Class 12 RD Sharma Solutions Chapter 4 Inverse Trigonometric Functions Exercise 4.1 (original) (raw)

Last Updated : 23 Jul, 2025

The Inverse trigonometric functions are critical for the solving equations where the angle is unknown and they help in the understanding of the various trigonometric properties. These functions, denoted as sin−1(x), cos−1(x) and tan−1(x) allow us to find the angle whose trigonometric function gives a specific value. They are the reverse of the primary trigonometric functions and play a vital role in the calculus and algebra.

Inverse Trigonometric Functions

The Inverse trigonometric functions are defined as the inverses of the basic trigonometric functions—sine, cosine and tangent. Specifically:

sin−1(x) or arcsine gives the angle whose sine is x where x is within [−1,1].
cos−1(x) or arccosine provides the angle whose cosine is x where x is within [−1,1].
tan−1(x) or arctangent gives the angle whose tangent is x with the x being any real number.

Question 1. Find the principal value of each of the following :

(i) Sin -1 (- √3/2)

****(ii) Sin** -1 (cos 2π/3)

****(iii)Sin** -1 (-√3 - 1/2√2)

(iv) Sin -1 (√3 + 1/2√2)

(v) Sin -1 (cos 3π/4)

(vi) Sin -1 (tan 5π/4)

Solution:

****(i)** **Sin -1 (-√3/ 2)

= Sin-1 [sin (- π / 3)]

= - π / 3 Ans.

(ii) Sin -1 (cos 2π / 3)

= Sin-1 (- 1 / 2)

= Sin-1 (- π / 6)

= - π / 6 Ans.

(iii) Sin -1 (√3 - 1/2√2)

= Sin-1 ( sin π / 12 )

= π / 12 Ans.

(iv) Sin -1 (√3 + 1/2 √2)

= Sin-1 (5π / 12)

= 5π / 12 Ans.

****(v)** Sin-1 (cos 3π / 4)

= Sin-1 (- √2 / 2)

= [Sin-1 (- π / 4)]

= - π / 4 Ans.

****(vi)** Sin-1 (tan 5π / 4)

= Sin-1 (1)

= Sin-1 [sin (π / 2)]

= π / 2 Ans.

Question 2.

****(i) Sin** -1 1/2 -2 Sin -1 1/√2

(ii) Sin -1 {cos (Sin -1 √3 / 2)}

Solution:

****(i)** Sin-1 1/2 -2 Sin-1 1/ √2

= Sin-1 1/2 - Sin-1 [ 2 x 1/ √2 √1- (1 /√2)2 ]

= Sin-1 1/2 - Sin-1 (1)

= π/6 - π /2

= π / 3 Ans.

****(ii)** Sin-1 { cos ( Sin-1 sin π / 3 )}

= Sin-1 { cos ( π / 3 ) }

= Sin-1 { 1/2 }

= Sin-1 { sin π / 6 }

= π / 6 Ans.

Question 3. Find the domain of each of the following functions :

****(i) f(x) = Sin** -1 x2

****(ii) f(x) = Sin** -1 x + sinx

****(iii) f(x) = Sin** -1 √x 2 - 1

****(iv) f(x) = Sin** -1 x + Sin-1 2x

**Solution:

****(i)** Domain ofSin-1 lies between the interval [ -1 , 1 ]

and x2 ∈ [ 0 , 1 ] as x2 can not be negative .

So, x ∈ [ -1 , 1 ]

Hence, the domain of the function f(x) = [ -1 , 1 ] Ans.

****(ii)** Let f(x) = g(x) + h(x) , where g(x) = Sin-1x and h(x) = sinx respectively.

Therefore , the domain of f(x) is given by the intersection of the domain g(x) & h(x) .

The domain of g(x) = [ -1 , 1 ]

The domain of h(x) = [ - ∞ , ∞ ]

Thus, the interaction of g(x) and h(x) is [ -1 , 1 ]

Hence , the domain of f(x) is [ -1 , 1 ] Ans.

****(iii)** As we know , the domain of Sin-1 x is [ -1 , 1 ]

Therefore , domain of Sin-1 √x2 - 1 will also lies in the interval [ -1 , 1 ]

:. x2 - 1 ∈ [ 0 , 1 ] as square root cannot be negative .

=> x2 ∈ [ 1 , 2 ]

=> x ∈ [ - √2 , -1 ] U [ 1 , √2 ]

Hence, the domain of function f(x) = [ - √2 , -1 ] U [ 1 , √2 ] Ans.

****(iv)** Let f(x) = g(x) + h(x), where g(x) = Sin-1 x x and h(x) = Sin-1 2x

Therefore, the domain of f(x) will be given by the intersection of g(x) and h(x) .

the domain of g(x) = [ -1 , 1 ]

lly , the domain of h(x) = [ -1/2 , 1/2 ]

g(x) ∩ h(x) = [ -1 , 1 ] ∩ [ -1/2 , 1/2 ]

Hence, the domain of the function f(x) = [- 1/ 2 , 1/2 ] Ans.

Question 4. If sin-1x + sin-1y + sin-1z + sin-1t = 2π, then find the value of x2 + y2 + z2 + t2 .

**Solution:

As we already know, **Range of sin-1 is [ - π / 2 , π / 2 ]

Given: (sin-1x) + (sin-1y) +(sin-1y)+(sin-1t) = 2 π

So, each takes the value of **π / 2

****:. x = 1, y = 1, z = 1 & t = 1**

Hence, x2 + y2+ z2 + t2 = 1+ 1 + 1 + 1 = 4 Ans.

Question 5. If (sin-1x)2 + ( sin-1y )2 + ( sin-1y )2 = 3π2/4, find the value of x2 + y2 + z2 .

**Solution:

As we already know , Range of sin-1 is [ - π / 2 , π / 2 ]

Given: ( sin-1x )2 + ( sin-1y )2 + ( sin-1y )2 = 3π2/4

****:.** each takes the value of **π / 2

**x = 1, y = 1 & z = 1 .

Hence , x2 + y2 + z2 = 1 + 1 + 1 = 3

Summary

Chapter 4 focuses on inverse trigonometric functions, also known as arcfunctions or antifunctions. The main topics covered in Exercise 4.1 typically include:

Definition and notation of inverse trigonometric functions
Domains and ranges of inverse trigonometric functions
Principal values of inverse trigonometric functions
Relationships between inverse trigonometric functions
Properties of inverse trigonometric functions
Solving equations involving inverse trigonometric functions

The chapter emphasizes understanding the behavior of these functions, their graphical representations, and how they relate to the standard trigonometric functions.