Inverse Trigonometric Functions — Practice Quiz

[-1, -1/√3] ∪ [1/√3, 1]

Step 1: For sin⁻¹(u) to be defined, we need -1 ≤ u ≤ 1. So -1 ≤ 2 - 3x² ≤ 1. Step 2: From 2 - 3x² ≥ -1: 3x² ≤ 3, so x² ≤ 1, meaning -1 ≤ x ≤ 1. Step 3: From 2 - 3x² ≤ 1: 3x² ≥ 1, so x² ≥ 1/3, meaning |x| ≥ 1/√3. Step 4: Combining both conditions: 1/3 ≤ x² ≤ 1, so 1/√3 ≤ |x| ≤ 1. Final Step: This gives x ∈ [-1, -1/√3] ∪ [1/√3, 1].

24/25

Step 1: Let sec⁻¹(5/4) = θ, so sec θ = 5/4, meaning cos θ = 4/5 and sin θ = 3/5. Step 2: Therefore sec⁻¹(5/4) = sin⁻¹(3/5), so the expression becomes sin[2sin⁻¹(3/5)]. Step 3: Use the identity sin(2α) = 2sinα cosα where α = sin⁻¹(3/5). Step 4: sin α = 3/5 and cos α = √(1 - 9/25) = 4/5. Final Step: sin[2sin⁻¹(3/5)] = 2 × (3/5) × (4/5) = 24/25.

π/3

Step 1: The range of cos⁻¹ is [0, π], and 13π/3 is not in this range. Step 2: Write 13π/3 in terms of a value in [0, π]. 13π/3 = 4π + π/3. Step 3: cos(13π/3) = cos(4π + π/3) = cos(π/3) [since cosine has period 2π]. Step 4: Now cos⁻¹(cos(π/3)) = π/3, since π/3 ∈ [0, π]. Final Step: Answer is π/3. Common error: Not reducing the angle to the principal range first.

π/2

Step 1: Use the property sin⁻¹x + cos⁻¹x = π/2 for x ∈ [-1, 1]. Step 2: From this: cos⁻¹x = π/2 - sin⁻¹x and cos⁻¹y = π/2 - sin⁻¹y. Step 3: Add them: cos⁻¹x + cos⁻¹y = π - (sin⁻¹x + sin⁻¹y). Step 4: Substitute sin⁻¹x + sin⁻¹y = π/2. Final Step: cos⁻¹x + cos⁻¹y = π - π/2 = π/2.