Inverse Trigonometric Functions — Domain, Range, Properties

Inverse Trigonometric Functions — Domain, Range, Properties

• 9 min read
Tags Inverse Trigonometric Functions

Why Inverse Trig Functions Matter

Regular trig functions take an angle and give a ratio. Inverse trig functions do the reverse — they take a ratio and return an angle. If sin⁡θ=12\sin\theta = \frac{1}{2}, then sin⁡−1(12)=π6\sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}. Simple idea, but the details around domain restrictions and principal values are where students lose marks.

This chapter carries consistent weightage in CBSE Class 12 boards (4-6 marks) and is a regular in JEE Main. The questions are typically formula-based — if you know the properties, you can solve them fast.

graph TD
    A[Inverse Trig Problem] --> B{What's given?}
    B -->|Expression to simplify| C[Apply identities]
    B -->|Find principal value| D[Check domain + range]
    B -->|Prove an identity| E[Convert and simplify]
    C --> F{Which identity?}
    F -->|Sum/difference| G["sin⁻¹x ± sin⁻¹y"]
    F -->|Double angle| H["2tan⁻¹x forms"]
    F -->|Complementary| I["sin⁻¹x + cos⁻¹x = π/2"]
    D --> J[Is input in valid domain?]
    J -->|Yes| K[Read from principal range]
    J -->|No| L[Undefined]

Key Terms & Definitions

Principal Value — Since trig functions are many-to-one, their inverses must be restricted to a specific range (the principal branch) to be well-defined. For example, sin⁡−1\sin^{-1} returns values only in [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right].

Domain — The set of valid inputs for the inverse function. For sin⁡−1x\sin^{-1}x, the domain is [−1,1][-1, 1].

Range (Principal Branch) — The set of possible output angles.

Domain and Range Table

FunctionDomainRange (Principal Value)
sin⁡−1x\sin^{-1}x[−1,1][-1, 1][−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]
cos⁡−1x\cos^{-1}x[−1,1][-1, 1][0,π][0, \pi]
tan⁡−1x\tan^{-1}x(−∞,∞)(-\infty, \infty)(−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)
cot⁡−1x\cot^{-1}x(−∞,∞)(-\infty, \infty)(0,π)(0, \pi)
sec⁡−1x\sec^{-1}x(−∞,−1]∪[1,∞)(-\infty, -1] \cup [1, \infty)[0,π]−{π2}[0, \pi] - \left\{\frac{\pi}{2}\right\}
csc⁡−1x\csc^{-1}x(−∞,−1]∪[1,∞)(-\infty, -1] \cup [1, \infty)[−π2,π2]−{0}\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] - \{0\}

Memorise these ranges as pairs: sin⁡−1\sin^{-1} and csc⁡−1\csc^{-1} share the same range [−π/2,π/2][-\pi/2, \pi/2]. cos⁡−1\cos^{-1} and sec⁡−1\sec^{-1} share [0,π][0, \pi]. tan⁡−1\tan^{-1} uses the open interval (−π/2,π/2)(-\pi/2, \pi/2) and cot⁡−1\cot^{-1} uses (0,π)(0, \pi).

Essential Properties

sin⁡−1x+cos⁡−1x=π2,∣x∣≤1\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}, \quad |x| \leq 1 tan⁡−1x+cot⁡−1x=π2,x∈R\tan^{-1}x + \cot^{-1}x = \frac{\pi}{2}, \quad x \in \mathbb{R} sec⁡−1x+csc⁡−1x=π2,∣x∣≥1\sec^{-1}x + \csc^{-1}x = \frac{\pi}{2}, \quad |x| \geq 1 sin⁡−1(−x)=−sin⁡−1x\sin^{-1}(-x) = -\sin^{-1}x cos⁡−1(−x)=π−cos⁡−1x\cos^{-1}(-x) = \pi - \cos^{-1}x tan⁡−1(−x)=−tan⁡−1x\tan^{-1}(-x) = -\tan^{-1}x tan⁡−1x+tan⁡−1y=tan⁡−1(x+y1−xy),xy<1\tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right), \quad xy < 1 tan⁡−1x−tan⁡−1y=tan⁡−1(x−y1+xy),xy>−1\tan^{-1}x - \tan^{-1}y = \tan^{-1}\left(\frac{x-y}{1+xy}\right), \quad xy > -1 2tan⁡−1x=sin⁡−1(2x1+x2)=cos⁡−1(1−x21+x2)=tan⁡−1(2x1−x2)2\tan^{-1}x = \sin^{-1}\left(\frac{2x}{1+x^2}\right) = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) = \tan^{-1}\left(\frac{2x}{1-x^2}\right)

