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sin⁻¹

Chapter 4

Inverse Trigonometric Functions

Sine/cosine/tangent inverses, principal values and their properties.

1. Why do we need inverse trigonometric functions?

An inverse function undoes what the original does: if ff sends xx to yy, then f−1f^{-1} sends yy back to xx. But a function can only be inverted when it is one-to-one — each output must come from exactly one input. The trigonometric functions fail this badly: because they are periodic, the equation sin⁡θ=12\sin\theta = \tfrac{1}{2} has infinitely many solutions (π6, 5π6, 13π6, … )\left(\tfrac{\pi}{6},\ \tfrac{5\pi}{6},\ \tfrac{13\pi}{6},\ \dots\right).

The fix is to chop the domain down to a single interval on which the function is one-to-one and still hits every value in its range. On that restricted interval an inverse exists, and the output it returns is called the principal value.

DefinitionOne-to-one (injective)

A function ff is one-to-one if f(a)=f(b)f(a) = f(b) forces a=ba = b — no two different inputs share an output. Graphically, every horizontal line meets the graph at most once.

Key idea: $\sin^{-1}$, $\cos^{-1}$, $\tan^{-1}$ are NOT the reciprocals $\tfrac{1}{\sin}$ etc. The $-1$ is inverse-function notation: $\sin^{-1}x$ is the angle whose sine is $x$. The reciprocal $\tfrac{1}{\sin x}$ is written $\operatorname{cosec} x$.

2. Principal value branches and ranges of all six inverses

To build each inverse we pick a standard restricted domain — the principal branch — chosen so the function covers its whole output range exactly once and (where possible) sits symmetrically about the origin. Memorise this table; almost every mistake in the chapter is a range slip.

Standard principal branches
  • sin⁡−1\sin^{-1}: domain [−1,1][-1,1], range [−π2,π2]\left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right].
  • cos⁡−1\cos^{-1}: domain [−1,1][-1,1], range [0,π][0, \pi].
  • tan⁡−1\tan^{-1}: domain R\mathbb{R}, range (−π2,π2)\left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right) (open — never reaches ±π2\pm\tfrac{\pi}{2}).
  • cot⁡−1\cot^{-1}: domain R\mathbb{R}, range (0,π)(0, \pi).
  • sec⁡−1\sec^{-1}: domain ∣x∣≥1|x|\ge 1, range [0,π]∖{π2}[0,\pi]\setminus\{\tfrac{\pi}{2}\}.
  • cosec⁡−1\operatorname{cosec}^{-1}: domain ∣x∣≥1|x|\ge 1, range [−π2,π2]∖{0}\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]\setminus\{0\}.
Note: Watch the range of $\cos^{-1}$ and $\cot^{-1}$: they live in $[0,\pi]$, not $\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$. This asymmetry is the source of most sign mistakes. When two candidate angles are numerically equal but opposite in sign, the principal value is the one inside the range above.

3. The inverse sine function in detail

DefinitionInverse sine (arcsine)

For x∈[−1,1]x \in [-1,1],  sin⁡−1x=y\ \sin^{-1}x = y means sin⁡y=x\sin y = x and y∈[−π2,π2]y \in \left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right].

The graph of y=sin⁡−1xy=\sin^{-1}x is the mirror image of y=sin⁡xy=\sin x (restricted to [−π2,π2]\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]) across the line y=xy=x. It rises steadily from (−1,−π2)\left(-1,-\tfrac{\pi}{2}\right) through the origin to (1,π2)\left(1,\tfrac{\pi}{2}\right), is increasing, continuous, and odd.

ExampleNegative argument
Find the principal value of sin⁡−1 ⁣(−12)\sin^{-1}\!\left(-\tfrac12\right) in radians and degrees.
ExampleWhen it does NOT exist
Find sin⁡−1(2)\sin^{-1}(2), if it exists.
ExampleFinding a domain
Find the domain of sin⁡−1(2−3x2)\sin^{-1}(2-3x^2).

4. The inverse cosine function in detail

DefinitionInverse cosine (arccosine)

For x∈[−1,1]x \in [-1,1],  cos⁡−1x=y\ \cos^{-1}x = y means cos⁡y=x\cos y = x and y∈[0,π]y \in [0, \pi].

The graph of y=cos⁡−1xy=\cos^{-1}x falls from (−1,π)(-1,\pi) to (1,0)(1,0): decreasing and continuous, with xx-intercept 11 and yy-intercept π2\tfrac{\pi}{2}. It is neither even nor odd. Because the range is [0,π][0,\pi], arccosine of a negative number is obtuse.

