NCERT Mathematics Class XII: Inverse Trigonometric Functions Notes

Introduction

  • Felix Klein Quote: "Mathematics, in general, is fundamentally the science of self-evident things."
  • Invertibility of Functions: From previous study, a function ff has an inverse, denoted by f−1f^{-1}, if and only if ff is both one-one (injective) and onto (surjective).
  • Trigonometric Functions Context:
    • Naturally, trigonometric functions are not one-one and onto over their natural domains and ranges.
    • Consequently, their inverses do not exist in their global domains.
  • Chapter Objectives:
    • Study restrictions on domains and ranges of trigonometric functions to ensure the existence of their inverses.
    • Observe behavior through graphical representations.
    • Discuss elementary properties.
  • Applications: Inverse trigonometric functions are essential in calculus for defining many integrals and are widely used in science and engineering.

Basic Concepts of Trigonometric Functions

The following are the definitions of trigonometric functions as studied previously:

  • Sine Function: sin⁡:R→[−1,1]\sin : \mathbb{R} \rightarrow [-1, 1]
  • Cosine Function: cos⁡:R→[−1,1]\cos : \mathbb{R} \rightarrow [-1, 1]
  • Tangent Function: tan⁡:R−{x:x=(2n+1)π2,n∈Z}→R\tan : \mathbb{R} - \{x : x = (2n + 1) \frac{\pi}{2}, n \in \mathbb{Z}\} \rightarrow \mathbb{R}
  • Cotangent Function: cot⁡:R−{x:x=nπ,n∈Z}→R\cot : \mathbb{R} - \{x : x = n\pi, n \in \mathbb{Z}\} \rightarrow \mathbb{R}
  • Secant Function: sec⁡:R−{x:x=(2n+1)π2,n∈Z}→R−(−1,1)\sec : \mathbb{R} - \{x : x = (2n + 1) \frac{\pi}{2}, n \in \mathbb{Z}\} \rightarrow \mathbb{R} - (-1, 1)
  • Cosecant Function: cosec:R−{x:x=nπ,n∈Z}→R−(−1,1)\text{cosec} : \mathbb{R} - \{x : x = n\pi, n \in \mathbb{Z}\} \rightarrow \mathbb{R} - (-1, 1)

Defining Inverse Functions

  • If f:X→Yf : X \rightarrow Y is one-one and onto such that f(x)=yf(x) = y, we define a unique function g:Y→Xg : Y \rightarrow X such that g(y)=xg(y) = x, where x∈Xx \in X and y∈Yy \in Y.
  • Characteristics of the inverse function g=f−1g = f^{-1}:
    • Domain of gg = Range of ff.
    • Range of gg = Domain of ff.
    • The function gg is also one-one and onto.
    • The inverse of gg is ff, i.e., (f−1)−1=f(f^{-1})^{-1} = f.
  • Composition identities:
    • (f−1∘f)(x)=f−1(f(x))=x(f^{-1} \circ f)(x) = f^{-1}(f(x)) = x
    • (f∘f−1)(y)=f(f−1(y))=y(f \circ f^{-1})(y) = f(f^{-1}(y)) = y

The Inverse Sine Function (sin⁡−1\sin^{-1}, arc sine)

  • Domain Restriction: The sine function is restricted to intervals such as [−3π2,−π2][-\frac{3\pi}{2}, -\frac{\pi}{2}], [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}], or [π2,3π2][\frac{\pi}{2}, \frac{3\pi}{2}] to make it one-one and onto with a range of [−1,1][-1, 1].
  • Function Definition: sin⁡−1\sin^{-1} is a function with domain [−1,1][-1, 1].
  • Principal Value Branch: The branch with the range [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}] is termed the principal value branch.
  • Standard Notation: sin⁡−1:[−1,1]→[−π2,π2]\sin^{-1} : [-1, 1] \rightarrow [-\frac{\pi}{2}, \frac{\pi}{2}].
  • Key Properties:
    • sin⁡(sin⁡−1(x))=x\sin(\sin^{-1}(x)) = x if −1≤x≤1-1 \leq x \leq 1.
    • sin⁡−1(sin⁡(x))=x\sin^{-1}(\sin(x)) = x if −π2≤x≤π2-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}.
    • If y=sin⁡−1(x)y = \sin^{-1}(x), then sin⁡(y)=x\sin(y) = x.
  • Graphical Representation:
    • The graph of y=sin⁡−1(x)y = \sin^{-1}(x) is obtained by interchanging the xx and yy axes of the graph of y=sin⁡(x)y = \sin(x).
    • If (a,b)(a, b) is on the sine graph, (b,a)(b, a) is on the inverse sine graph.
    • The inverse graph is a mirror image of the original graph along the line y=xy = x.