Valid for ∣x∣≤1|x| \leq 1 (for sin⁡−1\sin^{-1} form) and x≥0x \geq 0 (for cos⁡−1\cos^{-1} form).

Solved Examples — Easy to Hard

Example 1 (Easy — CBSE)

Find the principal value of sin⁡−1(−32)\sin^{-1}\left(-\frac{\sqrt{3}}{2}\right).

We need θ∈[−π/2,π/2]\theta \in [-\pi/2, \pi/2] such that sin⁡θ=−32\sin\theta = -\frac{\sqrt{3}}{2}.

Since sin⁡(−π/3)=−32\sin(-\pi/3) = -\frac{\sqrt{3}}{2} and −π/3∈[−π/2,π/2]-\pi/3 \in [-\pi/2, \pi/2]:

sin⁡−1(−32)=−π3\sin^{-1}\left(-\frac{\sqrt{3}}{2}\right) = \mathbf{-\frac{\pi}{3}}

Example 2 (Medium — JEE Main)

Simplify: tan⁡−112+tan⁡−113\tan^{-1}\frac{1}{2} + \tan^{-1}\frac{1}{3}.

Check: xy=16<1xy = \frac{1}{6} < 1, so we can use the addition formula.

tan⁡−112+tan⁡−113=tan⁡−1(12+131−16)=tan⁡−1(5656)=tan⁡−1(1)=π4\tan^{-1}\frac{1}{2} + \tan^{-1}\frac{1}{3} = \tan^{-1}\left(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{6}}\right) = \tan^{-1}\left(\frac{\frac{5}{6}}{\frac{5}{6}}\right) = \tan^{-1}(1) = \mathbf{\frac{\pi}{4}}

Example 3 (Medium — CBSE Board)

Prove: sin⁡−135+cos⁡−11213=sin⁡−15665\sin^{-1}\frac{3}{5} + \cos^{-1}\frac{12}{13} = \sin^{-1}\frac{56}{65}.

Let α=sin⁡−135\alpha = \sin^{-1}\frac{3}{5} and β=cos⁡−11213\beta = \cos^{-1}\frac{12}{13}.

Then sin⁡α=3/5\sin\alpha = 3/5, cos⁡α=4/5\cos\alpha = 4/5 and cos⁡β=12/13\cos\beta = 12/13, sin⁡β=5/13\sin\beta = 5/13.

sin⁡(α+β)=sin⁡αcos⁡β+cos⁡αsin⁡β=35⋅1213+45⋅513=36+2065=5665\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta = \frac{3}{5}\cdot\frac{12}{13} + \frac{4}{5}\cdot\frac{5}{13} = \frac{36+20}{65} = \frac{56}{65}

Hence α+β=sin⁡−15665\alpha + \beta = \sin^{-1}\frac{56}{65}. Proved.

Example 4 (Hard — JEE Advanced)

Solve: tan⁡−1(x−1)+tan⁡−1(x)+tan⁡−1(x+1)=tan⁡−1(3x)\tan^{-1}(x-1) + \tan^{-1}(x) + \tan^{-1}(x+1) = \tan^{-1}(3x).

Rearrange: tan⁡−1(x−1)+tan⁡−1(x+1)=tan⁡−1(3x)−tan⁡−1(x)\tan^{-1}(x-1) + \tan^{-1}(x+1) = \tan^{-1}(3x) - \tan^{-1}(x).

