Trigonometry & Inverse Trigonometry
Double Angles
Grade 11

Question:

<p>Maximum value of <span>\(y = 4\sin^2 \theta + 4\sin \theta \cos \theta + \cos^2 \theta\)</span> is</p>
<p>(P) 1</p>
<p>(Q) 0</p>
<p>(R) 7/8</p>
<p>(S) 5</p>
<p>(T) 6</p>

Step-by-Step Solution

Key Concept: Express the function entirely in terms of sin(2θ) and cos(2θ) using double angle formulas, then find the maximum using calculus or algebraic manipulation of the resulting linear combination.
<p><strong>Step 1:</strong> Rewrite the expression using basic identities.</p><p>y = 4sin²θ + 4sinθcosθ + cos²θ</p><p>y = 4sin²θ + 2(2sinθcosθ) + cos²θ</p><p>y = 4sin²θ + 2sin(2θ) + cos²θ</p><p><strong>Step 2:</strong> Express in terms of double angles. Use sin²θ = (1-cos2θ)/2 and cos²θ = (1+cos2θ)/2.</p><p>y = 4·(1-cos2θ)/2 + 2sin(2θ) + (1+cos2θ)/2</p><p>y = 2(1-cos2θ) + 2sin(2θ) + (1+cos2θ)/2</p><p>y = 2 - 2cos2θ + 2sin(2θ) + 1/2 + cos2θ/2</p><p>y = 5/2 - 3cos2θ/2 + 2sin(2θ)</p><p><strong>Step 3:</strong> Rearrange to standard form.</p><p>y = 5/2 + 2sin(2θ) - (3/2)cos(2θ)</p><p><strong>Step 4:</strong> Find the maximum of the expression a·sin(2θ) + b·cos(2θ). The maximum value of Asinφ + Bcosφ is √(A² + B²).</p><p>Here A = 2 and B = -3/2.</p><p>Maximum of [2sin(2θ) - (3/2)cos(2θ)] = √(4 + 9/4) = √(25/4) = 5/2</p><p><strong>Step 5:</strong> Calculate the overall maximum.</p><p>y_max = 5/2 + 5/2 = 5</p><p><strong>∴ Answer:</strong> T</p>
Correct Answer: T

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