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Learn9MA0Paper 2 · Pure Mathematics 2TrigonometryInverse trigonometric functions
Trigonometry · Subtopic 9 of 13

Inverse Trigonometric Functions

Why sin⁡x=12\sin x=\tfrac12 has infinitely many solutions but sin⁡−112\sin^{-1}\tfrac12 has only one, the restricted domains that make sine, cosine and tangent one-to-one, and the principal values your calculator gives.

Cambridge 9709 Paper 1Edexcel YMA01 WMA13Edexcel 9MA0 Paper 1Edexcel 9MA0 Paper 2AQA 7357 Paper 1AQA 7357 Paper 2≈ 35 minRevision 6 · updated 2026-10-27
Section 1 of 14

Learning objectives

  • Explain why sine, cosine and tangent need a restricted domain to have an inverse.
  • State the domain and range (principal values) of sin⁡−1x\sin^{-1}x, cos⁡−1x\cos^{-1}x and tan⁡−1x\tan^{-1}x.
  • Find exact principal values in degrees and radians.
  • Evaluate combinations such as sin⁡(cos⁡−135)\sin\left(\cos^{-1}\tfrac35\right) and cos⁡−1(cos⁡θ)\cos^{-1}(\cos\theta).
Pause and think

sin⁡30∘=12\sin30^\circ=\tfrac12 and sin⁡150∘=12\sin150^\circ=\tfrac12. If you ask "which angle has sine 12\tfrac12?", which answer should a calculator give?

Section 2 of 14

Core idea

Start hereWhich angle?

A ramp rises 11 m over a 22 m slope: sin⁡θ=0.5\sin\theta=0.5. What's θ\theta? 30∘30^\circ, obviously. But sin⁡150∘=0.5\sin150^\circ=0.5 too, and sin⁡390∘\sin390^\circ, and infinitely many more. A function must give one answer, so mathematicians agreed on a rule for which angle sin⁡−1\sin^{-1} returns.

The big idea

To invert sin, cos and tan we restrict their domains so they're one-one. The principal values: sin⁡−1x∈[−π2,π2]\sin^{-1}x\in[-\frac\pi2,\frac\pi2], cos⁡−1x∈[0,π]\cos^{-1}x\in[0,\pi], tan⁡−1x∈(−π2,π2)\tan^{-1}x\in(-\frac\pi2,\frac\pi2). Graphs are reflections in y=xy=x.

1. Restricting the domain

Restricting sine so it can be undone
−100100200300−1−0.50.510x (degrees)ysin x everywhereprincipal party = 0.5 meets sin x many times30°150°
Between −90∘-90^\circ and 90∘90^\circ, sine takes each value from −1-1 to 11 exactly once. That part is reversed to make sin⁡−1\sin^{-1}, which is why the calculator always gives 30∘30^\circ and never 150∘150^\circ.
functiondomainrange (principal values)sin⁡−1x−1≤x≤1−π2≤y≤π2cos⁡−1x−1≤x≤10≤y≤πtan⁡−1xx∈R−π2<y<π2\begin{array}{lll}\text{function}&\text{domain}&\text{range (principal values)}\\\hline\sin^{-1}x&-1\le x\le1&-\tfrac\pi2\le y\le\tfrac\pi2\\\cos^{-1}x&-1\le x\le1&0\le y\le\pi\\\tan^{-1}x&x\in\mathbb R&-\tfrac\pi2<y<\tfrac\pi2\end{array}

Each restricted piece is chosen to include all outputs from −1-1 to 11 (or all reals for tan) exactly once, and to include the acute angles.

Common misconceptionsin⁡−1x≠1sin⁡x\sin^{-1}x\neq\frac1{\sin x} The −1-1 means inverse function, written arcsin⁡x\arcsin x too. 1sin⁡x=cosec⁡x\frac1{\sin x}=\operatorname{cosec}x.

2. Evaluating

Worked exampleExact principal values

Find sin⁡−1(−12)\sin^{-1}(-\frac12), cos⁡−1(−12)\cos^{-1}(-\frac12), tan⁡−13\tan^{-1}\sqrt3.

Goal
Exact radians in the correct range.
Strategy
Use the exact-value table, then pick the angle inside the principal range.
  1. sin⁡−1(−12)=−π6\sin^{-1}(-\tfrac12)=-\tfrac\pi6
    Must be in [−π2,π2][-\frac\pi2,\frac\pi2]; negative input → negative angle.
  2. cos⁡−1(−12)=π−π3=2π3\cos^{-1}(-\tfrac12)=\pi-\tfrac\pi3=\tfrac{2\pi}3
    Must be in [0,π][0,\pi]: second quadrant.
  3. tan⁡−13=π3\tan^{-1}\sqrt3=\tfrac\pi3
    From the table.
Answer−π6-\frac\pi6, 2π3\frac{2\pi}3, π3\frac\pi3.
✓ Checkcos⁡2π3=cos⁡120∘=−12\cos\frac{2\pi}3=\cos120^\circ=-\frac12 ✓.
Worked examplesin⁡(cos⁡−135)\sin\left(\cos^{-1}\tfrac35\right)

Evaluate exactly.

Goal
A single number.
Strategy
Name the angle, draw its triangle, mind the sign.
  1. α=cos⁡−135⇒cos⁡α=35, 0≤α≤π\alpha=\cos^{-1}\tfrac35\Rightarrow\cos\alpha=\tfrac35,\ 0\le\alpha\le\pi
    Definition.
  2. opp=25−9=4\text{opp}=\sqrt{25-9}=4
    Triangle with adj 33, hyp 55.
  3. sin⁡α=+45\sin\alpha=+\tfrac45
    sin⁡\sin is non-negative on [0,π][0,\pi].
Answer45\frac45.
✓ Checksin⁡2α+cos⁡2α=16+925=1\sin^2\alpha+\cos^2\alpha=\frac{16+9}{25}=1 ✓.
Edge casesin⁡−1(sin⁡x)\sin^{-1}(\sin x) isn't always xx sin⁡−1(sin⁡5π6)=sin⁡−112=π6\sin^{-1}(\sin\frac{5\pi}6)=\sin^{-1}\frac12=\frac\pi6, because 5π6\frac{5\pi}6 is outside the principal range.
Pause and think

Domain of cos⁡−1(2x−1)\cos^{-1}(2x-1)?

Summary

  • Restrict to one-one, then invert.
  • Know the three principal ranges.
  • Graphs: reflect the restricted piece in y=xy=x.
Key words
Principal value
The single value an inverse trig function returns.
Historyarcsin and sin⁻¹. The notation sin⁡−1x\sin^{-1}x was introduced by John Herschel in 1813, and it has confused students ever since, because sin⁡−1x\sin^{-1}x is not 1sin⁡x\dfrac{1}{\sin x}. The other name, arcsin⁡x\arcsin x, means "the arc whose sine is xx": on the unit circle, the angle in radians is an arc length.
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