LHS: tan⁡−1(x−1)+(x+1)1−(x2−1)=tan⁡−12x2−x2\tan^{-1}\frac{(x-1)+(x+1)}{1-(x^2-1)} = \tan^{-1}\frac{2x}{2-x^2}

RHS: tan⁡−13x−x1+3x2=tan⁡−12x1+3x2\tan^{-1}\frac{3x-x}{1+3x^2} = \tan^{-1}\frac{2x}{1+3x^2}

Equating: 2x2−x2=2x1+3x2\frac{2x}{2-x^2} = \frac{2x}{1+3x^2}

Either 2x=0⇒x=02x = 0 \Rightarrow x = 0, or 2−x2=1+3x2⇒4x2=1⇒x=±122 - x^2 = 1 + 3x^2 \Rightarrow 4x^2 = 1 \Rightarrow x = \pm\frac{1}{2}.

Solutions: x∈{0,12,−12}x \in \left\{0, \frac{1}{2}, -\frac{1}{2}\right\}.

Exam-Specific Tips

CBSE Board: This chapter is worth 4-6 marks. Expect 1-2 questions: one on principal values and one on proving/simplifying an identity. The tan⁡−1\tan^{-1} addition formula and complementary property (sin⁡−1x+cos⁡−1x=π/2\sin^{-1}x + \cos^{-1}x = \pi/2) are tested almost every year.

JEE Main: 1 question per paper on average. Common types: simplify a sum of inverse trig terms, solve an equation involving inverse trig. The 2tan⁡−1x2\tan^{-1}x double-angle forms are JEE favourites.

Common Mistakes to Avoid

Mistake 1 — Ignoring domain conditions in the addition formula. The formula tan⁡−1x+tan⁡−1y=tan⁡−1x+y1−xy\tan^{-1}x + \tan^{-1}y = \tan^{-1}\frac{x+y}{1-xy} works only when xy<1xy < 1. When xy>1xy > 1, you need to add or subtract π\pi.

Mistake 2 — Writing sin⁡−1(−x)=π+sin⁡−1x\sin^{-1}(-x) = \pi + \sin^{-1}x. The correct property is sin⁡−1(−x)=−sin⁡−1x\sin^{-1}(-x) = -\sin^{-1}x. The π−\pi - pattern applies to cos⁡−1\cos^{-1}, not sin⁡−1\sin^{-1}.

Mistake 3 — Giving multiple values. Inverse trig functions return a single principal value. sin⁡−1(1/2)=π/6\sin^{-1}(1/2) = \pi/6 only, not 5π/65\pi/6 as well.

Mistake 4 — Confusing sin⁡−1x\sin^{-1}x with 1sin⁡x\frac{1}{\sin x}. sin⁡−1x\sin^{-1}x is the inverse function (arcsin). (sin⁡x)−1=csc⁡x(\sin x)^{-1} = \csc x is the reciprocal. Completely different things.

Mistake 5 — Wrong range for cos⁡−1\cos^{-1}. cos⁡−1(−1/2)=2π/3\cos^{-1}(-1/2) = 2\pi/3, not −π/3-\pi/3. The range of cos⁡−1\cos^{-1} is [0,π][0, \pi], so the answer is always non-negative.

Practice Questions

Q1. Find the principal value of cos⁡−1(−12)\cos^{-1}\left(-\frac{1}{2}\right).

cos⁡−1(−1/2)=π−cos⁡−1(1/2)=π−π/3=2π/3\cos^{-1}(-1/2) = \pi - \cos^{-1}(1/2) = \pi - \pi/3 = 2\pi/3.

Q2. Evaluate tan⁡−1(1)+tan⁡−1(2)+tan⁡−1(3)\tan^{-1}(1) + \tan^{-1}(2) + \tan^{-1}(3).

tan⁡−1(2)+tan⁡−1(3)=π+tan⁡−12+31−6=π+tan⁡−1(−1)=π−π/4=3π/4\tan^{-1}(2) + \tan^{-1}(3) = \pi + \tan^{-1}\frac{2+3}{1-6} = \pi + \tan^{-1}(-1) = \pi - \pi/4 = 3\pi/4. Adding tan⁡−1(1)=π/4\tan^{-1}(1) = \pi/4: total =π= \pi.

