Circular Representation of Inverse Trigonometric Functions

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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

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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

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  • 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.

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CBSE CLASS XII Related Questions

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    The motion of a particle is described as \( S = 2t^{3} - 6t^{2} + 8 \). Its velocity \( \frac{dS}{dt} \), when the acceleration \( \frac{d^{2}S}{dt^{2}} \) vanishes is :

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                    CBSE CLASS XII Previous Year Papers

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