INVERSE TRIGONOMETRY

Inverse trigonometric functions help us determine the angles when given the trigonometric ratio values. They are widely used in geometry, calculus, physics, and engineering.


1. Introduction to Inverse Trigonometric Functions

Trigonometric functions are many-to-one, meaning they don’t have a unique inverse unless their domain is restricted.

For example, sin⁡x\sin xsinx is periodic, meaning multiple angles have the same sine value. So, to define sin⁡−1x\sin^{-1} xsin−1x as a function, we restrict the domain of sin⁡x\sin xsinx to one principal branch where it is one-to-one.

Definition:

If y=f(x) is a function, then its inverse is x=f−1(y), meaning that f(f−1(y))=y.

For trigonometric functions:

  • siny=x⇒y=sin−1x

  • cos⁡y=x⇒y=cos⁡−1x\cos y = x \Rightarrow y = \cos^{-1} xcosy=x⇒y=cos−1x

  • tan⁡y=x⇒y=tan⁡−1x\tan y = x \Rightarrow y = \tan^{-1} xtany=x⇒y=tan−1x

  • And similarly for csc⁡−1x,sec⁡−1x,cot⁡−1x\csc^{-1} x, \sec^{-1} x, \cot^{-1} xcsc−1x,sec−1x,cot−1x.


2. Domains and Ranges of Inverse Trigonometric Functions

Each inverse function has a specific domain (input range) and range (output angle).

Function

Domain

Range

sin⁡−1x\sin^{-1} xsin−1x

[−1,1][-1,1][−1,1]

[−π2,π2][- \frac{\pi}{2}, \frac{\pi}{2}][−2π​,2π​]

cos⁡−1x\cos^{-1} xcos−1x

[−1,1][-1,1][−1,1]

[0,π][0, \pi][0,π]

tan⁡−1x\tan^{-1} xtan−1x

R\mathbb{R}R (all real numbers)

(−π2,π2)(- \frac{\pi}{2}, \frac{\pi}{2})(−2π​,2π​)

csc⁡−1x\csc^{-1} xcsc−1x

x≤−1x \leq -1x≤−1 or x≥1x \geq 1x≥1

[−π2,π2]∖{0}[- \frac{\pi}{2}, \frac{\pi}{2}] \setminus \{0\}[−2π​,2π​]∖{0}

sec⁡−1x\sec^{-1} xsec−1x

x≤−1x \leq -1x≤−1 or x≥1x \geq 1x≥1

[0,π]∖{π2}[0, \pi] \setminus \{ \frac{\pi}{2} \}[0,π]∖{2π​}

cot⁡−1x\cot^{-1} xcot−1x

R\mathbb{R}R

(0,π)(0, \pi)(0,π)

👉 Why is the range restricted?

  • To make the function one-to-one and well-defined, we pick one principal branch.


3. Properties of Inverse Trigonometric Functions

(a) Reciprocal Relations

  • sin⁡−1x+cos⁡−1x=π2\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}sin−1x+cos−1x=2π​

  • tan⁡−1x+cot⁡−1x=π2\tan^{-1} x + \cot^{-1} x = \frac{\pi}{2}tan−1x+cot−1x=2π​

  • sec⁡−1x+csc⁡−1x=π2\sec^{-1} x + \csc^{-1} x = \frac{\pi}{2}sec−1x+csc−1x=2π​

(b) Function Composition

  • sin⁡(sin⁡−1x)=x,x∈[−1,1]\sin(\sin^{-1} x) = x, \quad x \in [-1,1]sin(sin−1x)=x,x∈[−1,1]

  • cos⁡(cos⁡−1x)=x,x∈[−1,1]\cos(\cos^{-1} x) = x, \quad x \in [-1,1]cos(cos−1x)=x,x∈[−1,1]

  • tan⁡(tan⁡−1x)=x,x∈R\tan(\tan^{-1} x) = x, \quad x \in \mathbb{R}tan(tan−1x)=x,x∈R

  • sin⁡−1(sin⁡x)=x\sin^{-1} (\sin x) = xsin−1(sinx)=x, if x∈[−π2,π2]x \in [-\frac{\pi}{2}, \frac{\pi}{2}]x∈[−2π​,2π​]

  • cos⁡−1(cos⁡x)=x\cos^{-1} (\cos x) = xcos−1(cosx)=x, if x∈[0,π]x \in [0, \pi]x∈[0,π]


4. Graphs of Inverse Trigonometric Functions

Each function has a distinct graph showing its domain and range.

