Trigonometry & Inverse Trigonometry
Trigonometric values
Grade 11
Question:
<p>If \(\alpha = \dfrac{\pi}{3}\), find the value of \(\dfrac{\cos 2\alpha + \sec\alpha + 3\sqrt{3}}{\tan\alpha}\).</p>
<p>(a) \(\dfrac{\sqrt{3}}{2} + 1\)</p>
<p>(b) \(\dfrac{\sqrt{3}}{2} + 2\)</p>
<p>(c) \(\sqrt{3} + 2\)</p>
<p>(d) \(\dfrac{\sqrt{3}}{2} + 3\)</p>
Step-by-Step Solution
Key Concept: Substitute α = π/3 directly into the expression and evaluate using exact trigonometric values: cos(π/3) = 1/2, sec(π/3) = 2, tan(π/3) = √3, and cos(2π/3) = -1/2.
<p><strong>Step 1:</strong> Identify exact values for α = π/3:</p><p>• cos(π/3) = 1/2</p><p>• sec(π/3) = 1/cos(π/3) = 2</p><p>• tan(π/3) = √3</p><p>• cos(2π/3) = cos(2·π/3) = 2cos²(π/3) - 1 = 2(1/4) - 1 = <strong>-1/2</strong></p><p><strong>Step 2:</strong> Substitute into the numerator:</p><p>cos 2α + sec α + 3√3 = -1/2 + 2 + 3√3</p><p>= 3/2 + 3√3</p><p>= (3 + 6√3)/2</p><p><strong>Step 3:</strong> Divide by tan α:</p><p>Numerator/tan α = [(3 + 6√3)/2] ÷ √3</p><p>= (3 + 6√3)/(2√3)</p><p>= 3/(2√3) + 6√3/(2√3)</p><p>= (3√3)/6 + 3</p><p>= √3/2 + 3</p><p>= <strong>(6 + √3)/2</strong></p><p>∴ Answer: D</p>
Correct Answer: D