<p>If \( x = \sin^{-1}(\sin 10) \) and \( y = \cos^{-1}(\cos 10) \), then \( y - x \) is equal to:</p>
Step-by-Step Solution
Key Concept: Inverse trigonometric functions return values from their principal ranges: sin⁻¹ ∈ [-π/2, π/2] and cos⁻¹ ∈ [0, π]. Since 10 radians lies outside these ranges, we must first reduce it using periodicity and complementary angle properties.
<p><strong>Step 1:</strong> Find x = sin⁻¹(sin 10)</p><p>Since 10 radians: note that π ≈ 3.14, so 10 ≈ 3.18π. We have 3π < 10 < 3.5π.</p><p>Reduce to principal range [-π/2, π/2]: Since 10 = 3π + (10 - 3π) where 10 - 3π ≈ 0.575 rad,</p><p>sin(10) = sin(3π + (10-3π)) = -sin(10-3π) [using sin(3π + θ) = -sin θ]</p><p>Since 10 - 3π ≈ 0.575 ∈ (0, π/2), we have x = sin⁻¹(-sin(10-3π)) = -(10-3π) = 3π - 10</p><p><strong>Step 2:</strong> Find y = cos⁻¹(cos 10)</p><p>Since 10 = 3π + (10-3π) and 10-3π ≈ 0.575 ∈ [0, π],</p><p>cos(10) = cos(3π + (10-3π)) = -cos(10-3π) [using cos(3π + θ) = -cos θ]</p><p>But we need the value in [0, π]. Since cos(10) = -cos(10-3π), and 10-3π ∈ (0, π/2),</p><p>y = cos⁻¹(-cos(10-3π)) = π - (10-3π) = 4π - 10</p><p><strong>Step 3:</strong> Calculate y - x</p><p>y - x = (4π - 10) - (3π - 10) = 4π - 10 - 3π + 10 = π</p><p>∴ Answer: y - x = π</p>
Correct Answer: D