Q3. Simplify sin⁡−1(2x1−x2)\sin^{-1}(2x\sqrt{1-x^2}) for 0≤x≤120 \leq x \leq \frac{1}{\sqrt{2}}.

Put x=sin⁡θx = \sin\theta. Then 2sin⁡θcos⁡θ=sin⁡2θ2\sin\theta\cos\theta = \sin 2\theta. So the expression =2sin⁡−1x= 2\sin^{-1}x.

Q4. If sin⁡−1x+sin⁡−1y=π2\sin^{-1}x + \sin^{-1}y = \frac{\pi}{2}, find x2+y2x^2 + y^2.

sin⁡−1x=π2−sin⁡−1y=cos⁡−1y\sin^{-1}x = \frac{\pi}{2} - \sin^{-1}y = \cos^{-1}y. So x=cos⁡(sin⁡−1y)=1−y2x = \cos(\sin^{-1}y) = \sqrt{1-y^2}. Therefore x2+y2=1x^2 + y^2 = 1.

Q5. Find the value of tan⁡(cos⁡−145)\tan\left(\cos^{-1}\frac{4}{5}\right).

Let θ=cos⁡−1(4/5)\theta = \cos^{-1}(4/5). Then cos⁡θ=4/5\cos\theta = 4/5, sin⁡θ=3/5\sin\theta = 3/5. So tan⁡θ=3/4\tan\theta = 3/4.

Q6. Prove: 2tan⁡−113+tan⁡−117=π42\tan^{-1}\frac{1}{3} + \tan^{-1}\frac{1}{7} = \frac{\pi}{4}.

2tan⁡−1(1/3)=tan⁡−12/31−1/9=tan⁡−12/38/9=tan⁡−1(3/4)2\tan^{-1}(1/3) = \tan^{-1}\frac{2/3}{1-1/9} = \tan^{-1}\frac{2/3}{8/9} = \tan^{-1}(3/4). Then tan⁡−1(3/4)+tan⁡−1(1/7)=tan⁡−13/4+1/71−3/28=tan⁡−125/2825/28=tan⁡−1(1)=π/4\tan^{-1}(3/4) + \tan^{-1}(1/7) = \tan^{-1}\frac{3/4+1/7}{1-3/28} = \tan^{-1}\frac{25/28}{25/28} = \tan^{-1}(1) = \pi/4.

Q7. Solve: cos⁡−1x+cos⁡−12x=2π3\cos^{-1}x + \cos^{-1}2x = \frac{2\pi}{3}.

cos⁡−12x=2π3−cos⁡−1x\cos^{-1}2x = \frac{2\pi}{3} - \cos^{-1}x. Taking cosine: 2x=cos⁡2π3cos⁡(cos⁡−1x)+sin⁡2π3sin⁡(cos⁡−1x)=−x2+321−x22x = \cos\frac{2\pi}{3}\cos(\cos^{-1}x) + \sin\frac{2\pi}{3}\sin(\cos^{-1}x) = -\frac{x}{2} + \frac{\sqrt{3}}{2}\sqrt{1-x^2}. So 5x2=321−x2\frac{5x}{2} = \frac{\sqrt{3}}{2}\sqrt{1-x^2}. Squaring: 25x2=3(1−x2)25x^2 = 3(1-x^2), giving 28x2=328x^2 = 3, so x=327x = \frac{\sqrt{3}}{2\sqrt{7}}.

Q8. Find cot⁡(tan⁡−1a+cot⁡−1a)\cot(\tan^{-1}a + \cot^{-1}a).

tan⁡−1a+cot⁡−1a=π/2\tan^{-1}a + \cot^{-1}a = \pi/2. So cot⁡(π/2)=0\cot(\pi/2) = 0.

FAQs

Why do we restrict the domain/range?

Trig functions are periodic — sin⁡\sin gives the same value at infinitely many angles. To make the inverse a proper function (single output per input), we restrict to one branch (the principal branch).