ExamplePositive argument
Find the principal value of cos⁡−1 ⁣(32)\cos^{-1}\!\left(\tfrac{\sqrt3}{2}\right).
ExampleNegative argument
Find cos⁡−1 ⁣(−12)\cos^{-1}\!\left(-\tfrac12\right).

5. The inverse tangent function in detail

DefinitionInverse tangent (arctangent)

For x∈Rx \in \mathbb{R},  tan⁡−1x=y\ \tan^{-1}x = y means tan⁡y=x\tan y = x and y∈(−π2,π2)y \in \left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right).

Every real number has an arctangent. The graph passes through the origin, is increasing, continuous and odd, and flattens toward the horizontal asymptotes y=±π2y=\pm\tfrac{\pi}{2} without ever touching them.

ExampleStandard value
Find the principal value of tan⁡−1(3)\tan^{-1}(\sqrt3).
ExampleAngle outside the range
Evaluate tan⁡−1 ⁣(tan⁡3π5)\tan^{-1}\!\left(\tan\tfrac{3\pi}{5}\right).

6. Reciprocal inverses: cosec⁻¹, sec⁻¹ and cot⁻¹

The three reciprocal inverses are handled most easily by converting to sin⁡−1\sin^{-1}, cos⁡−1\cos^{-1} or a reference triangle.

Reciprocal identities
sin⁡−11x=cosec⁡−1x,cos⁡−11x=sec⁡−1x(∣x∣≥1)\sin^{-1}\tfrac1x = \operatorname{cosec}^{-1}x,\qquad \cos^{-1}\tfrac1x = \sec^{-1}x \qquad (|x|\ge 1)
Note: $\sec^{-1}$ has range $[0,\pi]\setminus\{\tfrac{\pi}{2}\}$ and $\operatorname{cosec}^{-1}$ has range $\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]\setminus\{0\}$. Both are defined only for $|x|\ge 1$; there is a gap on $(-1,1)$.
ExampleInverse cosecant
Find the principal value of cosec⁡−1(−1)\operatorname{cosec}^{-1}(-1).
ExampleInverse secant
Find the principal value of sec⁡−1(−2)\sec^{-1}(-2).
ExampleInverse cotangent via a triangle
If cot⁡−117=θ\cot^{-1}\tfrac17 = \theta, find cos⁡θ\cos\theta.

7. Complementary, reciprocal and negative-argument identities

These follow directly from the definitions and are used constantly.

Complementary pairs
sin⁡−1x+cos⁡−1x=π2,tan⁡−1x+cot⁡−1x=π2,sec⁡−1x+cosec⁡−1x=π2\sin^{-1}x + \cos^{-1}x = \tfrac{\pi}{2},\quad \tan^{-1}x + \cot^{-1}x = \tfrac{\pi}{2},\quad \sec^{-1}x + \operatorname{cosec}^{-1}x = \tfrac{\pi}{2}
Negative arguments
sin⁡−1(−x)=−sin⁡−1x,tan⁡−1(−x)=−tan⁡−1x,cos⁡−1(−x)=π−cos⁡−1x\sin^{-1}(-x) = -\sin^{-1}x,\quad \tan^{-1}(-x) = -\tan^{-1}x,\quad \cos^{-1}(-x) = \pi - \cos^{-1}x
ExampleComplementary identity
If sin⁡−1x=π5\sin^{-1}x = \tfrac{\pi}{5}, find cos⁡−1x\cos^{-1}x.
ExampleBounding a sum
Show that π2≤sin⁡−1x+cos⁡−1x+cos⁡−1x≤3π2\tfrac{\pi}{2} \le \sin^{-1}x + \cos^{-1}x + \cos^{-1}x \le \tfrac{3\pi}{2}... i.e. find the range of sin⁡−1x+cos⁡−1x\sin^{-1}x + \cos^{-1}x combined with an extra cos⁡−1x\cos^{-1}x.

