Inverse Trigonometric Functions Practice Problems

Inverse trigonometric functions are fundamental classes or types of mathematics instruments that are frequently employed in engineering and science to solve problems related to angles and trigonometric ratios.

In this article, we have covered, the basics of Inverse Trigonometric Functions, their practice formulas and others in detail.

What are Inverse trigonometric functions?

Inverse Trigonometric Functions are those that obtain the angles from trigonometric ratios and reverse the operations of standard trigonometric functions. Such functions are especially useful in solving equations wherein the angle is the unknown variable.

Using inverse trigonometric functions, we can change a complicated trigonometric expression into a simple equation containing only an unknown angle, solve it, and then see the results in the original context.

types Inverse Trigonometric functions

The six Inverse Trigonometric Functions are:

• sin⁻¹(x)
• cos⁻¹(x)
• tan⁻¹(x)
• cot⁻¹(x)
• sec⁻¹(x)
• cosec⁻¹(x)

Important Formulas of Inverse Trigonometric Functions

Basic Inverse Trigonometric Functions:

• sin⁻¹(sin θ) = θ, for -π/2 ≤ θ ≤ π/2
• cos⁻¹(cos θ) = θ, for 0 ≤ θ ≤ π
• tan⁻¹(tan θ) = θ, for -π/2 < θ < π/2

Domain and Range:

• Domain of sin⁻¹x: -1 ≤ x ≤ 1, Range: [-π/2, π/2]
• Domain of cos⁻¹x: -1 ≤ x ≤ 1, Range: [0, π]
• Domain of tan⁻¹x: all real numbers, Range: (-π/2, π/2)

Even-Odd Properties:

• sin⁻¹(-x) = -sin⁻¹x
• cos⁻¹(-x) = π - cos⁻¹x
• tan⁻¹(-x) = -tan⁻¹x

Complementary Angles:

• sin⁻¹x + cos⁻¹x = π/2

Negative Argument:

• cos⁻¹(-x) = π - cos⁻¹x
• tan⁻¹(-x) = -tan⁻¹x

Sum and Difference Formulas:

• sin⁻¹x ± sin⁻¹y = sin⁻¹[x√(1-y²) ± y√(1-x²)]
• tan⁻¹x ± tan⁻¹y = tan⁻¹[(x ± y)/(1 ∓ xy)]

Double Angle Formulas:

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

Half Angle Formulas:

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

Inverse Function Relationships:

• sin(sin⁻¹x) = x
• cos(cos⁻¹x) = x
• tan(tan⁻¹x) = x

Derivative Formulas:

• d/dx (sin⁻¹x) = 1/√(1-x²)
• d/dx (cos⁻¹x) = -1/√(1-x²)
• d/dx (tan⁻¹x) = 1/(1+x²)

Inverse Trigonometric Functions Practice Problems

Example 1: Evaluate sin⁻¹(1/2).

Solution:

Step 1: We know that sin(π/6) = 1/2

Step 2: Therefore, sin⁻¹(1/2) = π/6

Example 2: Simplify sin⁻¹(x) + cos⁻¹(x) for x ∈ [-1, 1].

Solution:

Step 1: We know the identity sin⁻¹(x) + cos⁻¹(x) = π/2 for x ∈ [-1, 1]

Step 2: Therefore, sin⁻¹(x) + cos⁻¹(x) = π/2

Example 3: Evaluate sin⁻¹(cos(π/4)).

Solution:

Step 1: We know that cos(π/4) = 1/√2

Step 2: Substitute this value into the expression: sin⁻¹(1/√2)

Step 3: Recognize that sin(π/4) = 1/√2

Step 4: Therefore, sin⁻¹(cos(π/4)) = π/4

Example 4: Solve tan⁻¹(x) + tan⁻¹(1/x) = π/2 for x > 0.

Solution:

Step 1: We know the identity tan⁻¹(x) + tan⁻¹(y) = π/2 when xy = 1

Step 2: In this case, y = 1/x

Step 3: Check if x × (1/x) = 1

Step 4: Indeed, x × (1/x) = 1 for all x ≠ 0

Step 5: Since we're given x > 0, the equation is true for all positive real numbers

Example 5: Evaluate cos⁻¹(-1/2) + sin⁻¹(1/2).

Solution:

Step 1: We know cos⁻¹(-1/2) = 2π/3

Step 2: We knowsin⁻¹(1/2) = π/6

Step 3: Add these values: 2π/3 + π/6 = 4π/6 + π/6 = 5π/6

Example 6: Simplify tan⁻¹(1/√3) + tan⁻¹(√3).