Is sin⁡−1(sin⁡x)=x\sin^{-1}(\sin x) = x always true?

Only when x∈[−π/2,π/2]x \in [-\pi/2, \pi/2]. For other values of xx, you need to find the equivalent angle in the principal range. For example, sin⁡−1(sin⁡5π6)=π−5π6=π6\sin^{-1}(\sin \frac{5\pi}{6}) = \pi - \frac{5\pi}{6} = \frac{\pi}{6}.

What are the most important formulas for JEE?

The tan⁡−1\tan^{-1} addition formula and the 2tan⁡−1x2\tan^{-1}x conversions. About 70% of JEE questions on this chapter use one of these.

How to handle equations with multiple inverse trig terms?

Move terms to isolate one inverse function on each side, then take trig of both sides. Alternatively, use substitution (x=sin⁡θx = \sin\theta) to convert the problem.

What’s the graph of y=sin⁡−1xy = \sin^{-1}x?

It’s the reflection of y=sin⁡xy = \sin x (restricted to [−π/2,π/2][-\pi/2, \pi/2]) across the line y=xy = x. An S-shaped curve from (−1,−π/2)(-1, -\pi/2) to (1,π/2)(1, \pi/2).

Additional Identities

sin⁡−11x=csc⁡−1x,cos⁡−11x=sec⁡−1x,tan⁡−11x=cot⁡−1x (for x>0)\sin^{-1}\frac{1}{x} = \csc^{-1}x, \quad \cos^{-1}\frac{1}{x} = \sec^{-1}x, \quad \tan^{-1}\frac{1}{x} = \cot^{-1}x \text{ (for } x > 0\text{)}

For x &lt; 0: tan⁡−11x=cot⁡−1x−π\tan^{-1}\frac{1}{x} = \cot^{-1}x - \pi

 sin⁡−1x+sin⁡−1y=sin⁡−1(x1−y2+y1−x2)\sin^{-1}x + \sin^{-1}y = \sin^{-1}(x\sqrt{1-y^2} + y\sqrt{1-x^2})

Valid when x2+y2≤1x^2 + y^2 \leq 1 or xy &lt; 0.

Converting between inverse trig forms

A common JEE technique: convert everything to tan⁡−1\tan^{-1} form and simplify.

If sin⁡−1x=θ\sin^{-1}x = \theta, then tan⁡θ=x1−x2\tan\theta = \frac{x}{\sqrt{1-x^2}}, so sin⁡−1x=tan⁡−1x1−x2\sin^{-1}x = \tan^{-1}\frac{x}{\sqrt{1-x^2}}.

Similarly: cos⁡−1x=tan⁡−11−x2x\cos^{-1}x = \tan^{-1}\frac{\sqrt{1-x^2}}{x} (for x>0x > 0).

This conversion is useful when a problem mixes different inverse trig functions.

Additional Solved Examples

Example 5 (JEE Main): Simplify tan⁡−112+tan⁡−115+tan⁡−118\tan^{-1}\frac{1}{2} + \tan^{-1}\frac{1}{5} + \tan^{-1}\frac{1}{8}.

First: tan⁡−112+tan⁡−115=tan⁡−11/2+1/51−1/10=tan⁡−17/109/10=tan⁡−179\tan^{-1}\frac{1}{2} + \tan^{-1}\frac{1}{5} = \tan^{-1}\frac{1/2 + 1/5}{1 - 1/10} = \tan^{-1}\frac{7/10}{9/10} = \tan^{-1}\frac{7}{9}

Then: tan⁡−179+tan⁡−118=tan⁡−17/9+1/81−7/72=tan⁡−165/7265/72=tan⁡−1(1)=π4\tan^{-1}\frac{7}{9} + \tan^{-1}\frac{1}{8} = \tan^{-1}\frac{7/9 + 1/8}{1 - 7/72} = \tan^{-1}\frac{65/72}{65/72} = \tan^{-1}(1) = \frac{\pi}{4}

Example 6 (CBSE Board): Find the value of cos⁡(sin⁡−135+cos⁡−11213)\cos(\sin^{-1}\frac{3}{5} + \cos^{-1}\frac{12}{13}).

