Trigonometry & Inverse Trigonometry
Trigonometric Equations
Grade 11
Question:
<p>The maximum value of \(5\sin\theta + 3\sin(\theta - \alpha)\) is 7, then the set of all possible values of \(\alpha\) is</p>
<p>(a) \(\left\{2n\pi \pm \frac{\pi}{3}\right\}\)</p>
<p>(b) \(\left\{2n\pi \pm \frac{2\pi}{3}\right\}\)</p>
<p>(c) \(\left[\frac{\pi}{3}, \frac{2\pi}{3}\right]\)</p>
<p>(d) None</p>
Step-by-Step Solution
Key Concept: The maximum value of a linear combination of sine functions asinθ + bsin(θ-α) occurs when the vectors are optimally aligned, and equals √(a² + b² + 2ab·cos(α)). Setting this equal to 7 gives us a constraint equation for α.
<p><strong>Step 1: Express the function using trigonometric identities.</strong></p><p>Let f(θ) = 5sinθ + 3sin(θ - α)</p><p>Expand: f(θ) = 5sinθ + 3(sinθ·cosα - cosθ·sinα)</p><p>= 5sinθ + 3sinθ·cosα - 3cosθ·sinα</p><p>= (5 + 3cosα)sinθ - 3sinα·cosθ</p><p></p><p><strong>Step 2: Find the maximum value using the standard form.</strong></p><p>Any expression of the form A·sinθ + B·cosθ has maximum value √(A² + B²).</p><p>Here: A = (5 + 3cosα) and B = -3sinα</p><p>Maximum = √[(5 + 3cosα)² + (-3sinα)²]</p><p>= √[(5 + 3cosα)² + 9sin²α]</p><p></p><p><strong>Step 3: Expand and simplify.</strong></p><p>= √[25 + 30cosα + 9cos²α + 9sin²α]</p><p>= √[25 + 30cosα + 9(cos²α + sin²α)]</p><p>= √[25 + 30cosα + 9]</p><p>= √[34 + 30cosα]</p><p></p><p><strong>Step 4: Set the maximum equal to 7.</strong></p><p>√[34 + 30cosα] = 7</p><p>34 + 30cosα = 49</p><p>30cosα = 15</p><p>cosα = 1/2</p><p></p><p><strong>Step 5: Solve for α.</strong></p><p>cosα = 1/2</p><p>α = 2nπ ± π/3, where n ∈ ℤ</p><p></p><p><strong>∴ Answer: A</strong></p>
Correct Answer: A