8. Composition properties and the classic fold-back pitfall

Composing a function with its inverse should give back the input — but only where the composition is genuinely defined. Feeding the inner value first is always safe:

sin⁡(sin⁡−1x)=x  (x∈[−1,1]),tan⁡(tan⁡−1x)=x  (x∈R)\sin(\sin^{-1}x) = x \ \ (x\in[-1,1]),\qquad \tan(\tan^{-1}x) = x \ \ (x\in\mathbb{R})

The reverse composition sin⁡−1(sin⁡θ)\sin^{-1}(\sin\theta) returns θ\theta only when θ\theta already lies in the principal range. Otherwise the answer is folded back into that range.

sin⁡−1(sin⁡θ)=θ  only if θ∈[−π2,π2]\sin^{-1}(\sin\theta) = \theta \ \text{ only if } \theta \in \left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]
Watch out: $\sin^{-1}\!\left(\sin\tfrac{2\pi}{3}\right) \ne \tfrac{2\pi}{3}$, because $\tfrac{2\pi}{3}$ is outside $\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$. Instead $\sin\tfrac{2\pi}{3} = \sin\!\left(\pi-\tfrac{2\pi}{3}\right) = \sin\tfrac{\pi}{3}$, so the value is $\tfrac{\pi}{3}$.
ExampleArccosine fold-back
Evaluate cos⁡−1 ⁣(cos⁡7π6)\cos^{-1}\!\left(\cos\tfrac{7\pi}{6}\right).
ExampleComposition via a triangle
Evaluate sin⁡ ⁣(cos⁡−135)\sin\!\left(\cos^{-1}\tfrac35\right).

9. Sum and difference of arctangents

The addition formula for arctangent lets two inverse tangents combine into one — provided the result stays in the principal range:

Addition / subtraction
tan⁡−1x+tan⁡−1y=tan⁡−1 ⁣(x+y1−xy) (xy<1),tan⁡−1x−tan⁡−1y=tan⁡−1 ⁣(x−y1+xy) (xy>−1)\tan^{-1}x + \tan^{-1}y = \tan^{-1}\!\left(\tfrac{x+y}{1-xy}\right)\ (xy<1),\qquad \tan^{-1}x - \tan^{-1}y = \tan^{-1}\!\left(\tfrac{x-y}{1+xy}\right)\ (xy>-1)
Note: When $xy > 1$ the true sum leaves the principal range, so you must correct it: for $x,y>0$ with $xy>1$, the sum is $\pi + \tan^{-1}\!\left(\tfrac{x+y}{1-xy}\right)$.
ExampleA sum that lands on a nice angle
Evaluate tan⁡−112+tan⁡−113\tan^{-1}\tfrac12 + \tan^{-1}\tfrac13.
ExampleProve a fixed identity
Prove that tan⁡−112+tan⁡−113=π4\tan^{-1}\tfrac12 + \tan^{-1}\tfrac13 = \tfrac{\pi}{4}.
ExampleSolving an equation
Solve tan⁡−1x−1x−2+tan⁡−1x+1x+2=π4\tan^{-1}\tfrac{x-1}{x-2} + \tan^{-1}\tfrac{x+1}{x+2} = \tfrac{\pi}{4}.

10. Double-angle (2 tan⁻¹) formulas

A single arctangent, doubled, can be written three ways. These power many integrals and half-angle substitutions.

2tan⁡−1x=tan⁡−1 ⁣(2x1−x2) (∣x∣<1)=sin⁡−1 ⁣(2x1+x2) (∣x∣≤1)=cos⁡−1 ⁣(1−x21+x2) (x≥0)2\tan^{-1}x = \tan^{-1}\!\left(\tfrac{2x}{1-x^2}\right)\ (|x|<1) = \sin^{-1}\!\left(\tfrac{2x}{1+x^2}\right)\ (|x|\le 1) = \cos^{-1}\!\left(\tfrac{1-x^2}{1+x^2}\right)\ (x\ge 0)
ExampleReduce a doubled arctangent
Express 2tan⁡−1132\tan^{-1}\tfrac13 as a single arctangent.
ExampleCombine sec⁻¹ with sin⁻¹
Evaluate sin⁡ ⁣(sin⁡−135+sec⁡−154)\sin\!\left(\sin^{-1}\tfrac35 + \sec^{-1}\tfrac54\right).

11. Graphs and a mental summary

Each inverse graph is the mirror image of the restricted original across the line y=xy=x. Arcsine rises from (−1,−π2)\left(-1,-\tfrac{\pi}{2}\right) to (1,π2)\left(1,\tfrac{\pi}{2}\right); arccosine falls from (−1,π)(-1,\pi) to (1,0)(1,0); arctangent flattens toward y=±π2y=\pm\tfrac{\pi}{2} without touching; arccotangent falls from π\pi toward 00; and arcsec / arccosec have a gap on (−1,1)(-1,1).

Key idea: Open the Visualize tab: drag the point past $\tfrac{\pi}{2}$ and watch $\sin^{-1}(\sin\theta)$ fold back — a live picture of the pitfall in Concept 8 — then switch to the graph view to see each principal-range band shaded.