Inverse trigonometric functions

sin⁡x\sin x takes the same value at infinitely many angles — sin⁡π6\sin\frac{\pi}{6}, sin⁡5π6\sin\frac{5\pi}{6}, sin⁡(π6+2π)\sin\left(\frac{\pi}{6}+2\pi\right) all equal 12\frac12 — so it is many-one and has no ordinary inverse on all of R\mathbb{R}. Inverse trigonometric functions repair this by restricting each trig function to one carefully chosen interval, its principal value branch, on which it becomes one-one and onto; the inverse of that restricted function is what ISC Class 12 studies as sin⁡−1x\sin^{-1}x, cos⁡−1x\cos^{-1}x, and so on for all six ratios. This is the third strand of Unit 1, Relations and Functions (Section A). You are expected to know each function's domain, principal value range, and elementary cancellation/complementary identities — this article's scope. A separate topic covers sum-and-difference identities between inverse trig functions; differentiating them belongs to Calculus and is excluded here.

The core insight

Each inverse trigonometric function is defined by restricting its trig function to one principal value branch on which it is one-one and onto, so y=sin⁡−1xy=\sin^{-1}x means both sin⁡y=x\sin y = x and yy lies in that fixed range — never just the first condition alone.

The core idea

A many-one function cannot be inverted directly: reversing the arrows from output back to input would send one value of xx to several different angles, which is not a function at all. The fix used throughout this topic is to shrink the domain of sin⁡x\sin x, cos⁡x\cos x, etc. down to one interval on which the function happens to be one-one (no repeats) and onto its full range — and only then is the inverse well defined, single-valued, on that shrunken piece.

sin⁡−1:[−1,1]→[−π2,π2],cos⁡−1:[−1,1]→[0,π],tan⁡−1:R→(−π2,π2)\sin^{-1}:[-1,1]\to\left[-\frac{\pi}{2},\frac{\pi}{2}\right], \qquad \cos^{-1}:[-1,1]\to[0,\pi], \qquad \tan^{-1}:\mathbb{R}\to\left(-\frac{\pi}{2},\frac{\pi}{2}\right)

where: the set on the left of each arrow is the domain (inputs the inverse function accepts) and the set on the right is the principal value range (the one interval of angles the output is forced to land in). The remaining three follow the same pattern: cot⁡−1:R→(0,π)\cot^{-1}:\mathbb{R}\to(0,\pi), sec⁡−1:R∖(−1,1)→[0,π]∖{π2}\sec^{-1}:\mathbb{R}\setminus(-1,1)\to[0,\pi]\setminus\left\{\frac{\pi}{2}\right\}, and csc⁡−1:R∖(−1,1)→[−π2,π2]∖{0}\csc^{-1}:\mathbb{R}\setminus(-1,1)\to\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\setminus\{0\}. By definition, y=sin⁡−1xy=\sin^{-1}x means both sin⁡y=x\sin y = x and yy lies in [−π2,π2]\left[-\frac{\pi}{2},\frac{\pi}{2}\right]; dropping the second condition is the single most common source of error in this topic, since infinitely many angles satisfy sin⁡y=x\sin y = x and only the range restriction picks out one.

Figure
Graph of y equals sine inverse x showing the restricted domain from minus one to one on the horizontal axis mapping to the principal range from minus pi over two to pi over two on the vertical axis as a single increasing curve
y = sin⁻¹x — one branch, one-one and onto x y -1 1 π/2 -π/2 (-1, -π/2) (1, π/2) Domain [-1,1] maps onto range [-π/2, π/2] — every x gives exactly one y

Two families of algebraic facts follow directly from this picture. First, cancellation properties, valid only on the stated domains:

sin⁡−1(sin⁡θ)=θ  for θ∈[−π2,π2],sin⁡(sin⁡−1x)=x  for x∈[−1,1]\sin^{-1}(\sin\theta) = \theta \ \text{ for } \theta \in \left[-\frac{\pi}{2},\frac{\pi}{2}\right], \qquad \sin(\sin^{-1}x) = x \ \text{ for } x \in [-1,1]

and analogously for the other five pairs, each restricted to its own principal branch — outside the stated domain, θ\theta must first be reduced to an equivalent angle inside the branch before the identity can be applied. Second, complementary and reciprocal relations, which hold across the full domain of each side:

sin⁡−1x+cos⁡−1x=π2,tan⁡−1x+cot⁡−1x=π2,sec⁡−1x+csc⁡−1x=π2,sin⁡−1x=csc⁡−11x\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}, \qquad \tan^{-1}x+\cot^{-1}x=\frac{\pi}{2}, \qquad \sec^{-1}x+\csc^{-1}x=\frac{\pi}{2}, \qquad \sin^{-1}x = \csc^{-1}\frac{1}{x}

where: each pair on the left is built from complementary trig ratios (sin⁡\sin and cos⁡\cos, and so on), and xx ranges over whichever domain makes both sides defined. These four relations are the main labour-savers of the topic: once one member of a complementary or reciprocal pair is known at some value, the other follows by a single subtraction or a reciprocal, with no need to re-identify a triangle or a standard angle from scratch.

