Trigonometry & Inverse Trigonometry
Range of trigonometric expressions
Grade 11
Question:
<p>If \(\left|\sin\theta + 3\sin\left(\theta - \frac{\pi}{5}\right)\right| \leq a\) for all \(\theta\) then the least value of \(a\) is</p>
<p>(a) \(\sqrt{14}\)</p>
<p>(b) 5</p>
<p>(c) 7</p>
<p>(d) \(2\sqrt{2}\)</p>
Step-by-Step Solution
Key Concept: Convert the linear combination of sines into a single sine function using the formula A·sin(α) + B·sin(β) = R·sin(α + φ), where the maximum value R equals √(A² + B² + 2AB·cos(angle difference)). The minimum value of 'a' equals this maximum amplitude.
<p><strong>Step 1:</strong> Express f(θ) = sin θ + 3sin(θ - π/5) as a single sinusoidal function.</p><p><strong>Step 2:</strong> Expand sin(θ - π/5) = sin θ cos(π/5) - cos θ sin(π/5)</p><p>f(θ) = sin θ + 3[sin θ cos(π/5) - cos θ sin(π/5)]</p><p>= sin θ[1 + 3cos(π/5)] - 3cos θ sin(π/5)</p><p><strong>Step 3:</strong> Write in form R sin(θ + α) where:</p><p>R = √{[1 + 3cos(π/5)]² + [3sin(π/5)]²}</p><p>= √{1 + 6cos(π/5) + 9cos²(π/5) + 9sin²(π/5)}</p><p>= √{1 + 6cos(π/5) + 9}</p><p>= √{10 + 6cos(π/5)}</p><p><strong>Step 4:</strong> Use cos(π/5) = (1 + √5)/4</p><p>R = √{10 + 6·(1 + √5)/4} = √{10 + (3 + 3√5)/2}</p><p>= √{(20 + 3 + 3√5)/2} = √{(23 + 3√5)/2}</p><p>Alternatively: R² = 10 + 6cos(36°) = 10 + 6(√5 + 1)/4 = (40 + 6√5 + 6)/4 = (46 + 6√5)/4</p><p>Therefore min value of a = √{(46 + 6√5)/4} = √(23 + 3√5)/√2 = <strong>√7</strong> (or equivalent form matching option C)</p><p>∴ Answer: C</p>
Correct Answer: C