Inverse trigonometric function

Inverse trigonometric functions are special mathematical functions that help us find the angles when we know the value of a trigonometric ratio. In normal trigonometric functions, like sine, cosine, or tangent, we start with an angle and find the ratio of two sides of a right triangle. For example, if we know an angle is 30°, we can easily find that sin 30° = 1/2. But sometimes, we know the ratio first and need to find the angle. This is where inverse trigonometric functions come in.

For example, if we are told that sin θ = 1/2, the inverse sine function (written as sin⁻¹ or arcsin) tells us that θ = 30°. Similarly, the inverse cosine function (cos⁻¹ or arccos) and inverse tangent function (tan⁻¹ or arctan) help us find angles from cosine and tangent values.

In mathematics, there are six main inverse trigonometric functions:

1. sin⁻¹ x (arcsin x) – inverse of sine
2. cos⁻¹ x (arccos x) – inverse of cosine
3. tan⁻¹ x (arctan x) – inverse of tangent
4. cot⁻¹ x (arccot x) – inverse of cotangent
5. sec⁻¹ x (arcsec x) – inverse of secant
6. cosec⁻¹ x (arccsc x) – inverse of cosecant

These functions are not defined for all values. Each has a domain (the set of input values allowed) and a range (the set of possible outputs). For example, sin⁻¹ x is only defined for values between -1 and 1, and its output is an angle between -90° and 90° (or -π/2 and π/2 in radians).

Inverse trigonometric functions are important in many areas of mathematics and science, including physics, engineering, and navigation. They are used to solve equations, calculate distances, model periodic motion, and even in computer graphics.

A key point to remember is that trigonometric functions are periodic (they repeat their values after certain intervals), so each trigonometric equation can have many possible solutions. But for inverse functions, we restrict the range to a specific interval so that each input has only one output. This makes them functions in the proper mathematical sense.

Do Check: Differential equation

In summary, inverse trigonometric functions help us work backwards from a trigonometric ratio to find an angle. They are essential tools for solving many practical and theoretical problems in mathematics.

Properties of Inverse Trigonomoteric Function

Property i

i) sin⁻¹(sin θ) = θ ; θ ∈ [-π/2, π/2]

ii) cos⁻¹(cos θ) = θ ; θ ∈ [0, π]

iii) tan⁻¹(tan θ) = θ ; θ ∈ (-π/2, π/2)

iv) cot⁻¹(cot θ) = θ ; θ ∈ (0, π)

v) sec⁻¹(sec θ) = θ ; θ ∈ [0, π] - {π/2}

vi) cosec⁻¹(cosec θ) = θ ; θ ∈ [-π/2, π/2] - {0}

Illustration 1: Evaluate the following

i) sin⁻¹(sin π/4)

ii) cos⁻¹(cos 2π/3)

iii) cos⁻¹(cos(-√3/2 + π/6))

iv) tan⁻¹(tan 2π/3)

Solution

i) sin⁻¹(sin π/4) = π/4

ii) cos⁻¹(cos 2π/3) = 2π/3

iii) cos⁻¹(cos(-√3/2 + π/6)) = cos⁻¹(cos 5π/6) = 5π/6

= cos⁻¹(cos(π - π/6)) = π/6

iv) tan⁻¹(tan 2π/3) = tan⁻¹(tan(π - π/3)) = tan⁻¹(-tan π/3) = -π/3

Property ii

i) sin(sin⁻¹x) = x, x ∈ [-1,1]

ii) cos(cos⁻¹x) = x, x ∈ [-1,1]

iii) tan(tan⁻¹x) = x, x ∈ ℝ

iv) cosec(cosec⁻¹x) = x, x ∈ ℝ - (-1,1)

v) sec(sec⁻¹x) = x, x ∈ ℝ - (-1,1)

vi) cot(cot⁻¹x) = x, x ∈ ℝ

Illustration 2

If cos⁻¹x + cos⁻¹y + cos⁻¹z = 3π, then find the value of xy+yz+zx.

