1. Basic concepts , Domain and Range ,Principal Branch

It is a long established fact that a reader will be distracted by the readable content of a page when looking at its layout. The point of using Lorem Ipsum is that it has a more-or-less normal distribution of letters, as opposed to using 'Content here, content here', making it look like readable English. Many desktop publishing packages and web page editors now use Lorem Ipsum as their default model text, and a search for 'lorem ipsum' will uncover many web sites still in their infancy. Various versions have evolved over the years, sometimes by accident, sometimes on purpose (injected humour and the like).

1. Basic concepts , Domain and Range ,Principal Branch

sudhanshu@kaysonseducation.co

Create professional doodle videos in minutes. No tech or design skills needed.

© Bryxen, Inc. All rights reserved.
Third-party components are the property of their respective owners.

1. Basic concepts , Domain and Range ,Principal Branch

Chapter-2

Inverse Trigonometric functions

Basic concepts , Domain and Range ,Principal Branch:

Let y = sinx        =>       x = sin-1y

  • https://media-mycbseguide.s3.amazonaws.com/images/static/revise/12/mathematics/ch02/image012.png should not be confused with . In fact   
  • i.e., https://media-mycbseguide.s3.amazonaws.com/images/static/revise/12/mathematics/ch02/image012.png ¹   
  • And similarly for other trigonometric functions.
  • The value of an inverse trigonometric functions which lies in its principal value branch is called the principal value of that inverse trigonometric functions.
  • The domains and ranges (principal value branches) of inverse trigonometric functions are given in the following table:
  • The functions

sin−1x & tan−1x

 are increasing functions in their domain.

  • The functions

cos−1x & cot−1x

 are decreasing functions in over domain.

For suitable values of domain, we have

• 

• 

2. Graphs of inverse trigonometric functions

Graphs of inverse trigonometric functions

 

3. Properties of inverse trigonometric functions and important results

Properties of inverse trigonometric functions and important results

Properties:

[1]

  • cos(cos-1x)=x
  • tan(tan-1x)=x
  • cot(cot-1x)=x
  • cosec(cosec-1x)=x
  • sec(sec-1x)=x
  • Cos-1(cosx)=x
  • tan-1(tanx)=x
  • cot-1 (cotx)=x
  • cosec-1 (cosecx)=x
  • sec-1 (secx)=x

[2]

[3]

  • sin–1 (–x) = – sin–1 x
  • tan–1 (–x) = – tan–1 x

[4]

[5]

  • tan–1 x + cot–1 x = π/2
  • sin–1 x + cos–1 x = π/2

[6]

Examples:

1. Find the domain of definition of f(x) = cos -1 (x2 – 4)

Solution:

Given f(x) = cos -1 (x2 – 4)

We know that domain of cos-1 (x2 – 4) lies in the interval [-1, 1]

Therefore, we can write as

-1 ≤ x2 – 4 ≤ 1

4 – 1 ≤ x2 ≤ 1 + 4

3 ≤ x2 ≤ 5

±√ 3 ≤ x ≤ ±√5

– √5 ≤ x ≤ – √3 and √3 ≤ x ≤ √5

Therefore domain of cos-1 (x2 – 4) is [- √5, – √3] [√3, √5]

2. Find the domain of f(x) = cos-1 2x + sin-1 x.

Solution:

Given that f(x) = cos-1 2x + sin-1 x.

Now we have to find the domain of f(x),

We know that domain of cos-1 x lies in the interval [-1, 1]

Also know that domain of sin-1 x lies in the interval [-1, 1]

Therefore, the domain of cos-1 (2x) lies in the interval [-1, 1]

Hence we can write as,

-1 ≤ 2x ≤ 1

– ½ ≤ x ≤ ½

Hence, domain of cos-1(2x) + sin-1 x lies in the interval [- ½, ½]

3. Find the principal value of each of the following:

(i) sec-1 (-√2)

(ii) sec-1 (2)

(iii) sec-1 (2 sin (3π/4))

(iv) sec-1 (2 tan (3π/4))

Solution:

(i) Given sec-1 (-√2)

Now let y = sec-1 (-√2)

