Inverse Trigonometric Identities

Inverse Trigonometric Identities: In mathematics, inverse trigonometric functions are also known as arcus functions or anti-trigonometric functions. The inverse trigonometric functions are the inverse functions of basic trigonometric functions, i.e., sine, cosine, tangent, cosecant, secant, and cotangent. It is used to find the angles with any trigonometric ratio. Inverse trigonometric functions are generally used in fields like geometry, engineering, etc. The representation of inverse trigonometric functions are:

If a = f(b), then the inverse function is

 b = f^-1(a) 

Examples of inverse inverse trigonometric functions are sin^-1x, cos^-1x, tan^-1x, etc.

The following table shows some trigonometric functions with their domain and range.

The following are the properties of inverse trigonometric functions:

Property 1:

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

Property 2:

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

Property 3

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

Property 4

1. sin^-1 x + cos^-1 x = π/2, for x ∈ [-1,1]
2. tan^-1 x + cot^-1 x = π/2, for x ∈ R
3. cosec^-1 x + sec^-1 x = π/2 , for |x| ≥ 1

Property 5

1. tan^-1 x + tan^-1 y = tan^-1 ( x + y )/(1 – xy), for xy < 1
2. tan^-1 x – tan^-1 y = tan^-1 (x – y)/(1 + xy), for xy > -1
3. tan^-1 x + tan^-1 y = π + tan^-1 (x + y)/(1 – xy), for xy >1 ; x, y >0

Property 6

1. 2tan^-1 x = sin^-1 (2x)/(1 + x²), for |x| ≤ 1
2. 2tan^-1 x = cos^-1 (1 – x²)/(1 + x²), for x ≥ 0
3. 2tan^-1 x = tan^-1 (2x)/(1 – x²), for -1 < x <1

The following are the identities of inverse trigonometric functions:

1. sin^-1 (sin x) = x provided -π/2 ≤ x ≤ π/2
2. cos^-1 (cos x) = x provided 0 ≤ x ≤ π
3. tan^-1 (tan x) = x provided -π/2 < x < π/2
4. sin(sin^-1 x) = x provided -1 ≤ x ≤ 1
5. cos(cos^-1 x) = x provided -1 ≤ x ≤ 1
6. tan(tan^-1 x) = x provided x ∈ R
7. cosec(cosec^-1 x) = x provided -1 ≤ x ≤ ∞ or -∞ < x ≤ 1
8. sec(sec^-1 x) = x provided 1 ≤ x ≤ ∞ or -∞ < x ≤ 1
9. cot(cot^-1 x) = x provided -∞ < x < ∞
10. [Tex]sin^{-1}(\frac{2x}{1 + x^2}) = 2 tan^{-1}x[/Tex]
11. [Tex]cos^{-1}(\frac{1 – x^2}{1 + x^2}) = 2 tan^{-1}x[/Tex]
12. [Tex]tan^{-1}(\frac{2x}{1 – x^2}) = 2 tan^{-1}x[/Tex]
13. 2cos^-1 x = cos^-1 (2x² – 1)
14. 2sin^-1x = sin^-1 2x√(1 – x²)
15. 3sin^-1x = sin^-1(3x – 4x³)
16. 3cos^-1 x = cos^-1 (4x³ – 3x)
17. 3tan^-1x = tan^-1((3x – x³/1 – 3x²))
18. sin^-1x + sin^-1y = sin^-1{ x√(1 – y²) + y√(1 – x²)}
19. sin^-1x – sin^-1y = sin^-1{ x√(1 – y²) – y√(1 – x²)}
20. cos^-1 x + cos^-1 y = cos^-1 [xy – √{(1 – x²)(1 – y²)}]
21. cos^-1 x – cos^-1 y = cos^-1 [xy + √{(1 – x²)(1 – y²)}
22. tan^-1 x + tan^-1 y = tan^-1(x + y/1 – xy)
23. tan^-1 x – tan^-1 y = tan^-1(x – y/1 + xy)
24. tan^-1 x + tan^-1 y +tan^-1 z = tan^-1 (x + y + z – xyz)/(1 – xy – yz – zx)

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Question 1: Prove sin^-1 x = sec^-1 1/√(1-x²)

Solution: 

Let sin^-1 x = y

⇒ sin y = x , (since sin y = perpendicular/hypotenuse ⇒ cos y = √(1- perpendicular² )/hypotenuse )

⇒ cos y = √(1 – x²), here hypotenuse = 1

⇒ sec y = 1/cos y

⇒ sec y = 1/√(1 – x²)

⇒ y = sec^-1 1/√(1 – x²)

⇒ sin^-1 x = sec^-1 1/√(1 – x²)

Hence, proved.

