Inverse Trigonometric Functions

• Definition of inverse trigonometric functions
• Range and principal value branch
• Graphs of inverse trigonometric functions
• Trigonometric equations and inverse trigonometric functions
• Evaluating inverse trigonometric functions using special triangles

Inverse Trigonometric Functions - Definition

• Inverse trigonometric functions are functions that undo the trigonometric functions.
• They allow us to find the angle from a known trigonometric ratio.
• Denoted as sin⁻¹, cos⁻¹, tan⁻¹, cot⁻¹, sec⁻¹, csc⁻¹.

Inverse Trigonometric Functions - Range and Principal Value Branch

• Range of inverse trigonometric functions depends on the principal value branch.
• Principal value branch is the specific interval for which the inverse function is defined.
• For sine and cosine, the principal value branch is [-π/2, π/2].
• For tangent, the principal value branch is (-π/2, π/2).
• For cotangent, secant, and cosecant, the principal value branch is [0, π].

Inverse Trigonometric Functions - Graphs

• Graph of y = sin⁻¹(x):
  • Domain: [-1, 1]
  • Range: [-π/2, π/2]
  • Symmetry: Odd function
  • Increasing from left to right
• Graph of y = cos⁻¹(x):
  • Domain: [-1, 1]
  • Range: [0, π]
  • Symmetry: Even function
  • Decreasing from left to right

Inverse Trigonometric Functions - Trigonometric Equations

• Trigonometric equations can be solved using inverse trigonometric functions.
• For example:
  • sin(x) = 1/2
  • cos(x) = 0
• Apply the inverse trigonometric function to both sides of the equation.
• Solve for x using the properties of inverse trigonometric functions.

Inverse Trigonometric Functions - Evaluating using Special Triangles

• Special triangles can be used to evaluate inverse trigonometric functions.
• Common angles from special triangles: 0°, 30°, 45°, 60°, 90°.
• By knowing the ratios for these angles, we can evaluate inverse trigonometric functions.
• For example:
  • sin⁻¹(1/2) = 30°
  • tan⁻¹√3 = 60°

Inverse Trigonometric Functions - Examples

1. Find the value of cos⁻¹(-1/2).

2. Solve the equation sec(x) = 2 for x.

3. Evaluate tan⁻¹(√3).

4. Find the angle sin(x) = 1.

5. Solve the equation csc(x) = -1/2 for x.

Inverse Trigonometric Functions - Solutions

1. cos⁻¹(-1/2): π/3

2. sec(x) = 2:
   • cos(x) = 1/2
   • x = π/3, 5π/3

3. tan⁻¹(√3): π/3

4. sin(x) = 1: π/2

5. csc(x) = -1/2:
   • sin(x) = -2
   • No solution in the given principal branch

Inverse Trigonometric Functions - IDENTITY

• Sum of two inverses of tangent:
  • tan⁻¹(x) + tan⁻¹(1/x) = π/2, x > 0
• Proof:
  • Let α = tan⁻¹(x), β = tan⁻¹(1/x).
  • tan(α + β) = (tan α + tan β) / (1 - tan α tan β)
  • tan(α + β) = (x + 1/x) / (1 - (x/x)(1/x))
  • tan(α + β) = (x + 1/x) / (1 - 1)
  • tan(α + β) = (x + 1/x) / 0
  • So (α + β) = π/2
  • Hence, tan⁻¹(x) + tan⁻¹(1/x) = π/2

Inverse Trigonometric Functions - IDENTITY

• Difference of two inverse trigonometric functions:
  • tan⁻¹(x) - tan⁻¹(y) = tan⁻¹((x - y)/(1 + xy))
• Proof:
  • Let α = tan⁻¹(x), β = tan⁻¹(y).
  • tan(α - β) = (tan α - tan β) / (1 + tan α tan β)
  • tan(α - β) = (x - y) / (1 + xy)
  • So α - β = tan⁻¹((x - y)/(1 + xy))
  • Hence, tan⁻¹(x) - tan⁻¹(y) = tan⁻¹((x - y)/(1 + xy))