Derivation AWhy sine inverse x plus cosine inverse x equals pi over two

This proof shows, using only the definition of sin⁡−1\sin^{-1} and one complementary-angle identity from earlier trigonometry, that sin⁡−1x+cos⁡−1x\sin^{-1}x+\cos^{-1}x is the constant π2\frac{\pi}{2} for every x∈[−1,1]x\in[-1,1] — not merely true at a few special values, but an identity that holds across the entire shared domain, because it follows from how the two principal branches are built rather than from any particular number.

Step 1. Fix any x∈[−1,1]x \in [-1,1] and let y=sin⁡−1xy = \sin^{-1}x. By definition this means sin⁡y=x\sin y = x and y∈[−π2,π2]y \in \left[-\frac{\pi}{2},\frac{\pi}{2}\right].

Step 2. Use the complementary-angle identity sin⁡y=cos⁡(π2−y)\sin y = \cos\left(\frac{\pi}{2}-y\right), which holds for every real yy. Combined with Step 1, this gives cos⁡(π2−y)=x\cos\left(\frac{\pi}{2}-y\right) = x.

Step 3. Check that π2−y\frac{\pi}{2}-y lands in the principal range of cos⁡−1\cos^{-1}. Since y∈[−π2,π2]y \in \left[-\frac{\pi}{2},\frac{\pi}{2}\right], multiplying by −1-1 and adding π2\frac{\pi}{2} throughout gives π2−y∈[0,π]\frac{\pi}{2}-y \in [0,\pi] — exactly the principal value range of cos⁡−1\cos^{-1}. So both conditions defining cos⁡−1x\cos^{-1}x are met by the angle π2−y\frac{\pi}{2}-y: it satisfies cos⁡(⋅)=x\cos(\cdot) = x and it lies in [0,π][0,\pi].

Step 4. Since cos⁡−1x\cos^{-1}x is defined as the unique angle in [0,π][0,\pi] whose cosine is xx, and π2−y\frac{\pi}{2}-y is exactly such an angle, uniqueness forces cos⁡−1x=π2−y\cos^{-1}x = \frac{\pi}{2}-y. Substituting back y=sin⁡−1xy=\sin^{-1}x gives cos⁡−1x=π2−sin⁡−1x\cos^{-1}x = \frac{\pi}{2}-\sin^{-1}x, i.e. sin⁡−1x+cos⁡−1x=π2\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}, for every x∈[−1,1]x\in[-1,1].

Worked example

Problem. Find the value of tan⁡−1(1)+cot⁡−1(1)+sec⁡−1(−2)\tan^{-1}(1) + \cot^{-1}(1) + \sec^{-1}(-\sqrt2).

Step 1. Evaluate tan⁡−1(1)\tan^{-1}(1) directly: we need y∈(−π2,π2)y \in \left(-\frac{\pi}{2},\frac{\pi}{2}\right) with tan⁡y=1\tan y = 1. Since tan⁡π4=1\tan\frac{\pi}{4}=1 and π4\frac{\pi}{4} lies in this range, tan⁡−1(1)=π4\tan^{-1}(1)=\frac{\pi}{4}.

Step 2. Evaluate cot⁡−1(1)\cot^{-1}(1) using the complementary relation instead of hunting a new angle: tan⁡−1x+cot⁡−1x=π2\tan^{-1}x+\cot^{-1}x=\frac{\pi}{2} with x=1x=1 gives cot⁡−1(1)=π2−π4=π4\cot^{-1}(1) = \frac{\pi}{2}-\frac{\pi}{4}=\frac{\pi}{4}.

Step 3. Evaluate sec⁡−1(−2)\sec^{-1}(-\sqrt2): we need y∈[0,π]∖{π2}y\in[0,\pi]\setminus\left\{\frac{\pi}{2}\right\} with sec⁡y=−2\sec y=-\sqrt2, i.e. cos⁡y=−12\cos y=-\frac{1}{\sqrt2}. Cosine is negative, so y∈(π2,π]y\in\left(\frac{\pi}{2},\pi\right]. Since cos⁡π4=12\cos\frac{\pi}{4}=\frac{1}{\sqrt2}, the supplementary relation gives cos⁡(π−π4)=−12\cos\left(\pi-\frac{\pi}{4}\right)=-\frac{1}{\sqrt2}, so y=3π4y=\frac{3\pi}{4}, which is valid. Hence sec⁡−1(−2)=3π4\sec^{-1}(-\sqrt2)=\frac{3\pi}{4}.

