Basics Concepts: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced PDF Download

What is an Inverse Function?

• A function accepts values, performs particular operations on these values, and generates an output. The inverse function agrees with the resultant, operates, and reaches back to the original function.
• If y=f(x) and x=g(y) are two functions such that f (g(y)) = y and g (f(y)) = x, then f and y are said to be inverse of each other.
  Example: If f(x) = 2x + 5 = y, then, g(y) = (y-5)/2 = x is the inverse of f(x).
• The inverse of f is denoted by f ^-1

Inverse Trigonometric Functions

• Trigonometric functions are many-one functions but we know that inverse of function exists if the function is bijective. 
• If we restrict the domain of trigonometric functions, then these functions become bijective and the inverse of trigonometric functions are defined within the restricted domain. 
• Example: y = f(x) = sin x, then its inverse is x = sin-1 y.

Inverse Trigonometric Formulas

Fig: Inverse Trigonometric Functions

The inverse trigonometric functions are the inverse functions of the trigonometric functions written as cos^-1 x, sin^-1 x, tan^-1 x, cot^-1 x, cosec^-1 x, sec^-1 x.

The inverse trigonometric functions are multivalued. For example, there are multiple values of ω such that z = sinω, so sin^-1z is not uniquely defined unless a principal value is defined. 

Such principal values are sometimes denoted with a capital letter so, for example, the principal value of the inverse sine may be variously denoted sin^-1z or arcsinz.

Let’s say, if y = sin x , then x = sin^-1 y, similarly for other trigonometric functions. This is one of the inverse trigonometric formulas. Now, y = sin^-1 (x), y ∈ [π/2 , π/2] and x ∈ [-1,1].

• Thus, sin^-1 x has infinitely many values for given x ∈ [-1, 1].
• There is only one value among these values which lies in the interval [π/2, π/2]. This value is called the principal value.

Domain and Range of Inverse Trigonometric Formulas

Solved Examples

Ques 1: Find the exact value of each expression without a calculator, in [0,2π).

1. sin^-1(−3√2)
2. cos^-1(−2√2)
3. tan^-1√3

Ans:

• Recall that −3√2 is from the 30−60−90 triangle. The reference angle for sin and 3√2 would be 60∘. Because this is sine and it is negative, it must be in the third or fourth quadrant. The answer is either 4π/3 or 5π/3.
• −2√2is from an isosceles right triangle. The reference angle is then 45∘. Because this is cosine and negative, the angle must be in either the second or third quadrant. The answer is either 3π/4 or 5π/4.
• √3is also from a 30−60−90 triangle. Tangent is √3 for the reference angle 60∘. Tangent is positive in the first and third quadrants, so the answer would be π/3 or 4π/3.

Notice how each one of these examples yields two answers. This poses a problem when finding a singular inverse for each of the trig functions. Therefore, we need to restrict the domain in which the inverses can be found.

Ques 2: Find the value of tan^-1(1.1106).
Ans: Let A=tan⁻¹(1.1106)
Then, tanA = 1.1106
A = 48°
tan48 = 1.1106
[Use calculator in degree mode]
tan⁻¹ 1.1106=48°

Properties of Inverse

Here are the properties of the inverse trigonometric functions with proof.

Property 1

1. sin^-1 (1/x) = cosec^-1x , x ≥ 1 or x ≤ -1
2. cos^-1 (1/x) = sec^-1x , x ≥ 1 or x ≤ -1
3. tan^-1 (1/x) = cot^-1x , x > 0

Proof : sin^-1 (1/x) = cosec^-1x , x ≥ 1 or x ≤ -1,

Let  sin⁻¹x=y

i.e. x = cosec y

1/π = sin y

Property 2

1. sin^-1(-x) = – sin^-1(x),    x ∈ [-1,1]
2. tan^-1(-x) = -tan^-1(x),   x ∈ R
3. cosec^-1(-x) = -cosec^-1(x), |x| ≥ 1

Proof: sin^-1(-x) = -sin^-1(x),    x ∈ [-1,1] 
Let,  Sin^-1(-x) = y

Then −x=sin y

x=−sin y

x=sin(−y)

sin⁻¹=sin⁻¹(sin(−y))

sin⁻¹x=y

sin⁻¹x=−sin⁻¹(−x)

Hence,sin⁻¹(−x)=−sin⁻¹ x ∈ [-1,1]

Property 3

1. cos^-1(-x) = π – cos^-1 x, x ∈ [-1,1]
2. sec^-1(-x) = π – sec^-1x, |x| ≥ 1
3. cot^-1(-x) = π – cot^-1x, x ∈ R

Proof : cos^-1(-x) = π – cos^-1 x, x ∈ [-1,1]

Let cos⁻¹(−x)=y

cos y=−x   x=−cos y

x=cos(π−y)

Since,  cos π−q=−cos q

cos⁻¹ x=π−y

cos⁻¹ x=π–cos−1–x

Hence, cos⁻¹−x=π–cos⁻¹x

Solved Example

Ques 1: Prove that “sin^-1(-x) = – sin^-1(x),    x ∈ [-1,1]”

Ans: Let, sin⁻¹(−x)=y

Then −x=sin y

x=−sin y

x=sin(−y)

sin⁻¹ x=arcsin(sin(−y))

sin⁻¹ x=y

sin⁻¹ x=−sin⁻¹(−x)

Hence, sin⁻¹(−x)=−sin⁻¹ x, x ∈ [-1,1]

This concludes our discussion on the topic of trigonometric inverse functions.
Ques 2: sin^-1(cos π/3)=?
Ans: sin⁻¹  [substitute cos(π/3)=1/2]
= π/6 [substitute sin^-1 (1/2) = π/6]

Ques 3: Find the value of sin (π/4+Cos^-1(√2/2)).
Ans: 
Let y=sin
Then, cosA= √2/2
Multiplying the numerator as well as denominator by √2 we get:

cosA=1/√2
A = π/4
Therefore
y = sin
y = sin (π/2)
hence, y=1.