Sec y = -√2

We know that sec π/4 = √2

Therefore, -sec (π/4) = -√2

= sec (π – π/4)

= sec (3π/4)

Thus the range of principal value of sec-1 is [0, π] – {π/2}

And sec (3π/4) = – √2

Hence the principal value of sec-1 (-√2) is 3π/4

(ii) Given sec-1 (2)

Let y = sec-1 (2)

Sec y = 2

= Sec π/3

Therefore the range of principal value of sec-1 is [0, π] – {π/2} and sec π/3 = 2

Thus the principal value of sec-1 (2) is π/3

(iii) Given sec-1 (2 sin (3π/4))

But we know that sin (3π/4) = 1/√2

Therefore 2 sin (3π/4) = 2 × 1/√2

2 sin (3π/4) = √2

Therefore by substituting above values in sec-1 (2 sin (3π/4)), we get

Sec-1 (√2)

Let Sec-1 (√2) = y

Sec y = √2

Sec (π/4) = √2

Therefore range of principal value of sec-1 is [0, π] – {π/2} and sec (π/4) = √2

Thus the principal value of sec-1 (2 sin (3π/4)) is π/4.

(iv) Given sec-1 (2 tan (3π/4))

But we know that tan (3π/4) = -1

Therefore, 2 tan (3π/4) = 2 × -1

2 tan (3π/4) = -2

By substituting these values in sec-1 (2 tan (3π/4)), we get

Sec-1 (-2)

Now let y = Sec-1 (-2)

Sec y = – 2

– sec (π/3) = -2

= sec (π – π/3)

= sec (2π/3)

Therefore the range of principal value of sec-1 is [0, π] – {π/2} and sec (2π/3) = -2

Thus, the principal value of sec-1 (2 tan (3π/4)) is (2π/3).

4. Evaluate:

(i) Cot (sin-1 (3/4) + sec-1 (4/3))

(ii) Sin (tan-1 x + tan-1 1/x) for x < 0

Solution:

(i) Given Cot (sin-1 (3/4) + sec-1 (4/3))

(ii) Given Sin (tan-1 x + tan-1 1/x) for x < 0

5. If cos-1 (x/2) + cos-1 (y/3) = α, then prove that 9x2 – 12xy cos α + 4y2 = 36 sin2 α

Solution:

Given cos-1 (x/2) + cos-1 (y/3) = α

6. If sin-1 (2a/1 + a2) – cos-1(1 – b2/1 + b2) = tan-1(2x/1 – x2), then prove that x = (a – b)/ (1 + a b)

Solution:

Given sin-1 (2a/1 + a2) – cos-1(1 – b2/1 + b2) = tan-1(2x/1 – x2)

7. If sin-1 x + sin-1 y = π/3 and cos-1 x – cos-1 y = π/6, find the values of x and y.

Solution:

Given sin-1 x + sin-1 y = π/3 ……. Equation (i)

And cos-1 x – cos-1 y = π/6 ……… Equation (ii)

8. Which of the following is the principal value branch of cos-1 x?

(a) [-π/2, π/2] (b) (0, π) (c) [0. π] (d) [0, π] – {π/2}

Solution:

(c) [0. π]

As we know that the principal value branch cos-1 x is [0, π].

9. Which of the following is the principal value branch of cosec-1 x?

(a) (-π/2, π/2) (b) [0, π] – {π/2} (c) [-π/2, π/2] (d) [-π/2, π/2] – {0}

Solution:

(d) [-π/2, π/2] – {0}

As the principal branch of cosec-1 x is [-π/2, π/2] – {0}.

10. If 3 tan-1 x + cot-1 x = π, then x equals

(a) 0 (b) 1 (c) -1 (d) ½

Solution:

(b) 1

Given, 3 tan-1 x + cot-1 x = π

2 tan-1 x + tan-1 x + cot-1 x = π

2 tan-1 x + π/2 = π (As tan-1 + cot-1 = π/2)

2 tan-1 x = π/2

tan-1 x = π/4

x = 1

11. The value of sin-1 cos 33π/5 is

(a) 3π/5 (b) -7π/5 (c) π/10 (d) -π/10

Solution:

(d) -π/10