Trigonometry & Inverse Trigonometry
Inverse Trigonometric Functions
Grade 12
Question:
<p>If <span class="math">\sin^{-1}: [-1,1] \to \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]</span> and <span class="math">\cos^{-1}: [-1,1] \to [0, \pi]</span> be two bijective functions, respectively inverses of bijective functions <span class="math">\sin: \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \to [-1,1]</span> and <span class="math">\cos: [0,\pi] \to [-1,1]</span>, then <span class="math">\sin^{-1}x + \cos^{-1}x</span> is</p>
<p>(a) <span class="math">\frac{\pi}{2}</span></p>
<p>(b) <span class="math">\pi</span></p>
<p>(c) <span class="math">\left[0, \frac{3\pi}{4}\right]</span></p>
<p>(d) not a constant</p>
Step-by-Step Solution
Key Concept: Use the fundamental identity relating inverse sine and cosine functions: for any x in [-1,1], the angles sin⁻¹(x) and cos⁻¹(x) are complementary angles that sum to π/2.
<p><strong>Step 1:</strong> Let sin⁻¹(x) = α, where α ∈ [-π/2, π/2].</p><p>Then by definition: sin(α) = x</p><p><strong>Step 2:</strong> Let cos⁻¹(x) = β, where β ∈ [0, π].</p><p>Then by definition: cos(β) = x</p><p><strong>Step 3:</strong> From sin(α) = x and cos(β) = x, we have sin(α) = cos(β).</p><p><strong>Step 4:</strong> Using the cofunction identity: sin(α) = cos(π/2 - α).</p><p>Since sin(α) = cos(β), we can write: cos(π/2 - α) = cos(β).</p><p><strong>Step 5:</strong> For α ∈ [-π/2, π/2], we have (π/2 - α) ∈ [0, π].</p><p>Since cosine is one-to-one on [0, π], and both (π/2 - α) and β lie in this interval:</p><p>β = π/2 - α</p><p><strong>Step 6:</strong> Therefore: sin⁻¹(x) + cos⁻¹(x) = α + β = α + (π/2 - α) = π/2</p><p><strong>Step 7:</strong> This result is independent of x for all x ∈ [-1, 1].</p><p><strong>Verification:</strong> Check with x = 0: sin⁻¹(0) + cos⁻¹(0) = 0 + π/2 = π/2 ✓</p><p>Check with x = 1: sin⁻¹(1) + cos⁻¹(1) = π/2 + 0 = π/2 ✓</p><p>∴ Answer: A</p>
Correct Answer: A