Inverse Trigonometric Functions

Inverse Trigonometric Functions – Complete Solutions

2.1 Introduction

Inverse trigonometric functions are the inverse functions of the trigonometric functions sin, cos, tan, cot, sec, and cosec. They are used to find the angle when the trigonometric value is known.

2.2 Principal Values and Ranges

Function	Domain	Range (Principal Value)
sin⁻¹(x)	[-1, 1]	[-π/2, π/2]
cos⁻¹(x)	[-1, 1]	[0, π]
tan⁻¹(x)	(-∞, ∞)	(-π/2, π/2)
cot⁻¹(x)	(-∞, ∞)	(0, π)
sec⁻¹(x)	(-∞, -1] ∪ [1, ∞)	[0, π] – {π/2}
cosec⁻¹(x)	(-∞, -1] ∪ [1, ∞)	[-π/2, π/2] – {0}

2.3 Important Properties

Property 1: Complementary Functions

sin⁻¹(x) + cos⁻¹(x) = π/2 for x ∈ [-1, 1]

Proof: Let sin⁻¹(x) = α, so sin α = x and α ∈ [-π/2, π/2]

Let cos⁻¹(x) = β, so cos β = x and β ∈ [0, π]

Since sin α = x = cos β and sin α = cos(π/2 – α)

We have π/2 – α = β, therefore α + β = π/2

Hence sin⁻¹(x) + cos⁻¹(x) = π/2

Property 2:

tan⁻¹(x) + cot⁻¹(x) = π/2 for all x ∈ R

Property 3: Odd Functions

sin⁻¹(-x) = -sin⁻¹(x)

tan⁻¹(-x) = -tan⁻¹(x)

cosec⁻¹(-x) = -cosec⁻¹(x)

2.4 Addition Formulas

For tan⁻¹:

tan⁻¹(a) + tan⁻¹(b) = tan⁻¹((a+b)/(1-ab)) when ab < 1

tan⁻¹(a) + tan⁻¹(b) = π + tan⁻¹((a+b)/(1-ab)) when ab > 1 and a > 0

Example:

Find tan⁻¹(1) + tan⁻¹(2) + tan⁻¹(3)

tan⁻¹(1) + tan⁻¹(2) = tan⁻¹((1+2)/(1-2)) = tan⁻¹(-3)

tan⁻¹(-3) + tan⁻¹(3) = 0 (since they are negatives)

Therefore, the sum = π

2.5 Solved Examples

Example 1: Find the principal value of sin⁻¹(√3/2)

Solution:

Let sin⁻¹(√3/2) = θ, where θ ∈ [-π/2, π/2]

Then sin θ = √3/2

We know sin(π/3) = √3/2 and π/3 ∈ [-π/2, π/2]

Therefore, sin⁻¹(√3/2) = π/3

Example 2: Find cos⁻¹(cos(2π/3))

Solution:

cos(2π/3) = -1/2

cos⁻¹(-1/2) = 2π/3 (since 2π/3 ∈ [0, π])

Therefore, cos⁻¹(cos(2π/3)) = 2π/3

Example 3: Evaluate sin(tan⁻¹(3/4))

Solution:

Let tan⁻¹(3/4) = θ, where θ ∈ (-π/2, π/2)

Then tan θ = 3/4

Drawing a right triangle: opposite = 3, adjacent = 4

Hypotenuse = √(9+16) = 5

sin θ = 3/5

Therefore, sin(tan⁻¹(3/4)) = 3/5

2.6 Practice Problems with Solutions

Problem 1: Find tan⁻¹(1) + tan⁻¹(1/2) + tan⁻¹(1/3)

Solution:

tan⁻¹(1) = π/4

tan⁻¹(1/2) + tan⁻¹(1/3) = tan⁻¹((1/2 + 1/3)/(1 – 1/6)) = tan⁻¹((5/6)/(5/6)) = tan⁻¹(1) = π/4

Therefore, total = π/4 + π/4 = π/2

Problem 2: Solve: tan⁻¹(x) + tan⁻¹(2x) = π/4

Solution:

Using addition formula: tan⁻¹((x + 2x)/(1 – 2x²)) = π/4

tan⁻¹((3x)/(1 – 2x²)) = π/4

(3x)/(1 – 2x²) = 1

3x = 1 – 2x²

2x² + 3x – 1 = 0

x = (-3 ± √(9+8))/4 = (-3 ± √17)/4

Since x must satisfy |x| < 1, x = (√17 - 3)/4