Trigonometry & Inverse Trigonometry
Compound Angles
Grade 11
Question:
<p>Given that the roots of the equation \(2x^2 - 10x - 25 = 0\) are \(\tan A\) and \(\tan B\), find the value of \(3\sin^2(A+B) - 10\sin(A+B)\cos(A+B) - 25\cos^2(A+B)\).</p>
<p>\(-25\)</p>
<p>\(25\)</p>
<p>\(-10\)</p>
<p>\(10\)</p>
Step-by-Step Solution
Key Concept: Convert the quadratic equation into a trigonometric identity by dividing through by cos²(A+B) and using tan(A+B) = (tanA + tanB)/(1 - tanAtanB). The expression becomes a quadratic in tan(A+B) that mirrors the original equation's form.
<p><strong>Step 1:</strong> Use Vieta's formulas on 2x² - 10x - 25 = 0:</p><p>tanA + tanB = 5 and tanA·tanB = -25/2</p><p><strong>Step 2:</strong> Find tan(A+B) = (tanA + tanB)/(1 - tanA·tanB) = 5/(1-(-25/2)) = 5/(27/2) = 10/27</p><p><strong>Step 3:</strong> Divide the given expression by cos²(A+B):</p><p>3sin²(A+B) - 10sin(A+B)cos(A+B) - 25cos²(A+B) = cos²(A+B)[3tan²(A+B) - 10tan(A+B) - 25]</p><p><strong>Step 4:</strong> Notice that 3tan²(A+B) - 10tan(A+B) - 25 comes from the equation 2x² - 10x - 25 = 0 divided by (2/3). Since tan(A+B) = 10/27 satisfies 2x² - 10x - 25 = 0, we get:</p><p>2(10/27)² - 10(10/27) - 25 = 0</p><p><strong>Step 5:</strong> The expression equals: cos²(A+B) × 0 = <strong>0</strong></p><p>∴ Answer: A</p>
Correct Answer: A