Graph of y=sin⁡−1xy = \sin^{-1} xy=sin−1x

  • It is increasing in [−1,1][-1,1][−1,1].

  • Symmetric about the origin.

Graph of y=cos⁡−1xy = \cos^{-1} xy=cos−1x

  • It is decreasing in [−1,1][-1,1][−1,1].

Graph of y=tan⁡−1xy = \tan^{-1} xy=tan−1x

  • It is increasing in (−∞,∞)(-\infty, \infty)(−∞,∞).

📌 Graphing tip:

  • The inverse function graph can be obtained by reflecting the normal function graph across the line y=xy = xy=x.


5. Differentiation of Inverse Trigonometric Functions

The derivatives are crucial for calculus.

Function

Derivative

ddxsin⁡−1x\frac{d}{dx} \sin^{-1} xdxd​sin−1x

11−x2\frac{1}{\sqrt{1 - x^2}}1−x2​1​

ddxcos⁡−1x\frac{d}{dx} \cos^{-1} xdxd​cos−1x

−11−x2-\frac{1}{\sqrt{1 - x^2}}−1−x2​1​

ddxtan⁡−1x\frac{d}{dx} \tan^{-1} xdxd​tan−1x

11+x2\frac{1}{1 + x^2}1+x21​

ddxcsc⁡−1x\frac{d}{dx} \csc^{-1} xdxd​csc−1x

( -\frac{1}{

ddxsec⁡−1x\frac{d}{dx} \sec^{-1} xdxd​sec−1x

( \frac{1}{

ddxcot⁡−1x\frac{d}{dx} \cot^{-1} xdxd​cot−1x

−11+x2-\frac{1}{1 + x^2}−1+x21​


6. Integration of Inverse Trigonometric Functions

Common integrals include:

∫dx1−x2=sin⁡−1x+C\int \frac{dx}{\sqrt{1 - x^2}} = \sin^{-1} x + C∫1−x2​dx​=sin−1x+C∫dx1+x2=tan⁡−1x+C\int \frac{dx}{1 + x^2} = \tan^{-1} x + C∫1+x2dx​=tan−1x+C∫dxxx2−1=sec⁡−1x+C\int \frac{dx}{x \sqrt{x^2 - 1}} = \sec^{-1} x + C∫xx2−1​dx​=sec−1x+C

👉 Application in Definite Integrals:
Inverse trigonometric functions are used in evaluating integrals in calculus.


7. Applications of Inverse Trigonometric Functions

Inverse trigonometric functions appear in:

  • Physics (Wave motion, oscillations)

  • Navigation & GPS (Finding direction angles)

  • Architecture (Angles in bridges and buildings)

  • Robotics & AI (Motion planning)

  • Computer Graphics (3D transformations)


8. Important Questions

  1. Solve: sin⁡−1(12)+cos⁡−1(12)\sin^{-1} (\frac{1}{2}) + \cos^{-1} (\frac{1}{2})sin−1(21​)+cos−1(21​).

  2. Prove that tan⁡−1x+tan⁡−1y=tan⁡−1(x+y1−xy)\tan^{-1} x + \tan^{-1} y = \tan^{-1} \left( \frac{x + y}{1 - xy} \right)tan−1x+tan−1y=tan−1(1−xyx+y​), for xy<1xy < 1xy<1.

  3. Find the derivative of y=tan⁡−1(1+x2−1)y = \tan^{-1} (\sqrt{1 + x^2} - 1)y=tan−1(1+x2​−1).