Step 4. Add the three principal values: π4+π4+3π4=5π4\frac{\pi}{4}+\frac{\pi}{4}+\frac{3\pi}{4} = \frac{5\pi}{4}.

tan⁡−1(1)+cot⁡−1(1)+sec⁡−1(−2)=5π4\tan^{-1}(1) + \cot^{-1}(1) + \sec^{-1}(-\sqrt2) = \mathbf{\frac{5\pi}{4}}

Competitive stretch

Using the same three angles, verify sin⁡−1(sin⁡(π/4))+cos⁡−1(cos⁡(3π/4))\sin^{-1}(\sin(\pi/4)) + \cos^{-1}(\cos(3\pi/4)): since π/4∈[−π2,π2]\pi/4 \in \left[-\frac{\pi}{2},\frac{\pi}{2}\right] the cancellation applies directly, giving π/4\pi/4; since 3π/4∈[0,π]3\pi/4 \in [0,\pi] it also applies directly, giving 3π/43\pi/4; the sum is π\pi — a reminder that the cancellation shortcut only fires when the angle is already inside the correct branch, which both angles here happen to be.

Common traps

  • Writing sin⁡−1x=1sin⁡x\sin^{-1}x=\dfrac{1}{\sin x} — the −1-1 denotes the inverse function, not a reciprocal; the reciprocal of sin⁡x\sin x is csc⁡x\csc x, a different object entirely.
  • Applying sin⁡−1(sin⁡θ)=θ\sin^{-1}(\sin\theta)=\theta for a θ\theta outside [−π2,π2]\left[-\frac{\pi}{2},\frac{\pi}{2}\right] without first reducing it to an equivalent angle inside that range.
  • Picking the wrong sign or "quadrant" for the output: cos⁡−1\cos^{-1}'s range [0,π][0,\pi] never goes negative, so a negative input gives a positive obtuse output, while sin⁡−1\sin^{-1}'s and tan⁡−1\tan^{-1}'s ranges are symmetric about 00, so a negative input there gives a negative output — mixing these two patterns up is the commonest evaluation mistake.
  • Evaluating sec⁡−1x\sec^{-1}x or csc⁡−1x\csc^{-1}x for x∈(−1,1)x\in(-1,1): this interval is excluded from their domain R∖(−1,1)\mathbb{R}\setminus(-1,1), so no real value exists there.

Ask the tutor: If choosing the right principal-value quadrant for a negative input still trips you up, ask the AI tutor to walk through a few more evaluation examples with you.

Why cant sin inverse x just be defined as any angle whose sine is x?

Because then it would not be a function at all. Infinitely many angles share the same sine value — π/6\pi/6, 5π/65\pi/6, and infinitely many angles found by adding multiples of 2π2\pi to either all satisfy sin⁡θ=1/2\sin\theta = 1/2. A function must assign exactly one output to each input, so mathematicians agree in advance on one interval, the principal value range [−π2,π2]\left[-\frac{\pi}{2},\frac{\pi}{2}\right], and sin⁡−1x\sin^{-1}x is defined to be the unique angle inside that interval whose sine equals xx. Every other valid angle is deliberately excluded, not because it is wrong, but because the definition needs a single answer.

How do I remember which range goes with which inverse function?

Anchor on two functions and build the rest from the complementary relations. Memorize sin⁡−1:[−π2,π2]\sin^{-1}:\left[-\frac{\pi}{2},\frac{\pi}{2}\right] (symmetric about zero, since sine is odd) and cos⁡−1:[0,π]\cos^{-1}:[0,\pi] (starts at zero, never negative, since cosine is not odd). Then cot⁡−1x=π2−tan⁡−1x\cot^{-1}x = \frac{\pi}{2}-\tan^{-1}x forces cot⁡−1\cot^{-1}'s range to be (0,π)(0,\pi), and csc⁡−1x=π2−sec⁡−1x\csc^{-1}x=\frac{\pi}{2}-\sec^{-1}x pattern similarly links the reciprocal pair. Practising a handful of standard-angle evaluations does more for retention than memorizing the table cold.

Is sin inverse x the same topic as the general idea of an invertible function?

They are related but the syllabus treats them separately. The general theory of when a function has an inverse (one-one and onto conditions) is developed elsewhere in this unit for algebraic functions. Inverse trigonometric functions are introduced here as their own special, self-contained case: since sin⁡x\sin x, cos⁡x\cos x, and so on are never one-one on all of R\mathbb{R}, this article restricts each one to a chosen principal branch first, and only then talks about its inverse — you do not need the general invertibility machinery to follow this topic.

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Where this leads