Inverse Trigonometric Functions - IDENTITY

• Product of two inverse trigonometric functions:
  • tan⁻¹(x) * tan⁻¹(y) = tan⁻¹((xy - 1)/(x + y))
• Proof:
  • Let α = tan⁻¹(x), β = tan⁻¹(y).
  • tan(α + β) = (tan α + tan β) / (1 - tan α tan β)
  • tan(α + β) = (x + y) / (1 - xy)
  • tan(α - β) = (tan α - tan β) / (1 + tan α tan β)
  • tan(α - β) = (x - y) / (1 + xy)
  • tan(α + β) * tan(α - β) = ((x + y) / (1 - xy)) * ((x - y) / (1 + xy))
  • tan(α + β) * tan(α - β) = (xy - 1) / (x + y)
  • So tan⁻¹(x) * tan⁻¹(y) = tan⁻¹((xy - 1)/(x + y))

Inverse Trigonometric Functions - IDENTITY

• Sum of inverse trigonometric functions:
  • tan⁻¹(x) + tan⁻¹(y) = tan⁻¹((x + y)/(1 - xy))
• Proof:
  • Let α = tan⁻¹(x), β = tan⁻¹(y).
  • tan(α + β) = (tan α + tan β) / (1 - tan α tan β)
  • tan(α + β) = (x + y) / (1 - xy)
  • So tan⁻¹(x) + tan⁻¹(y) = tan⁻¹((x + y)/(1 - xy))

Inverse Trigonometric Functions - IDENTITY

• Division of inverse trigonometric functions:
  • tan⁻¹(x) / tan⁻¹(y) = (xy - 1) / (x - y)
• Proof:
  • Let α = tan⁻¹(x), β = tan⁻¹(y).
  • tan(α - β) = (tan α - tan β) / (1 + tan α tan β)
  • tan(α - β) = (x - y) / (1 + xy)
  • So tan⁻¹(x) / tan⁻¹(y) = (xy - 1) / (x - y)

Inverse Trigonometric Functions - IDENTITY

• Sum of inverse trigonometric functions:
  • sin⁻¹(x) + sin⁻¹(y) = sin⁻¹((x√(1 - y²) + y√(1 - x²))/√(1 - x²)(1 - y²))
• Proof:
  • Let α = sin⁻¹(x), β = sin⁻¹(y).
  • sin(α + β) = sin α cos β + cos α sin β
  • sin(α + β) = (x/sqrt(1 - x²)) * (sqrt(1 - y²)/sqrt(1 - y²)) + (y/sqrt(1 - y²)) * (sqrt(1 - x²)/sqrt(1 - x²))
  • sin(α + β) = (x√(1 - y²) + y√(1 - x²))/√(1 - x²)(1 - y²)
  • So sin⁻¹(x) + sin⁻¹(y) = sin⁻¹((x√(1 - y²) + y√(1 - x²))/√(1 - x²)(1 - y²))

Inverse Trigonometric Functions - IDENTITY

• Difference of inverse trigonometric functions:
  • sin⁻¹(x) - sin⁻¹(y) = sin⁻¹((x√(1 - y²) - y√(1 - x²))/√(1 - x²)(1 - y²))
• Proof:
  • Let α = sin⁻¹(x), β = sin⁻¹(y).
  • sin(α - β) = sin α cos β - cos α sin β
  • sin(α - β) = (x/sqrt(1 - x²)) * (sqrt(1 - y²)/sqrt(1 - y²)) - (y/sqrt(1 - y²)) * (sqrt(1 - x²)/sqrt(1 - x²))
  • sin(α - β) = (x√(1 - y²) - y√(1 - x²))/√(1 - x²)(1 - y²)
  • So sin⁻¹(x) - sin⁻¹(y) = sin⁻¹((x√(1 - y²) - y√(1 - x²))/√(1 - x²)(1 - y²))

Inverse Trigonometric Functions - IDENTITY - Sum of two inverses of tangent

• Sum of two inverses of tangent:
  • tan⁻¹(x) + tan⁻¹(1/x) = π/2, x > 0
• Proof:
  • Let α = tan⁻¹(x), β = tan⁻¹(1/x).
  • tan(α + β) = (tan α + tan β) / (1 - tan α tan β)
  • tan(α + β) = (x + 1/x) / (1 - (x/x)(1/x))
  • tan(α + β) = (x + 1/x) / (1 - 1)
  • tan(α + β) = (x + 1/x) / 0
  • So (α + β) = π/2
  • Hence, tan⁻¹(x) + tan⁻¹(1/x) = π/2

Inverse Trigonometric Functions - IDENTITY - Difference of two inverse trigonometric functions

