Inverse Trigonometric Functions Formula

Inverse trigonometric functions are functions that reverse the trigonometric functions: given a ratio, they return the corresponding angle.

The Formula

arcsin⁡(sin⁡θ)=θ\arcsin(\sin\theta) = \theta for θ∈[−π2,π2]\theta \in [-\frac{\pi}{2}, \frac{\pi}{2}]; arccos⁡(cos⁡θ)=θ\arccos(\cos\theta) = \theta for θ∈[0,π]\theta \in [0, \pi]; arctan⁡(tan⁡θ)=θ\arctan(\tan\theta) = \theta for θ∈(−π2,π2)\theta \in (-\frac{\pi}{2}, \frac{\pi}{2})

When to use: Regular trig functions answer: 'Given an angle, what's the ratio?' Inverse trig functions answer the reverse: 'Given a ratio, what's the angle?' Since sin⁡\sin and cos⁡\cos are many-to-one (many angles give the same ratio), we must restrict their domains to make the inverse a proper function. Think of it like this: if you know the slope of a ramp is 0.50.5, arcsin⁡(0.5)=30°\arcsin(0.5) = 30° tells you the angle.

Quick Example

arcsin⁡ ⁣(12)=π6becausesin⁡π6=12\arcsin\!\left(\frac{1}{2}\right) = \frac{\pi}{6} \quad \text{because} \quad \sin\frac{\pi}{6} = \frac{1}{2}
arctan⁡(1)=π4becausetan⁡π4=1\arctan(1) = \frac{\pi}{4} \quad \text{because} \quad \tan\frac{\pi}{4} = 1

Notation

arcsin⁡x=sin⁡−1x\arcsin x = \sin^{-1} x, arccos⁡x=cos⁡−1x\arccos x = \cos^{-1} x, arctan⁡x=tan⁡−1x\arctan x = \tan^{-1} x. The −1-1 superscript means inverse, NOT reciprocal.

What This Formula Means

Functions that reverse the trigonometric functions: given a ratio, they return the corresponding angle. arcsin⁡\arcsin, arccos⁡\arccos, and arctan⁡\arctan are the inverses of sin⁡\sin, cos⁡\cos, and tan⁡\tan on restricted domains.

Regular trig functions answer: 'Given an angle, what's the ratio?' Inverse trig functions answer the reverse: 'Given a ratio, what's the angle?' Since sin⁡\sin and cos⁡\cos are many-to-one (many angles give the same ratio), we must restrict their domains to make the inverse a proper function. Think of it like this: if you know the slope of a ramp is 0.50.5, arcsin⁡(0.5)=30°\arcsin(0.5) = 30° tells you the angle.

Formal View

arcsin⁡ ⁣:[−1,1]→[−π2,π2]\arcsin\colon [-1,1] \to [-\tfrac{\pi}{2}, \tfrac{\pi}{2}]; arccos⁡ ⁣:[−1,1]→[0,π]\arccos\colon [-1,1] \to [0, \pi]; arctan⁡ ⁣:R→(−π2,π2)\arctan\colon \mathbb{R} \to (-\tfrac{\pi}{2}, \tfrac{\pi}{2})

Worked Examples

Example 1

easy
Evaluate arcsin⁡ ⁣(12)\arcsin\!\left(\frac{1}{2}\right), arccos⁡ ⁣(−22)\arccos\!\left(-\frac{\sqrt{2}}{2}\right), and arctan⁡(1)\arctan(1). State the range of each inverse trig function.

Answer

arcsin⁡(12)=π6\arcsin(\frac{1}{2})=\frac{\pi}{6}; arccos⁡(−22)=3π4\arccos(-\frac{\sqrt{2}}{2})=\frac{3\pi}{4}; arctan⁡(1)=π4\arctan(1)=\frac{\pi}{4}

First step

1
arcsin⁡(12)\arcsin(\frac{1}{2}): find θ∈[−π/2,π/2]\theta\in[-\pi/2,\pi/2] with sin⁡θ=12\sin\theta=\frac{1}{2}. Answer: θ=π6\theta=\frac{\pi}{6}.