Question 2: Prove tan^-1 x = cosec^-1 √(1 + x²)/x

Solution:

Let tan^-1 x = y

⇒ tan y = x , perpendicular = x and base = 1

⇒ sin y = x/√(x² + 1) , (since hypotenuse = √(perpendicular² + base² ) )

⇒ cosec y = 1/sin y

⇒ cosec y = √(x² + 1)/x

⇒ y = cosec^-1 √(x² + 1)/x

⇒ tan^-1 x = cosec^-1 √(x² + 1)/x

Hence, proved.

Question 3: Evaluate tan(cos^-1 x)

Solution: 

Let cos^-1 x = y

⇒ cos y = x , base = x and hypotenuse = 1 therefore sin y = √(1 – x²)/1

⇒ tan y = sin y/ cos y

⇒ tan y = √(1 – x²)/x

⇒ y = tan^-1 √(1 – x²)/x

⇒ cos^-1 x = tan^-1 √(1 – x²)/x

Therefore, tan(cos^-1 x) = tan(tan^-1 √(1 – x²)/x ) = √(1 – x²)/x.

Question 4: tan^-1 √(sin x) + cot^-1 √(sin x) = y. Find cos y.

Solution: 

We know that tan^-1 x + cot^-1 x = /2 therefore comparing this identity with the equation given in the question we get y = π/2

Thus, cos y = cos π/2 = 0.

Question 5: tan^-1 (1 – x)/(1 + x) = (1/2)tan^-1 x, x > 0. Solve for x.

Solution: 

tan^-1 (1 – x)/(1 + x) = (1/2)tan^-1 x

⇒ 2tan^-1 (1 – x)/(1 + x) = tan^-1 x     …(1)

We know that, 2tan^-1 x = tan^-1 2x/(1 – x²).

Therefore, LHS of equation (1) can be written as

tan^-1 [ { 2(1 – x)/(1 + x)}/{ 1 – [(1 – x)(1 + x)]²}]

= tan^-1 [ {2(1 – x)(1 + x)} / { (1 + x)² – (1 – x)² }]

= tan^-1 [ 2(1 – x²)/(4x)]

= tan^-1 (1 – x²)/(2x)

Since, LHS = RHS therefore

tan^-1 (1 – x²)/(2x) = tan^-1 x

⇒ (1 – x²)/2x = x

⇒ 1 – x² = 2x²

⇒ 3x² = 1

⇒ x = ± 1/√3

Since, x must be greater than 0 therefore x = 1/√3 is the acceptable answer.

Question 6: Prove tan^-1 √x = (1/2)cos^-1 (1 – x)/(1 + x)

Solution: 

Let tan^-1 √x = y

⇒ tan y = √x

⇒ tan² y = x

Therefore,

RHS = (1/2)cos^-1 ( 1- tan² y)/(1 + tan² y)

= (1/2)cos^-1 (cos² y – sin² y)/(cos² y + sin² y)

= (1/2)cos^-1 (cos² y – sin² y)

= (1/2)cos^-1 (cos 2y)

= (1/2)(2y)

= y

= tan^-1 √x

= LHS

Hence, proved.

Question 7: tan^-1 (2x)/(1 – x²) + cot^-1 (1 – x²)/(2x) = π/2, -1 < x < 1. Solve for x.

Solutions: 

tan^-1 (2x)/(1 – x²) + cot^-1 (1 – x²)/(2x) = π/2

⇒ tan^-1 (2x)/(1 – x²) + tan^-1 (2x)/(1 – x²) = π/2

⇒ 2tan^-1 (2x)/(1 – x²) = ∏/2

⇒ tan^-1 (2x)/(1 – x²) = ∏/4

⇒ (2x)/(1 – x²) = tan ∏/4

⇒ (2x)/(1 – x²) = 1

⇒ 2x = 1 – x²

⇒ x² + 2x -1 = 0

⇒ x = [-2 ± √(2² – 4(1)(-1))] / 2

⇒ x = [-2 ± √8] / 2

⇒ x = -1 ± √2

⇒ x = -1 + √2 or x = -1 – √2

But according to the question x ∈ (-1, 1) therefore for the given equation the solution set is x ∈ ∅.