• Difference of two inverse trigonometric functions:
  • tan⁻¹(x) - tan⁻¹(y) = tan⁻¹((x - y)/(1 + xy))
• Proof:
  • Let α = tan⁻¹(x), β = tan⁻¹(y).
  • tan(α - β) = (tan α - tan β) / (1 + tan α tan β)
  • tan(α - β) = (x - y) / (1 + xy)
  • So α - β = tan⁻¹((x - y)/(1 + xy))
  • Hence, tan⁻¹(x) - tan⁻¹(y) = tan⁻¹((x - y)/(1 + xy))

Inverse Trigonometric Functions - IDENTITY - Product of two inverse trigonometric functions

• Product of two inverse trigonometric functions:
  • tan⁻¹(x) * tan⁻¹(y) = tan⁻¹((xy - 1)/(x + y))
• Proof:
  • Let α = tan⁻¹(x), β = tan⁻¹(y).
  • tan(α + β) = (tan α + tan β) / (1 - tan α tan β)
  • tan(α + β) = (x + y) / (1 - xy)
  • tan(α - β) = (tan α - tan β) / (1 + tan α tan β)
  • tan(α - β) = (x - y) / (1 + xy)
  • tan(α + β) * tan(α - β) = ((x + y) / (1 - xy)) * ((x - y) / (1 + xy))
  • tan(α + β) * tan(α - β) = (xy - 1) / (x + y)
  • So tan⁻¹(x) * tan⁻¹(y) = tan⁻¹((xy - 1)/(x + y))

Inverse Trigonometric Functions - IDENTITY - Sum of inverse trigonometric functions

• Sum of inverse trigonometric functions:
  • tan⁻¹(x) + tan⁻¹(y) = tan⁻¹((x + y)/(1 - xy))
• Proof:
  • Let α = tan⁻¹(x), β = tan⁻¹(y).
  • tan(α + β) = (tan α + tan β) / (1 - tan α tan β)
  • tan(α + β) = (x + y) / (1 - xy)
  • So tan⁻¹(x) + tan⁻¹(y) = tan⁻¹((x + y)/(1 - xy))

Inverse Trigonometric Functions - IDENTITY - Division of inverse trigonometric functions

• Division of inverse trigonometric functions:
  • tan⁻¹(x) / tan⁻¹(y) = (xy - 1) / (x - y)
• Proof:
  • Let α = tan⁻¹(x), β = tan⁻¹(y).
  • tan(α - β) = (tan α - tan β) / (1 + tan α tan β)
  • tan(α - β) = (x - y) / (1 + xy)
  • So tan⁻¹(x) / tan⁻¹(y) = (xy - 1) / (x - y)

Inverse Trigonometric Functions - IDENTITY - Sum of inverse trigonometric functions

• Sum of inverse trigonometric functions:
  • sin⁻¹(x) + sin⁻¹(y) = sin⁻¹((x√(1 - y²) + y√(1 - x²))/√(1 - x²)(1 - y²))
• Proof:
  • Let α = sin⁻¹(x), β = sin⁻¹(y).
  • sin(α + β) = sin α cos β + cos α sin β
  • sin(α + β) = (x/sqrt(1 - x²)) * (sqrt(1 - y²)/sqrt(1 - y²)) + (y/sqrt(1 - y²)) * (sqrt(1 - x²)/sqrt(1 - x²))
  • sin(α + β) = (x√(1 - y²) + y√(1 - x²))/√(1 - x²)(1 - y²)
  • So sin⁻¹(x) + sin⁻¹(y) = sin⁻¹((x√(1 - y²) + y√(1 - x²))/√(1 - x²)(1 - y²))

Inverse Trigonometric Functions - IDENTITY - Difference of inverse trigonometric functions

• Difference of inverse trigonometric functions:
  • sin⁻¹(x) - sin⁻¹(y) = sin⁻¹((x√(1 - y²) - y√(1 - x²))/√(1 - x²)(1 - y²))
• Proof:
  • Let α = sin⁻¹(x), β = sin⁻¹(y).
  • sin(α - β) = sin α cos β - cos α sin β
  • sin(α - β) = (x/sqrt(1 - x²)) * (sqrt(1 - y²)/sqrt(1 - y²)) - (y/sqrt(1 - y²)) * (sqrt(1 - x²)/sqrt(1 - x²))
  • sin(α - β) = (x√(1 - y²) - y√(1 - x²))/√(1 - x²)(1 - y²)
  • So sin⁻¹(x) - sin⁻¹(y) = sin⁻¹((x√(1 - y²) - y√(1 - x²))/√(1 - x²)(1 - y²))