Solution:

Step 1: We know the identity tan⁻¹(x) + tan⁻¹(y) = π/2 when xy = 1

Step 2: Check if (1/√3) × √3 = 1

Step 3: Indeed, (1/√3) × √3 = 1

Step 4: Therefore, tan⁻¹(1/√3) + tan⁻¹(√3) = π/2

Example 7: Evaluate sin⁻¹(sin(5π/4)).

Solution:

Step 1: We know that sin(5π/4) = -√2/2

Step 2: Substitute this value: sin⁻¹(-√2/2)

Step 3: Recognize that sin(5π/4) and sin(-π/4) both equal -√2/2

Step 4: Since sin⁻¹ is defined to give values in [-π/2, π/2], we choose -π/4

Step 5: Therefore, sin⁻¹(sin(5π/4)) = -π/4

Example 8: Solve the equation 2sin⁻¹(x) = π/3.

Solution:

Step 1: Divide both sides by 2: sin⁻¹(x) = π/6

Step 2: Apply sin to both sides: sin(sin⁻¹(x)) = sin(π/6)

Step 3: Simplify: x = 1/2

Step 4: Therefore, the solution is x = 1/2

Example 9: Evaluate cos⁻¹(cos(7π/6)).

Solution:

Step 1: We know that cos(7π/6) = -√3/2

Step 2: Substitute this value: cos⁻¹(-√3/2)

Step 3: Recognize that cos(7π/6) and cos(5π/6) both equal -√3/2

Step 4: Since cos⁻¹ is defined to give values in [0, π], we choose 5π/6

Step 5: Therefore, cos⁻¹(cos(7π/6)) = 5π/6

Example 10: Evaluate tan⁻¹(-1).

Solution:

Step 1: We know that tan(3π/4) = -1

Step 2: However, tan⁻¹ is defined to give values in (-π/2, π/2)

Step 3: The angle in this range with tangent -1 is -π/4

Step 4: Therefore, tan⁻¹(-1) = -π/4

Example 11: Simplify 2tan⁻¹(1/3).

Solution:

Step 1: We know the identity tan⁻¹(x) + tan⁻¹(y) = tan⁻¹((x+y)/(1-xy)) when xy < 1

Step 2: Here, we have tan⁻¹(1/3) + tan⁻¹(1/3)

Step 3: Let x = y = 1/3. Check if xy < 1: (1/3)(1/3) = 1/9 < 1, so we can use the identity

Step 4: tan⁻¹(1/3) + tan⁻¹(1/3) = tan⁻¹((1/3+1/3)/(1-(1/3)(1/3))) = tan⁻¹((2/3)/(8/9)) = tan⁻¹(3/4)

Step 5: We know that tan(π/3) = √3 ≈ 1.732, which is greater than 3/4

Step 6: Therefore, 2tan⁻¹(1/3) = tan⁻¹(3/4) = π/3

Example 12: Evaluate sin⁻¹(-√3/2).

Solution:

Step 1: We know that sin(-π/3) = -√3/2

Step 2: Since sin⁻¹ is defined to give values in [-π/2, π/2], and -π/3 is in this range

Step 3: Therefore, sin⁻¹(-√3/2) = -π/3

Example 13: Solve the equation cos⁻¹(x) + sin⁻¹(x) = π for x ∈ [-1, 1].

Solution:

Step 1: We know the identity cos⁻¹(x) + sin⁻¹(x) = π/2 for x ∈ [-1, 1]

Step 2: The given equation cos⁻¹(x) + sin⁻¹(x) = π is inconsistent with this identity

Step 3: Therefore, there is no solution in the given domain

Example 14: Evaluate tan⁻¹(sin(π/3)/cos(π/3)).

Solution:

Step 1: We know that sin(π/3) = √3/2 and cos(π/3) = 1/2

Step 2: Substitute these values: tan⁻¹((√3/2)/(1/2))

Step 3: Simplify: tan⁻¹(√3) = tan⁻¹(tan(π/3))

Step 4: Since π/3 is in the range (-π/2, π/2) where tan⁻¹ is defined

Step 5: Therefore, tan⁻¹(sin(π/3)/cos(π/3)) = π/3

Worksheet on Inverse Trigonometric Functions

You can Download the worksheet on Inverse Trigonometric Functions and their solution from- Inverse Trigonometric Functions

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