∑TN 12th MathsLearn · Visualize · Practice
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 function 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.

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: $\sin^{-1}$, $\cos^{-1}$ and $\tan^{-1}$ are not the reciprocals $\tfrac{1}{\sin}$ etc. The $-1$ is inverse-function notation: $\sin^{-1}x$ is the angle whose sine is $x$.

2. Principal value branches and ranges

To build each inverse we pick a standard restricted domain — the principal branch. It is chosen so the function covers its whole output range exactly once, and (where possible) sits symmetrically about the origin.

Standard principal branches
  • sin⁡−1\sin^{-1}: domain [−1,1][-1,1], range (principal values) [−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}\}.
  • csc⁡−1\csc^{-1} (cosec): 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 many sign mistakes.

3. Formal definitions of the three main inverses

DefinitionInverse sine (arcsine)

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

DefinitionInverse cosine (arccosine)

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

DefinitionInverse tangent (arctangent)

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

So computing an inverse trig value is always the same two-part question: find the angle with the required ratio, and make sure that angle lands inside the principal range.

4. Key identities

These complementary and negative-argument identities follow directly from the definitions and are used constantly in problems.

Complementary pairs
sin⁡−1x+cos⁡−1x=π2,tan⁡−1x+cot⁡−1x=π2\sin^{-1}x + \cos^{-1}x = \dfrac{\pi}{2},\qquad \tan^{-1}x + \cot^{-1}x = \dfrac{\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,\qquad \tan^{-1}(-x) = -\tan^{-1}x,\qquad \cos^{-1}(-x) = \pi - \cos^{-1}x

The addition formula for arctangent lets two inverse tangents combine into one:

Arctangent addition
tan⁡−1x+tan⁡−1y=tan⁡−1 ⁣(x+y1−xy),valid when xy<1\tan^{-1}x + \tan^{-1}y = \tan^{-1}\!\left(\dfrac{x+y}{1-xy}\right),\quad \text{valid when } xy < 1
Note: When $xy > 1$ the true sum leaves the principal range, so you must add or subtract $\pi$ to correct it: e.g. for $x,y>0$ with $xy>1$, the sum is $\pi + \tan^{-1}\!\left(\tfrac{x+y}{1-xy}\right)$.
More useful relations
  • cos⁡−1(−x)=π−cos⁡−1x\cos^{-1}(-x) = \pi - \cos^{-1}x and cot⁡−1(−x)=π−cot⁡−1x\cot^{-1}(-x) = \pi - \cot^{-1}x.
  • sin⁡−1x=cos⁡−11−x2\sin^{-1}x = \cos^{-1}\sqrt{1-x^2} for 0≤x≤10 \le x \le 1.
  • tan⁡−1x=sin⁡−1x1+x2=cos⁡−111+x2\tan^{-1}x = \sin^{-1}\dfrac{x}{\sqrt{1+x^2}} = \cos^{-1}\dfrac{1}{\sqrt{1+x^2}} for x≥0x \ge 0.
Example
If sin⁡−1x=π5\sin^{-1}x = \tfrac{\pi}{5}, find cos⁡−1x\cos^{-1}x.

5. Evaluating principal values

To evaluate an inverse at a standard value, recall the exact angle whose ratio you want, then slide it into the correct principal range using the sign rules above.

Standard reference values
  • sin⁡π6=12, sin⁡π4=12, sin⁡π3=32\sin\tfrac{\pi}{6} = \tfrac{1}{2},\ \sin\tfrac{\pi}{4} = \tfrac{1}{\sqrt2},\ \sin\tfrac{\pi}{3} = \tfrac{\sqrt3}{2}.
  • cos⁡π6=32, cos⁡π3=12, cos⁡π2=0\cos\tfrac{\pi}{6} = \tfrac{\sqrt3}{2},\ \cos\tfrac{\pi}{3} = \tfrac{1}{2},\ \cos\tfrac{\pi}{2} = 0.
  • tan⁡π6=13, tan⁡π4=1, tan⁡π3=3\tan\tfrac{\pi}{6} = \tfrac{1}{\sqrt3},\ \tan\tfrac{\pi}{4} = 1,\ \tan\tfrac{\pi}{3} = \sqrt3.
ExamplePositive argument
Evaluate sin⁡−1 ⁣(12)\sin^{-1}\!\left(\tfrac{1}{2}\right).
ExampleNegative argument for arccos
Evaluate cos⁡−1 ⁣(−12)\cos^{-1}\!\left(-\tfrac{1}{2}\right).
ExampleArctangent
Evaluate tan⁡−1 ⁣(3)\tan^{-1}\!\left(\sqrt3\right).

6. Composition properties and the classic 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 $\sin\tfrac{2\pi}{3} = \sin\tfrac{\pi}{3}$, so the value is $\tfrac{\pi}{3}$.
Example
Evaluate cos⁡−1 ⁣(cos⁡7π6)\cos^{-1}\!\left(\cos\tfrac{7\pi}{6}\right).

7. Graphs and a mental summary

Each inverse graph is the mirror image of the restricted original across the line y=xy = x. The arcsine graph 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); and arctangent flattens toward the horizontal asymptotes y=±π2y = \pm\tfrac{\pi}{2} without ever touching them.

Key idea: Open the Visualize tab: drag the point past $\tfrac{\pi}{2}$ and watch $\arcsin(\sin\theta)$ fold back — a live picture of the pitfall in Concept 6.