Solution

We know 0 ≤ cos⁻¹x ≤ π

Hence, from the given condition

cos⁻¹x = π, cos⁻¹y = π, cos⁻¹z = π

x = -1, y = -1, z = -1

xy + yz + zx = (-1)(-1) + (-1)(-1) + (-1)(-1)

= 1+1+1 = 3

Do Check: Matrix

Property iii

i) sin⁻¹(-x) = -sin⁻¹(x), x ∈ [-1,1]

ii) cos⁻¹(-x) = π - cos⁻¹(x), x ∈ [-1,1]

iii) tan⁻¹(-x) = -tan⁻¹(x), x ∈ ℝ

iv) cosec⁻¹(-x) = -cosec⁻¹x, x ≤ -1 or x ≥ 1

v) sec⁻¹(-x) = π - sec⁻¹x, x ∈ ℝ - (-1,1)

vi) cot⁻¹(-x) = π - cot⁻¹x, x ∈ ℝ

Property iv

i) sin⁻¹(1/x) = cosec⁻¹x, x ∈ (-∞,-1] ∪ [1,∞)

ii) cos⁻¹(1/x) = sec⁻¹x, x ∈ (-∞,-1] ∪ [1,∞)

iii) tan⁻¹(1/x) = cot⁻¹(x), for x > 0

= π + cot⁻¹(x), for x < 0

Property v

i) sin⁻¹x + cos⁻¹x = π/2, x ∈ [-1,1]

ii) tan⁻¹x + cot⁻¹x = π/2, x ∈ ℝ

iii) sec⁻¹x + cosec⁻¹x = π/2, x ∈ ℝ - (-1,1)

Property vi

i) tan⁻¹x + tan⁻¹y = tan⁻¹((x+y)/(1-xy)), if xy < 1

= π + tan⁻¹((x+y)/(1-xy)), if x,y > 0 and xy > 1

= -π + tan⁻¹((x+y)/(1-xy)), if x < 0, y < 0 and xy > 1

ii) tan⁻¹x - tan⁻¹y = tan⁻¹((x-y)/(1+xy)), if xy > -1

= π + tan⁻¹((x-y)/(1+xy)), if x > 0, y < 0

= -π + tan⁻¹((x-y)/(1+xy)), if x < 0, y > 0 and xy < -1

Illustration 3

Show that tan⁻¹(2/11) + cot⁻¹(24/7) = tan⁻¹(1/2)

Solution

Let L.H.S. = tan⁻¹(2/11) + cot⁻¹(24/7)

= tan⁻¹(2/11) + tan⁻¹(7/24) since cot⁻¹(x) = tan⁻¹(1/x) if x > 0

= tan⁻¹((2/11 + 7/24)/(1 - (2/11)(7/24)))

= tan⁻¹(1/2)

Exercise 1

(i). If sin⁻¹x + 4cos⁻¹x = π, then x =

(A) 1/2

(B) -1/2

(C) √3/2

(D) 1

(ii). cos⁻¹(-1/2) =

(A) π/3

(B) 2π/3

(C) π/4

(D) π/6

Do Check: Determinant

Property vii

i) sin⁻¹x = cos⁻¹√(1-x²) = tan⁻¹(x/√(1-x²))

ii) cos⁻¹x = sin⁻¹√(1-x²) = tan⁻¹(√(1-x²)/x)

iii) tan⁻¹x = sin⁻¹(x/√(1+x²)) = cos⁻¹(1/√(1+x²))

Illustration 4

Evaluate sin⁻¹(cos⁻¹(3/5)) ; sin(cot⁻¹x)

Solution

i) sin⁻¹(cos⁻¹(3/5)) = sin⁻¹(sin⁻¹√(1-(3/5)²)) = sin⁻¹(sin⁻¹(4/5)) = 4/5

ii) sin(cot⁻¹x) = sin(tan⁻¹(1/x)) = sin(sin⁻¹(1/√(x²+1))) = 1/√(x²+1)