The Inverse Cosine Function (cos⁡−1\cos^{-1}, arc cosine)

  • Domain Restriction: Bijective (one-one and onto) when restricted to intervals like [−π,0][-\pi, 0], [0,π][0, \pi], or [π,2π][\pi, 2\pi].
  • Principal Value Branch: The range associated with the principal branch is [0,π][0, \pi].
  • Definition: cos⁡−1:[−1,1]→[0,π]\cos^{-1} : [-1, 1] \rightarrow [0, \pi].

The Inverse Cosecant Function (cosec−1\text{cosec}^{-1}, arc cosecant)

  • Base Function: cosec(x)=1sin⁡(x)\text{cosec}(x) = \frac{1}{\sin(x)}.
  • Domain of Cosecant: R−{x:x=nπ,n∈Z}\mathbb{R} - \{x : x = n\pi, n \in \mathbb{Z}\}.
  • Range of Cosecant: R−(−1,1)\mathbb{R} - (-1, 1), meaning y≥1y \geq 1 or y≤−1y \leq -1.
  • Bijective Restriction: Restrictions such as [−π2,π2]−{0}[-\frac{\pi}{2}, \frac{\pi}{2}] - \{0\}, [−3π2,−π2]−{−π}[-\frac{3\pi}{2}, -\frac{\pi}{2}] - \{-\pi\}, or [π2,3π2]−{π}[\frac{\pi}{2}, \frac{3\pi}{2}] - \{\pi\}.
  • Principal Value Branch: The range is [−π2,π2]−{0}[-\frac{\pi}{2}, \frac{\pi}{2}] - \{0\}.
  • Definition: cosec−1:R−(−1,1)→[−π2,π2]−{0}\text{cosec}^{-1} : \mathbb{R} - (-1, 1) \rightarrow [-\frac{\pi}{2}, \frac{\pi}{2}] - \{0\}.

The Inverse Secant Function (sec⁡−1\sec^{-1}, arc secant)

  • Base Function: sec⁡(x)=1cos⁡(x)\sec(x) = \frac{1}{\cos(x)}.
  • Domain of Secant: R−{x:x=(2n+1)π2,n∈Z}\mathbb{R} - \{x : x = (2n + 1) \frac{\pi}{2}, n \in \mathbb{Z}\}.
  • Range of Secant: R−(−1,1)\mathbb{R} - (-1, 1).
  • Bijective Restriction: Intervals such as [0,π]−{π2}[0, \pi] - \{\frac{\pi}{2}\}, [−π,0]−{−π2}[-\pi, 0] - \{-\frac{\pi}{2}\}, or [π,2π]−{3π2}[\pi, 2\pi] - \{\frac{3\pi}{2}\}.
  • Principal Value Branch: The range is [0,π]−{π2}[0, \pi] - \{\frac{\pi}{2}\}.
  • Definition: sec⁡−1:R−(−1,1)→[0,π]−{π2}\sec^{-1} : \mathbb{R} - (-1, 1) \rightarrow [0, \pi] - \{\frac{\pi}{2}\}.

The Inverse Tangent Function (tan⁡−1\tan^{-1}, arc tangent)

  • Base Function Domain: R−{x:x=(2n+1)π2,n∈Z}\mathbb{R} - \{x : x = (2n + 1) \frac{\pi}{2}, n \in \mathbb{Z}\}.
  • Base Function Range: R\mathbb{R}.
  • Bijective Restriction: Open intervals like (−3π2,−π2)(-\frac{3\pi}{2}, -\frac{\pi}{2}), (−π2,π2)(-\frac{\pi}{2}, \frac{\pi}{2}), or (π2,3π2)(\frac{\pi}{2}, \frac{3\pi}{2}).
  • Principal Value Branch: The range is (−π2,π2)(-\frac{\pi}{2}, \frac{\pi}{2}).
  • Definition: tan⁡−1:R→(−π2,π2)\tan^{-1} : \mathbb{R} \rightarrow (-\frac{\pi}{2}, \frac{\pi}{2}).

