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Inverse Trigonometric Functions

Sin, cos and tan are many-to-one, so to invert them we restrict their domains. The resulting inverse functions arcsin, arccos and arctan each have a specific domain and range you must know.

25 min Trigonometric Functions
Edexcel A level Maths: 6.5 Inverse Trigonometric FunctionsCurated by Kepler, not produced by us. Watch it before the notes below.
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What you'll be able to do

  • Understand why the domains are restricted
  • State the domain and range of each inverse
  • Sketch arcsin, arccos, arctan
  • Evaluate inverse trig values
1

Restricting the domain

A function only has an inverse if it is one-to-one. We restrict sin⁡\sin to [−π2,π2]\left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right], cos⁡\cos to [0,π][0, \pi], and tan⁡\tan to (−π2,π2)\left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right). The inverse swaps domain and range.

arcsin⁡:[−1,1]→[−π2,π2]\arcsin: [-1,1] \to [-\tfrac{\pi}{2}, \tfrac{\pi}{2}]
Domain and range of arcsin.
2

Domains and ranges

arcsin⁡x\arcsin x: domain [−1,1][-1,1], range [−π2,π2]\left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]. arccos⁡x\arccos x: domain [−1,1][-1,1], range [0,π][0,\pi]. arctan⁡x\arctan x: domain all reals, range (−π2,π2)\left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right).

Example 1Find arcsin⁡12\arcsin\tfrac12.
1We need the angle in [−π2,π2][-\tfrac{\pi}{2}, \tfrac{\pi}{2}] with sine 12\tfrac12.
2That is π6\tfrac{\pi}{6}.
Answer
π6\tfrac{\pi}{6}

Tip — arccos has range [0, π] — its answers are never negative.

Formula recap

arcsin⁡x:[−1,1]→[−π2,π2]\arcsin x: [-1,1] \to [-\tfrac{\pi}{2}, \tfrac{\pi}{2}]
Inverse sine.
arccos⁡x:[−1,1]→[0,π]\arccos x: [-1,1] \to [0, \pi]
Inverse cosine.
arctan⁡x:R→(−π2,π2)\arctan x: \mathbb{R} \to (-\tfrac{\pi}{2}, \tfrac{\pi}{2})
Inverse tangent.

Common mistakes to avoid

Giving arccos a negative answer.
arccos has range [0, π]; answers are in that interval.
Using domain [−1,1] for arctan.
arctan accepts all real numbers.

Key takeaways

  • Restrict domains to make sin, cos, tan one-to-one.
  • arcsin/arccos domain [−1,1]; arctan domain all reals.
  • Ranges: arcsin [−π/2, π/2], arccos [0, π], arctan (−π/2, π/2).

Test yourself

Answer a few questions, then review anything you missed.

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Answer each question before opening its answer. This self-check does not record a quiz score.

  1. We restrict domains of trig functions so that they are:

    • one-to-one
    • periodic
    • continuous
    • bounded
    Show answer and explanation

    Answer: one-to-one

    Needed for an inverse.

  2. The range of arccos⁡x\arccos x is:

    • [0,π][0, \pi]
    • [−π2,π2][-\tfrac{\pi}{2}, \tfrac{\pi}{2}]
    • (−π2,π2)(-\tfrac{\pi}{2}, \tfrac{\pi}{2})
    • [−1,1][-1,1]
    Show answer and explanation

    Answer: [0,π][0, \pi]

    Non-negative angles.

  3. The domain of arctan⁡x\arctan x is:

    • all reals
    • [−1,1][-1,1]
    • [0,π][0,\pi]
    • [−π2,π2][-\tfrac{\pi}{2}, \tfrac{\pi}{2}]
    Show answer and explanation

    Answer: all reals

    tan covers all values.

  4. arcsin⁡12\arcsin\tfrac12 equals:

    • π6\tfrac{\pi}{6}
    • π3\tfrac{\pi}{3}
    • π2\tfrac{\pi}{2}
    • 5π6\tfrac{5\pi}{6}
    Show answer and explanation

    Answer: π6\tfrac{\pi}{6}

    In [−π/2, π/2].

  5. The range of arcsin⁡x\arcsin x is:

    • [−π2,π2][-\tfrac{\pi}{2}, \tfrac{\pi}{2}]
    • [0,π][0, \pi]
    • all reals
    • [−1,1][-1,1]
    Show answer and explanation

    Answer: [−π2,π2][-\tfrac{\pi}{2}, \tfrac{\pi}{2}]

    Restricted sine range.

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