the inverse a proper function (one output per input), we restrict to a principal value range. For arcsin this is [-pi/2, pi/2], for arccos it is [0, pi], and for arctan it is (-pi/2, pi/2).

No. sin^(-1)(x) means arcsin(x), the inverse function. The reciprocal 1/sin(x) is written as csc(x) (cosecant). This is a common source of confusion due to the ambiguous exponent notation.

Arcsin and arccos only accept inputs between -1 and 1 inclusive, since sine and cosine never exceed those bounds. Arctan accepts any real number as input since tangent can produce any real value.

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