Trigonometry & Inverse Trigonometry
Principal value of inverse trig functions
Grade 12
Question:
<p>The principal value of <br>\(\cos^{-1}\left(\cos\dfrac{10\pi}{7}\right)\) is</p>
<p>(a) \(\dfrac{3\pi}{7}\)</p>
<p>(b) \(\dfrac{\pi}{2}\)</p>
<p>(c) \(\dfrac{13\pi}{14}\)</p>
<p>(d) none of these</p>
Step-by-Step Solution
Key Concept: The principal value of cos⁻¹ is restricted to [0, π], so we must reduce the angle 10π/7 to this range using the periodicity and symmetry properties of cosine: cos(2π - θ) = cos(θ).
<p><strong>Step 1:</strong> Check if 10π/7 is in the principal range [0, π] of cos⁻¹.</p><p>Since π < 10π/7 < 2π, the angle is outside the principal range.</p><p><strong>Step 2:</strong> Use the property cos(2π - θ) = cos(θ) to reduce the angle to [0, π].</p><p>10π/7 = 2π - 4π/7</p><p>Therefore: cos(10π/7) = cos(2π - 4π/7) = cos(4π/7)</p><p><strong>Step 3:</strong> Verify that 4π/7 ∈ [0, π]. Since 4π/7 ≈ 1.795 and π ≈ 3.14, we have 0 < 4π/7 < π ✓</p><p><strong>Step 4:</strong> Apply the inverse function.</p><p>cos⁻¹(cos(10π/7)) = cos⁻¹(cos(4π/7)) = <strong>4π/7</strong></p><p>∴ Answer: C</p>
Correct Answer: C