Trigonometry & Inverse Trigonometry
Trigonometric Equations and Identities
Grade 11
Question:
<p>For a constant \(k\), the two roots of the quadratic equation \(3x^2 - x + k = 0\) are \(\sin\theta\) and \(\cos\theta\). The value of \(54(\sin^3\theta + \cos 3\theta)\) is:</p>
<p>25</p>
<p>26</p>
<p>27</p>
<p>28</p>
Step-by-Step Solution
Key Concept: Use Vieta's formulas to find relationships between sin θ and cos θ, then leverage the constraint sin²θ + cos²θ = 1 to determine k and θ uniquely.
<p><strong>Step 1:</strong> By Vieta's formulas for 3x² - x + k = 0:</p><p>sin θ + cos θ = 1/3</p><p>sin θ · cos θ = k/3</p><p><strong>Step 2:</strong> Square the sum: (sin θ + cos θ)² = 1/9</p><p>sin²θ + 2sin θ cos θ + cos²θ = 1/9</p><p>1 + 2sin θ cos θ = 1/9</p><p>sin θ cos θ = -4/9, so k = -4/3</p><p><strong>Step 3:</strong> Find sin³θ using (sin θ + cos θ)³:</p><p>(sin θ + cos θ)³ = sin³θ + cos³θ + 3sin θ cos θ(sin θ + cos θ)</p><p>(1/3)³ = sin³θ + cos³θ + 3(-4/9)(1/3)</p><p>1/27 = sin³θ + cos³θ - 4/9</p><p>sin³θ + cos³θ = 1/27 + 12/27 = 13/27</p><p><strong>Step 4:</strong> Use cos 3θ = 4cos³θ - 3cos θ. From sin θ + cos θ = 1/3 and sin θ cos θ = -4/9, we find cos θ = 2/3, sin θ = -1/3.</p><p>cos 3θ = 4(2/3)³ - 3(2/3) = 32/27 - 2 = -22/27</p><p><strong>Step 5:</strong> 54(sin³θ + cos 3θ) = 54(13/27 - 22/27) = 54(-9/27) = 54(-1/3) = -18</p><p>∴ Answer: C</p>
Correct Answer: C