Question 8: tan^-1 1/(1 + 1.2) + tan^-1 1/(1 + 2.3) + … + tan^-1 1/(1 + n(n + 1)) = tan^-1 x. Solve for x.

Solution:  

tan^-1 1/(1 + 1.2) + tan^-1 1/(1 + 2.3) + … + tan^-1 1/(1 + n(n + 1)) = tan^-1 x  

⇒ tan^-1 (2 – 1)/(1 + 1.2) + tan^-1 (3 – 2)/(1 + 2.3) + … + tan^-1 (n + 1 – n)/(1 + n(n + 1)) = tan^-1 x

⇒ (tan^-1 2 – tan^-1 1) + (tan^-1 3 – tan^-1 2) + … + (tan^-1 (n + 1) – tan^-1 n) = tan^-1 x

⇒ tan^-1 (n + 1) – tan^-1 1 = tan^-1 x

⇒ tan^-1 n/(1 + (n + 1).1) = tan^-1 x

⇒ tan^-1 n/(n + 2) = tan^-1 x

⇒ x = n/(n + 2)

Question 9: If 2tan^-1 (sin x) = tan^-1 (2sec x) then solve for x.

Solution: 

2tan^-1 (sin x) = tan^-1 (2sec x)

⇒ tan^-1 (2sin x)/(1 – sin² x) = tan^-1 (2/cos x)

⇒ (2sin x)/(1 – sin² x) = 2/cos x

⇒ sin x/cos² x = 1/cos x

⇒ sin x cos x = cos² x

⇒ sin x cos x – cos² x = 0

⇒ cos x(sin x – cos x) = 0

⇒ cos x = 0 or sin x – cos x = 0

⇒ cos x = cos π/2 or tan x = tan π/4

⇒ x = π/2 or x = π/4

But at x = π/2 the given equation does not exist hence x = π/4 is the only solution.

Question 10: Prove that cot^-1 [ {√(1 + sin x) + √(1 – sin x)}/{√(1 + sin x) – √(1 – sin x)}] = x/2, x ∈ (0, π/4)

Solution: 

Let x = 2y therefore

LHS = cot^-1 [{√(1+sin 2y) + √(1-sin 2y)}/{√(1+sin 2y) – √(1-sin 2y)}]

= cot^-1 [{√(cos² y + sin² y + 2sin y cos y) + √(cos² y + sin² y – 2sin y cos y)}/{√(cos² y + sin² y + 2sin y cos y) – √(cos² y + sin² y – 2sin y cos y)} ] 

= cot^-1 [{√(cos y + sin y)² + √(cos y – sin y)²} / {√(cos y + sin y)² – √(cos y – sin y)²}] 

= cot^-1 [( cos y + sin y + cos y – sin y )/(cos y + sin y – cos y + sin y)] 

= cot^-1 (2cos y)/(2sin y)

= cot^-1 (cot y)

= y

= x/2.

Problem 1: Solve for x in the equation sin^-1(x) + cos^-1(x) = π/2

Problem 2: Prove that tan^-1(1) + tan^-1(2) + tan^-1(3) = π

Problem 3: Evaluate cos⁡(sin^-1(0.5))

Problem 4: If tan^-1(x) + tan^-1(2x) = π/4, then find x

What are inverse trigonometric functions?

Inverse trigonometric functions are the inverse functions of the basic trigonometric functions (sine, cosine, tangent, cosecant, secant, and cotangent). They are used to find the angles corresponding to given trigonometric ratios.

Why are inverse trigonometric functions important?

Inverse trigonometric functions are essential in various fields like geometry, engineering, and physics because they help determine angles from trigonometric ratios, which is crucial for solving many practical problems.

What are the domains and ranges of inverse trigonometric functions?

Each inverse trigonometric function has specific domains and ranges:

sin^-1(x) : Domain [-1, 1] and Range [- π/2, π/2]

cos^-1(x) : Domain [-1, 1] and Range [ 0, π]

tan⁡^-1(x) : Domain R and Range (- π/2, π/2)

Can inverse trigonometric functions be used in calculus?

Yes, inverse trigonometric functions are frequently used in calculus for integration and differentiation. They are particularly useful for integrating functions that involve trigonometric expressions.