Chapter 2: Inverse Trigonometric Functions

Chapter-2

Inverse Trigonometric functions

Basic concepts , Domain and Range ,Principal Branch:

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

•  should not be confused with . In fact   
• i.e.,  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

• 

• 

Graphs of inverse trigonometric functions

 

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