Circular Representation of Inverse Trigonometric Functions

Collegedunia Team logo

Collegedunia Team

Content Curator

Consider a function f. The inverse of this function f exists only if f is one to one function. It is given by

f(x)=y and f-1(y)=x

For the sine function sin: R→R. For all θ € R the sinθ = x is a one-to-one function. Here for the domain R the sine function doesn’t have an inverse. But when the domain is considered in [-π/2, π/2] there are infinite values for various angles of θ. In this domain, the equation sinθ = x can be satisfied. From the equation it can be written that θ = sin-1(x).

Read Also: Inverse Trigonometric Formulas

Key Terms: Single-valued function, multiple-valued function, one to one function, sine function, properties of inverse trigonometric function


Circular Representation of Inverse Trigonometric Functions

[Click Here for Sample Questions]

The most important thing to be noted here is that is θ is an angle and it represents the value sin-1(x). In short, θ is the angle and x is the value of the arcsine function of the angle. For different angles, the arcsine function has different values ranging from -1 to 1 and the angle between -π/2 and π/2. So, sin-1(x) is a multiple-valued function.

Read More: Class 10 Introduction to Trigonometry

But the sine function has a certain value for a certain angle. For every sin θ there is a value x. So, sinθ is a single-valued function. All real numbers are the domain of sine function and the range is [-1,1].

sin-1(x)= θ

x=sin θ

Where, -π/2 ≤ x ≤ π/2 and -1 ≤ x ≤ 1

The sine function is considered as one to one function during the intervals -π/2 and π/2. But other intervals like [3π/2,5π/2], [-5π/2, -3π/2] also define the inverse of sine function. Hence sin−1x is a function with domain [-1, 1] = {x ∈ R: - 1 ≤ x ≤ 1} and range [- π2, π2] or [3π2, 5π2] or [- 5π2, -3π2] and etc.,

Read More: Trigonometric Identities

In the same way, cos θ=x(-1≤x≤1) then θ=cos-1(x). Here θ is the angle and the value of cosine function of this angle is x. For all the tan-1x, cot-1x, sec-1x, cosec-1x there is an angle θ and the value of inverse functions of these angles is x.

Also Read:


Properties

[Click Here for Sample Questions]

  • If sin θ=x (-1≤x≤1) then θ=sin-1(x).
  • If cos θ=x (-1≤x≤1) then θ=cos-1(x).
  • If tan θ=x (-∞<x </x) then θ=tan-1x).
  • If cot θ=x (-∞<x </x) then θ=cot-1(x).
  • If cosec θ=x (|x|≥ 1) then θ=cosec-1(x).
  • If sec θ=x (|x|≥ 1) then θ=sec-1(x)
Vice-versa
  • sin-1(x)=θ then sin θ=x
  • cos-1(x)= θ then cos θ=x
  • tan-1(x)= θ then tan θ=x
  • cot-1(x)= θ then cot θ=x
  • cosec-1(x)= θ then cosec θ=x
  • sec-1(x)= θ then sec θ=x

Also Read:


Sample Questions

Ques. Evaluate sin-1(-√3/2). (3 marks)

Ans. θ = sin-1(-√3/2)

sin θ = -√3/2

We need to find the angle between -π/2 and π/2 that equals to -√3/2.

Sin(-π/3) = -√3/2

Hence θ = sin-1(-√3/2) = -π/3

Ques. Evaluate arc tan(- 1 ) (3 marks)

Ans. Let y = arc tan(- 1 )

tan y = - 1 with - π / 2 < y < π / 2

We know that tan (π / 4) = 1.

And tan(- x) = - tan x. So

tan (-π / 4) = - 1

Compare the last statement with tan y = - 1 to obtain

y = - π/4

Ques. Simplify cos(arc sin x ) (3 marks)

Ans. Let z = cos ( arc sin x )

y = arc sin x

So that z = cos y

y = arc sin x may also be written as

sin y = x with - π / 2 ≤ y ≤ π / 2

Also

sin2y + cos2y = 1

cos y = + or - √ (1 - x2)

But - π / 2 ≤ y ≤ π / 2 so that cos y is positive

z = cos y = cos(arc sin x) = √ (1 - x 2)

Ques. arc sin(sin 7π/4) (3 marks)

Ans. arc sin(sin ( y ) ) = y only for - π / 2 ≤ y ≤ π /2

sin (7 π / 4) = sin (-π / 4) as follows

arc sin(sin (7 π / 4)) = arc sin( sin (- π / 4))- π / 4 at - π / 2 ≤ y ≤ π / 2.

Hence

arc sin(sin (7 π / 4)) = - π / 4

Ques. Find the values of sec csc−1 (2/√3) (3 marks)

Ans. sec csc−1 (2/√3)

= sec csc−1 (csc π/3)

= sec (csc−1csc π/3)

= sec π/2

= 2

Therefore, sec csc−1 (2/√3) = 2

Ques. Find the principal value of sin-1(2) if it exists. (2 marks)

Ans. The domain of y=sin−1(x) is [-1,1]

And 2 doesn't belong to this domain.

So, sin−1(2) doesn't exist.

For Latest Updates on Upcoming Board Exams, Click Here: https://t.me/class_10_12_board_updates


Check-Out: 

CBSE CLASS XII Related Questions

  • 1.
    A function \[ f:\mathbb{R}-\left\{\frac{3}{5}\right\} \to \mathbb{R}-\left\{\frac{3}{5}\right\} \] is defined as \[ f(x)=\frac{3x+2}{5x-3}. \] Show that \(f\) is one-one and onto.


      • 2.
        Assertion (A) : In an experiment of throwing an unbiased die, the probability of getting a prime number given that number appearing on the die being odd is \( \frac{2}{3} \).
        Reason (R) : For any two events \( A \) and \( B \), \( P(A|B) = \frac{P(A \cup B)}{P(B)} \).

          • Both Assertion (A) and Reason (R) are true and the Reason (R) is the correct explanation of the Assertion (A).
          • Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
          • Assertion (A) is true and Reason (R) is false.
          • Assertion (A) is false and Reason (R) is true.

        • 3.

          The domain of \[ f(x)=\cos^{-1}(2x-5) \] is: 

            • \([-1, 1]\)
            • \([4, 6]\)
            • \([-7, -3]\)
            • \([2, 3]\)

          • 4.

            Check whether \[ f:\mathbb{R}-\{3\}\rightarrow\mathbb{R} \] defined as \[ f(x)=\frac{x-2}{x-3} \] is onto or not. 


              • 5.
                Find: \[ \int \frac{x^2}{(x^2+9)(x^2+16)}\,dx \]


                  • 6.
                    For \[ f(x)=x+\frac{1}{x}, \quad x\neq 0. \]

                      • local maximum value is 2
                      • local minimum value is \( -2 \)
                      • local maximum value is \( -2 \)
                      • local minimum value \( < \) local maximum value
                    CBSE CLASS XII Previous Year Papers

                    Comments


                    No Comments To Show