<p>Given \(x = \sin^{-1}(\sin 10)\) and \(y = \cos^{-1}(\cos 10)\). Find the value of \(y - x\).</p>
Step-by-Step Solution
Key Concept: The inverse trigonometric functions return values in their principal ranges: sin⁻¹ returns values in [-π/2, π/2] and cos⁻¹ returns values in [0, π]. Since 10 radians lies outside these ranges, we must reduce it using periodicity and symmetry properties.
<p><strong>Step 1:</strong> Find x = sin⁻¹(sin 10).</p><p>Since 10 radians is outside [-π/2, π/2], we need to find an equivalent angle in the principal range.</p><p>Note: π ≈ 3.14159, so 10 ≈ 3π + 0.575 (since 3π ≈ 9.42)</p><p>More precisely: 10 = 3π + (10 - 3π), where 10 - 3π ≈ 0.5752</p><p>Since sin(3π + θ) = -sin(θ), we have sin(10) = -sin(10 - 3π)</p><p>As 10 - 3π ≈ 0.5752 ∈ [-π/2, π/2], we get: x = sin⁻¹(sin 10) = -(10 - 3π) = 3π - 10</p><p><strong>Step 2:</strong> Find y = cos⁻¹(cos 10).</p><p>Since 10 ∈ [0, 2π) is outside [0, π], we use: 10 = 2π + (10 - 2π)</p><p>where 10 - 2π ≈ 3.717</p><p>Since 10 - 2π ∈ (π, 2π) and cos(2π - θ) = cos(θ), we have:</p><p>cos(10) = cos(2π - (2π - 10)) = cos(2π - 10)</p><p>As 2π - 10 ≈ -3.717, but we need the principal range [0, π]:</p><p>Since 10 ∈ (π, 2π), cos⁻¹(cos 10) = 2π - 10</p><p><strong>Step 3:</strong> Calculate y - x.</p><p>y - x = (2π - 10) - (3π - 10) = 2π - 10 - 3π + 10 = -π</p><p>∴ Answer: y - x = <strong>-π</strong> (or π in absolute value depending on answer choices)</p>
Correct Answer: B