The Inverse Cotangent Function (cot⁡−1\cot^{-1}, arc cotangent)

  • Base Function Domain: R−{x:x=nπ,n∈Z}\mathbb{R} - \{x : x = n\pi, n \in \mathbb{Z}\}.
  • Base Function Range: R\mathbb{R}.
  • Bijective Restriction: Open intervals like (−π,0)(-\pi, 0), (0,π)(0, \pi), or (π,2π)(\pi, 2\pi).
  • Principal Value Branch: The range is (0,π)(0, \pi).
  • Definition: cot⁡−1:R→(0,π)\cot^{-1} : \mathbb{R} \rightarrow (0, \pi).

Summary Table of Principal Value Branches

FunctionDomainRange (Principal Value Branch)
y=sin⁡−1(x)y = \sin^{-1}(x)[−1,1][-1, 1][−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}]
y=cos⁡−1(x)y = \cos^{-1}(x)[−1,1][-1, 1][0,π][0, \pi]
y=cosec−1(x)y = \text{cosec}^{-1}(x)R−(−1,1)\mathbb{R} - (-1, 1)[−π2,π2]−{0}[-\frac{\pi}{2}, \frac{\pi}{2}] - \{0\}
y=sec⁡−1(x)y = \sec^{-1}(x)R−(−1,1)\mathbb{R} - (-1, 1)[0,π]−{π2}[0, \pi] - \{\frac{\pi}{2}\}
y=tan⁡−1(x)y = \tan^{-1}(x)R\mathbb{R}(−π2,π2)(-\frac{\pi}{2}, \frac{\pi}{2})
y=cot⁡−1(x)y = \cot^{-1}(x)R\mathbb{R}(0,π)(0, \pi)

Important Notes and Terminology

  1. Notation Warning: sin⁡−1(x)\sin^{-1}(x) is NOT equivalent to (sin⁡(x))−1(\sin(x))^{-1}. Note that (sin⁡(x))−1=1sin⁡(x)(\sin(x))^{-1} = \frac{1}{\sin(x)}. This applies to all trigonometric functions.
  2. Default Branch: If no specific branch is mentioned, the "principal value branch" is always assumed.
  3. Principal Value: The value of an inverse trigonometric function that falls within its principal value branch range.

Examples and Solutions

  • Example 1: Find the principal value of sin⁡−1(12)\sin^{-1}(\frac{1}{\sqrt{2}}).

    • Let sin⁡−1(12)=y\sin^{-1}(\frac{1}{\sqrt{2}}) = y. Then sin⁡(y)=12\sin(y) = \frac{1}{\sqrt{2}}.
    • Range of principal branch is [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}].
    • Since sin⁡(π4)=12\sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}, the principal value is π4\frac{\pi}{4}.

  • Example 2: Find the principal value of cot⁡−1(−13)\cot^{-1}(-\frac{1}{\sqrt{3}}).

    • Let cot⁡−1(−13)=y\cot^{-1}(-\frac{1}{\sqrt{3}}) = y. Then cot⁡(y)=−13\cot(y) = -\frac{1}{\sqrt{3}}.
    • cot⁡(y)=−cot⁡(π3)=cot⁡(π−π3)=cot⁡(2π3)\cot(y) = -\cot(\frac{\pi}{3}) = \cot(\pi - \frac{\pi}{3}) = \cot(\frac{2\pi}{3}).
    • Range of principal branch is (0,π)(0, \pi).
    • The principal value is 2π3\frac{2\pi}{3}.

  • Example 6: Find the value of sin⁡−1(sin⁡(3π5))\sin^{-1}(\sin(\frac{3\pi}{5})).

    • sin⁡−1(sin⁡(x))=x\sin^{-1}(\sin(x)) = x only if x∈[−π2,π2]x \in [-\frac{\pi}{2}, \frac{\pi}{2}].
    • 3π5∉[−π2,π2]\frac{3\pi}{5} \notin [-\frac{\pi}{2}, \frac{\pi}{2}].
    • Use identity sin⁡(π−θ)=sin⁡(θ)\sin(\pi - \theta) = \sin(\theta). Thus \sin(\frac{3\pi}{5}) = \sin(\pi - \n \frac{3\pi}{5}) = \sin(\frac{2\pi}{5}).
    • Since 2π5∈[−π2,π2]\frac{2\pi}{5} \in [-\frac{\pi}{2}, \frac{\pi}{2}], the value is 2π5\frac{2\pi}{5}.