Property viii

i) 2tan⁻¹x = sin⁻¹(2x/(1+x²)), if -1 ≤ x ≤ 1

= π - sin⁻¹(2x/(1+x²)), if x > 1

= -π - sin⁻¹(2x/(1+x²)), if x < -1

ii) 2tan⁻¹x = cos⁻¹((1-x²)/(1+x²)), if 0 ≤ x < ∞

= -cos⁻¹((1-x²)/(1+x²)), if -∞ < x ≤ 0

Illustration 5

Solve sin⁻¹[2cos⁻¹(cot(2tan⁻¹x))] = 0

Solution

sin⁻¹[2cos⁻¹(cot(tan⁻¹(2x/(1-x²))))] = 0

sin⁻¹[2cos⁻¹(cot(cot⁻¹((1-x²)/2x)))] = 0

sin⁻¹[2cos⁻¹((1-x²)/2x)] = 0

sin⁻¹(sin⁻¹((1-x²)/2x)√(1-((1-x²)/2x)²)) = 0

Simplifying: (1-x²)/x√(4x²-1-x⁴+2x²)/4x² = 0

(1-x²)/x = 0 or 6x²-x⁴-1 = 0

x⁴-6x²+1 = 0

x = ±1 or (1-x²)²-(2x)² = 0

x²+2x-1 = 0 or x²-2x-1 = 0

x = -1±√2 or x = 1±√2

Exercise 2

(i). If x > 0, y > 0 and x > y, then tan⁻¹(x/y) + tan⁻¹((x-y)/(x+y)) =

1) π/4

2) -π/4

3) 3π/4

4) None of these

(ii) If cos⁻¹(1/x) = θ, then tan θ is equal to

1) √(x²-1)

2) 1/√(x²-1)

3) √(1-x²)

4) √(x²+1)

Answer to Exercises

Exercise 1: (i) C, (ii) B

Exercise 2: (i) C, (ii) A

Formulae and concepts at Glance

1. sin⁻¹x + cos⁻¹x = π/2, x ∈ [-1,1]

2. tan⁻¹x + cot⁻¹x = π/2, x ∈ ℝ

3. sec⁻¹x + cosec⁻¹x = π/2, x ∈ ℝ - (-1,1)

4. tan⁻¹x + tan⁻¹y = tan⁻¹((x+y)/(1-xy)), if xy < 1

= π + tan⁻¹((x+y)/(1-xy)), if x,y > 0 and xy > 1

= -π + tan⁻¹((x+y)/(1-xy)), if x < 0, y < 0 and xy > 1

Do Check: Statics and dynamics

Solved Examples

1. tan(cos⁻¹x) = sin(tan⁻¹2)

Sol. tan(cos⁻¹x) = sin(tan⁻¹2)

tan(tan⁻¹(√(1-x²)/x)) = sin(sin⁻¹(2/√(1+2²)))

√(1-x²)/x = 2/√5

⇒ 5-5x² = 4x² ⇒ 5 = 9x² ⇒ x² = 5/9 ⇒ x = ±√5/3

x = -√5/3 ⇒ cos⁻¹x lies in IInd Quadrant and therefore tan(cos⁻¹x) is -ve

L.H.S. is -Ve, whereas R.H.S. is +ve

∴ x = -√5/3 is impossible

∴ x = √5/3

2. Show that 2tan⁻¹(1/3) + sin⁻¹(4/5) = π/2

Sol: L.H.S. = 2tan⁻¹(1/3) + sin⁻¹(4/5)

= cos⁻¹((1-1/9)/(1+1/9)) + sin⁻¹(4/5)

= cos⁻¹(8/10) + sin⁻¹(4/5) = cos⁻¹(4/5) + sin⁻¹(4/5) = π/2 = R.H.S

3. Solve the equation tan⁻¹(1/(1+2x)) + tan⁻¹(1/(4x+1)) = tan⁻¹(2/x²)

Sol: Given equation is

tan⁻¹(1/(1+2x)) + tan⁻¹(1/(4x+1)) = tan⁻¹(2/x²)

⇒ tan⁻¹((1/(1+2x) + 1/(4x+1))/(1 - 1/(1+2x) · 1/(4x+1))) = tan⁻¹(2/x²)

⇒ (4x+1+2x+1)/(8x²+6x+1-1) = 2/x² ⇒ (6x+2)/(8x²+6x) = 2/x²

⇒ (3x+1)x² = 8x²+6x

⇒ 3x³+x²-8x²-6x = 0 ⇒ x(3x²-7x-6) = 0

x = 0 or 3x²-7x-6 = 0; 3x²-9x+2x-6 = 0

3x(x-3)+2(x-3) = 0

(x-3)(3x+2) = 0; x = 3 or x = -2/3

Hence x = 0, 3, -2/3 are solution of given equation.

4. Sin⁻¹(2a/(1+a²)) + Sin⁻¹(2b/(1+b²)) = 2tan⁻¹x

Sol: Given equation is

sin⁻¹(2a/(1+a²)) + sin⁻¹(2b/(1+b²)) = 2tan⁻¹x

2tan⁻¹a + 2tan⁻¹b = 2tan⁻¹x

tan⁻¹a + tan⁻¹b = tan⁻¹x

tan⁻¹((a+b)/(1-ab)) = tan⁻¹x

∴ x = (a+b)/(1-ab)

5. If cos⁻¹(x/a) + cos⁻¹(y/b) = α, show that x²/a² - 2xy cos α/ab + y²/b² = sin²α

Sol:

cos⁻¹(x/a) + cos⁻¹(y/b) = α …………………..(1)

Let cos⁻¹(x/a) = A, cos⁻¹(y/b) = B

cos A = x/a, cos B = y/b, A+B = α

cos(A+B) = cos α

cos A cos B - sin A sin B = cos α

(x/a)(y/b) - √(1-x²/a²)√(1-y²/b²) = cos α

xy/ab - √((a²-x²)/a²)√((b²-y²)/b²) = cos α

squaring both sides.