Let α=sin⁡−135\alpha = \sin^{-1}\frac{3}{5}, β=cos⁡−11213\beta = \cos^{-1}\frac{12}{13}. Then sin⁡α=3/5\sin\alpha = 3/5, cos⁡α=4/5\cos\alpha = 4/5, cos⁡β=12/13\cos\beta = 12/13, sin⁡β=5/13\sin\beta = 5/13.

cos⁡(α+β)=cos⁡αcos⁡β−sin⁡αsin⁡β=45⋅1213−35⋅513=48−1565=3365\cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta = \frac{4}{5}\cdot\frac{12}{13} - \frac{3}{5}\cdot\frac{5}{13} = \frac{48-15}{65} = \frac{33}{65}

Graph Properties Summary

FunctionDomainRangeNature
sin⁡−1x\sin^{-1}x[−1,1][-1,1][−π/2,π/2][-\pi/2, \pi/2]Increasing, odd
cos⁡−1x\cos^{-1}x[−1,1][-1,1][0,π][0, \pi]Decreasing, neither odd nor even
tan⁡−1x\tan^{-1}xR\mathbb{R}(−π/2,π/2)(-\pi/2, \pi/2)Increasing, odd

sin⁡−1\sin^{-1} and tan⁡−1\tan^{-1} are odd functions: f(−x)=−f(x)f(-x) = -f(x). But cos⁡−1\cos^{-1} is NOT odd. Instead, cos⁡−1(−x)=π−cos⁡−1x\cos^{-1}(-x) = \pi - \cos^{-1}x. This asymmetry is a frequent source of errors in JEE.

When xy>1xy > 1 in the tan⁡−1\tan^{-1} addition formula

The standard formula tan⁡−1x+tan⁡−1y=tan⁡−1x+y1−xy\tan^{-1}x + \tan^{-1}y = \tan^{-1}\frac{x+y}{1-xy} works only when xy &lt; 1.

When xy>1xy > 1 and both x,y>0x, y > 0:

tan⁡−1x+tan⁡−1y=π+tan⁡−1x+y1−xy\tan^{-1}x + \tan^{-1}y = \pi + \tan^{-1}\frac{x+y}{1-xy}

When xy>1xy > 1 and both x, y &lt; 0:

tan⁡−1x+tan⁡−1y=−π+tan⁡−1x+y1−xy\tan^{-1}x + \tan^{-1}y = -\pi + \tan^{-1}\frac{x+y}{1-xy}

Forgetting the π\pi correction when xy>1xy > 1 is the most common JEE error on this topic. Always check whether xy &lt; 1 before applying the formula.

Substitution techniques

Many inverse trig problems simplify dramatically with the right substitution:

ExpressionSubstituteResult
1−x2\sqrt{1-x^2}x=sin⁡θx = \sin\thetacos⁡θ\cos\theta
1+x2\sqrt{1+x^2}x=tan⁡θx = \tan\thetasec⁡θ\sec\theta
x2−1\sqrt{x^2-1}x=sec⁡θx = \sec\thetatan⁡θ\tan\theta
2x1+x2\frac{2x}{1+x^2}x=tan⁡θx = \tan\thetasin⁡2θ\sin 2\theta
1−x21+x2\frac{1-x^2}{1+x^2}x=tan⁡θx = \tan\thetacos⁡2θ\cos 2\theta

Master these substitutions and most JEE inverse trig problems become one-line solutions.

Practice Questions

Evaluate arcsin(1/2) + arccos(1/2)

hard

Find sin⁻¹(1/2) — Basic Inverse Trig

easy NCERT Class 12

Prove sin⁻¹x + cos⁻¹x = π/2

medium CBSE 2024 Board Exam

Prove that arctan(1) + arctan(2) + arctan(3) = pi

easy

Simplify tan⁻¹(1) + tan⁻¹(2) + tan⁻¹(3)

hard JEE Main 2024