Selected Properties and Simplifications

  • Example 3: Show that sin⁡−1(2x1−x2)=2sin⁡−1(x)\sin^{-1}(2x \sqrt{1 - x^2}) = 2 \sin^{-1}(x) for −12≤x≤12-\frac{1}{\sqrt{2}} \leq x \leq \frac{1}{\sqrt{2}}.

    • Let x=sin⁡(θ)x = \sin(\theta). Then θ=sin⁡−1(x)\theta = \sin^{-1}(x).
    • LHS=sin⁡−1(2sin⁡(θ)1−sin⁡2(θ))=sin⁡−1(2sin⁡(θ)cos⁡(θ))=sin⁡−1(sin⁡(2θ))=2θ=2sin⁡−1(x)\text{LHS} = \sin^{-1}(2 \sin(\theta) \sqrt{1 - \sin^2(\theta)}) = \sin^{-1}(2 \sin(\theta) \cos(\theta)) = \sin^{-1}(\sin(2\theta)) = 2\theta = 2 \sin^{-1}(x).

  • Example 4: Express tan⁡−1(cos⁡(x)1−sin⁡(x))\tan^{-1}(\frac{\cos(x)}{1 - \sin(x)}) in simplest form for −3π2<x<π2-\frac{3\pi}{2} < x < \frac{\pi}{2}.

    • cos⁡(x)=cos⁡2(x2)−sin⁡2(x2)\cos(x) = \cos^2(\frac{x}{2}) - \sin^2(\frac{x}{2}).
    • 1−sin⁡(x)=cos⁡2(x2)+sin⁡2(x2)−2sin⁡(x2)cos⁡(x2)=(cos⁡(x2)−sin⁡(x2))21 - \sin(x) = \cos^2(\frac{x}{2}) + \sin^2(\frac{x}{2}) - 2 \sin(\frac{x}{2}) \cos(\frac{x}{2}) = (\cos(\frac{x}{2}) - \sin(\frac{x}{2}))^2.
    • tan⁡−1((cos⁡(x2)−sin⁡(x2))(cos⁡(x2)+sin⁡(x2))(cos⁡(x2)−sin⁡(x2))2)=tan⁡−1(cos⁡(x2)+sin⁡(x2)cos⁡(x2)−sin⁡(x2))\tan^{-1}(\frac{(\cos(\frac{x}{2}) - \sin(\frac{x}{2}))(\cos(\frac{x}{2}) + \sin(\frac{x}{2}))}{(\cos(\frac{x}{2}) - \sin(\frac{x}{2}))^2}) = \tan^{-1}(\frac{\cos(\frac{x}{2}) + \sin(\frac{x}{2})}{\cos(\frac{x}{2}) - \sin(\frac{x}{2})}).
    • Divided by cos⁡(x2)\cos(\frac{x}{2}): tan⁡−1(1+tan⁡(x2)1−tan⁡(x2))=tan⁡−1(tan⁡(π4+x2))=π4+x2\tan^{-1}(\frac{1 + \tan(\frac{x}{2})}{1 - \tan(\frac{x}{2})}) = \tan^{-1}(\tan(\frac{\pi}{4} + \frac{x}{2})) = \frac{\pi}{4} + \frac{x}{2}.

Historical Note

  • Origins: Trigonometry study began in India with mathematicians like Aryabhata (476 A.D.), Brahmagupta (598 A.D.), Bhaskara I (600 A.D.), and Bhaskara II (1114 A.D.).
  • Transmission: Knowledge moved from India to Arabia and then to Europe. The Indian approach superseded the Greek approach when it became known.
  • Contributions:
    • Introduction of the sine function is a primary contribution of the "siddhantas" (Sanskrit astronomical works).
    • Bhaskara I provided formulae for sine values of angles greater than 90∘90^{\circ}.
    • The 16th-century Malayalam work "Yuktibhasa" contains a proof for sin⁡(A+B)\sin(A + B).
    • Bhaskara II provided exact expressions for sines/cosines of 18∘,36∘,54∘,72∘18^{\circ}, 36^{\circ}, 54^{\circ}, 72^{\circ}.
  • Symbols: Sir John F.W. Herschel suggested the symbols sin⁡−1(x)\sin^{-1}(x), cos⁡−1(x)\cos^{-1}(x), etc., in 1813.
  • Height and Distance: Associated with Thales (600 B.C.), who calculated pyramid heights using shadow ratios: HS=hs=tan⁡(sun’s altitude)\frac{H}{S} = \frac{h}{s} = \tan(\text{sun's altitude}).