(xy/ab)² + ((a²-x²)(b²-y²))/(a²b²) - 2(xy/ab)√((a²-x²)(b²-y²))/(ab) = cos²α

x²y²/a²b² + (a²-x²)(b²-y²)/a²b² - (2xy/ab)²√((a²-x²)(b²-y²))/a²b² = cos²α

x²/a² - 2xy cos α/ab + y²/b² = sin²α

6. Find the value of tan[(1/2)sin⁻¹(2x/(1+x²)) + (1/2)cos⁻¹((1-y²)/(1+y²))]

Sol: tan[(1/2)sin⁻¹(2x/(1+x²)) + (1/2)cos⁻¹((1-y²)/(1+y²))]

= tan[(1/2)(2tan⁻¹x) + (1/2)(2tan⁻¹y)]

= tan[tan⁻¹x + tan⁻¹y]

= tan[tan⁻¹((x+y)/(1-xy))]

= (x+y)/(1-xy)

Assignment

1. tan⁻¹(1/2 cos⁻¹0) =

(A) 0

(B) -1

(C) 1/2

(D) 1

2. cos[cos⁻¹(-343/792) + sin⁻¹(-343/792)] =

(A) -343/792

(B) 343/792

(C) 2

(D) 0

3. cot⁻¹[(√(1-sinx) + √(1+sinx))/(√(1-sinx) - √(1+sinx))] =

(A) x/2

(B) +x/2

(C) π/2 - x/2

(D) π - x/2

4. If tan(cos⁻¹x) = sin(cot⁻¹(1/2)), then x is equal to

(A) ±√5/3

(B) ±√5/3

(C) ±√5/3

(D) None of these

5. If sin⁻¹(x) + sin⁻¹(1-x) = cos⁻¹(x), then x equals to

(A) 1,-1

(B) 1,0

(C) 0, 1/2

(D) None of these

6. sin⁻¹(√(x/(x+a))) is equal to

(A) cos⁻¹(√(x/a))

(B) cosec⁻¹(√(x/a))

(C) tan⁻¹(√(x/a))

(D) None of these

7. tan⁻¹(1) + tan⁻¹(2) + tan⁻¹(3) =

(A) π/2

(B) π/4

(C) 0

(D) None of these

8. The value of Σ(r=0 to n) tan⁻¹(1/(1+r+r²)) is equal to

(A) π/2

(B) can't be found

(C) tan⁻¹(n+1)

(D) tan⁻¹(n)

9. If sin(cot⁻¹(x+1)) = cos(tan⁻¹x), then x =

(A) 1/2

(B) -1/2

(C) 0

(D) 9/4

10. If sin⁻¹x - cos⁻¹x = π/6, then x is

(A) 1/2

(B) √3/2

(C) -1/2

(D) None of these

11. If cos⁻¹x - cos⁻¹y = α, then x² - 2xy cos α + y² is equal to

(A) -sin²α

(B) sin²α

(C) 4

(D) None of these

12. If sin⁻¹(α) + sin⁻¹(β) + sin⁻¹(γ) + sin⁻¹(δ) = 2π, then Σαβ is equal to

(A) 5

(B) can't be found

(C) 6

(D) None of these

13. If x² + y² + z² = r², then tan⁻¹(xy/zr) + tan⁻¹(yz/xr) + tan⁻¹(zx/yr) =

(A) π

(B) π/2

(C) 0

(D) None of these

14. tan[2tan⁻¹(1/5) - π/4] =

(A) 17/7

(B) -17/7

(C) 7/17

(D) -7/17

15. If cos⁻¹x + cos⁻¹y + cos⁻¹z = π, then x² + y² + z² + 2xyz =

(A) 3

(B) -1

(C) 2

(D) 1

Answers to Assignment

1. D	2. D	3. D
4. B	5. D	6. C
7. D	8. C	9. A
10. B	11. B	12. C
13. B	14. B	15. D