Full solution

  1. 2
    arccos⁡(−22)\arccos(-\frac{\sqrt{2}}{2}): find θ∈[0,π]\theta\in[0,\pi] with cos⁡θ=−22\cos\theta=-\frac{\sqrt{2}}{2}. Answer: θ=3π4\theta=\frac{3\pi}{4} (Q2).
  2. 3
    arctan⁡(1)\arctan(1): find θ∈(−π/2,π/2)\theta\in(-\pi/2,\pi/2) with tan⁡θ=1\tan\theta=1. Answer: θ=π4\theta=\frac{\pi}{4}.
Inverse trig functions return angles in their restricted ranges: arcsin⁡\arcsin in [−π/2,π/2][-\pi/2,\pi/2]; arccos⁡\arccos in [0,π][0,\pi]; arctan⁡\arctan in (−π/2,π/2)(-\pi/2,\pi/2). These restrictions ensure the inverse is a function (one-to-one).

Example 2

hard
Simplify sin⁡(arccos⁡(x))\sin(\arccos(x)) for x∈[−1,1]x\in[-1,1] without trigonometric functions in the final answer.

Example 3

easy
A right triangle has opposite side 11 and adjacent side 3\sqrt{3}. Find the angle opposite to the side of length 11.

Common Mistakes

  • Treating sin⁡−1\sin^{-1} as 1sin⁡\frac{1}{\sin} - the −1-1 superscript means inverse function, and reciprocal is csc⁡\csc.
  • Reporting an angle outside the principal range - arcsin⁡\arcsin returns only [−π2,π2][-\frac{\pi}{2},\frac{\pi}{2}], arccos⁡\arccos only [0,π][0,\pi].
  • Assuming arcsin⁡(sin⁡θ)=θ\arcsin(\sin\theta)=\theta for every θ\theta - it only holds when θ\theta is already in the restricted domain.

Why This Formula Matters

They turn measured ratios back into directions and angles — the angle of elevation to a plane, the launch angle for a given trajectory. Because each gives only ONE angle from a restricted range, students who forget the range report an angle the calculator never intended (e.g. expecting arcsin⁡\arcsin to return 150°150°). Recognizing it by "Am I starting from a ratio and asking for the angle, with the answer pinned to one restricted range?" — rather than by familiar numbers — is what lets a student tell it apart from reciprocal trig functions and forward trig functions and general inverse function in a mixed problem set.

Frequently Asked Questions

What is the Inverse Trigonometric Functions formula?

Functions that reverse the trigonometric functions: given a ratio, they return the corresponding angle. arcsin⁡\arcsin, arccos⁡\arccos, and arctan⁡\arctan are the inverses of sin⁡\sin, cos⁡\cos, and tan⁡\tan on restricted domains.

How do you use the Inverse Trigonometric Functions formula?

Regular trig functions answer: 'Given an angle, what's the ratio?' Inverse trig functions answer the reverse: 'Given a ratio, what's the angle?' Since sin⁡\sin and cos⁡\cos are many-to-one (many angles give the same ratio), we must restrict their domains to make the inverse a proper function. Think of it like this: if you know the slope of a ramp is 0.50.5, arcsin⁡(0.5)=30°\arcsin(0.5) = 30° tells you the angle.

What do the symbols mean in the Inverse Trigonometric Functions formula?

arcsin⁡x=sin⁡−1x\arcsin x = \sin^{-1} x, arccos⁡x=cos⁡−1x\arccos x = \cos^{-1} x, arctan⁡x=tan⁡−1x\arctan x = \tan^{-1} x. The −1-1 superscript means inverse, NOT reciprocal.

Why is the Inverse Trigonometric Functions formula important in Math?

They turn measured ratios back into directions and angles — the angle of elevation to a plane, the launch angle for a given trajectory. Because each gives only ONE angle from a restricted range, students who forget the range report an angle the calculator never intended (e.g. expecting arcsin⁡\arcsin to return 150°150°). Recognizing it by "Am I starting from a ratio and asking for the angle, with the answer pinned to one restricted range?" — rather than by familiar numbers — is what lets a student tell it apart from reciprocal trig functions and forward trig functions and general inverse function in a mixed problem set.

What do students get wrong about Inverse Trigonometric Functions?

The procedure for inverse trigonometric functions is the easy part; the trap is treating sin⁡−1\sin^{-1} as 1sin⁡\frac{1}{\sin}. Asking "Am I starting from a ratio and asking for the angle, with the answer pinned to one restricted range?" first is what keeps a correct-looking calculation from being attached to the wrong concept.

What should I learn before the Inverse Trigonometric Functions formula?

Before studying the Inverse Trigonometric Functions formula, you should understand: trigonometric functions, inverse function, domain.

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