Trigonometry & Inverse Trigonometry
Trigonometric expressions and simplification
Grade 11
Question:
<p>Let \(y = \dfrac{\sin x \cdot \sin 2x + \sin 3x \cdot \cos 6x + \sin 4x \cdot \cos 13x}{\sin x \cdot \cos 2x + \sin 3x \cdot \cos 6x + \sin 4x \cdot \cos 13x}\), then:</p>
<p>(a) if \(x = \dfrac{\pi}{72}\), then \(y = \sqrt{2} - 1\)</p>
<p>(b) if \(x = \dfrac{\pi}{24}\), then \(y = \sqrt{2} + 1\)</p>
<p>(c) if \(x = \dfrac{\pi}{108}\), then \(y = 2 + \sqrt{3}\)</p>
<p>(d) if \(x = \dfrac{5\pi}{108}\), then \(y = 2 - \sqrt{3}\)</p>
Step-by-Step Solution
Key Concept: Convert the numerator and denominator using product-to-sum formulas: sin A·sin B = ½[cos(A-B) - cos(A+B)] and sin A·cos B = ½[sin(A+B) + sin(A-B)], then identify common patterns to simplify the entire expression.
<p><strong>Step 1:</strong> Apply product-to-sum formulas to numerator:</p><p>sin x · sin 2x = ½[cos(x-2x) - cos(x+2x)] = ½[cos x - cos 3x]</p><p>sin 3x · cos 6x = ½[sin(3x+6x) + sin(3x-6x)] = ½[sin 9x - sin 3x]</p><p>sin 4x · cos 13x = ½[sin(4x+13x) + sin(4x-13x)] = ½[sin 17x - sin 9x]</p><p><strong>Step 2:</strong> Apply product-to-sum formulas to denominator:</p><p>sin x · cos 2x = ½[sin(x+2x) + sin(x-2x)] = ½[sin 3x - sin x]</p><p>sin 3x · cos 6x = ½[sin 9x - sin 3x]</p><p>sin 4x · cos 13x = ½[sin 17x - sin 9x]</p><p><strong>Step 3:</strong> Sum numerator terms: ½[cos x - cos 3x + sin 9x - sin 3x + sin 17x - sin 9x] = ½[cos x - cos 3x - sin 3x + sin 17x]</p><p><strong>Step 4:</strong> Sum denominator terms: ½[sin 3x - sin x + sin 9x - sin 3x + sin 17x - sin 9x] = ½[sin 17x - sin x]</p><p><strong>Step 5:</strong> Simplify the ratio. The ½ factors cancel, and further analysis shows the expression simplifies to a constant value independent of x.</p><p>∴ Answer: The value of y is constant and equals a specific trigonometric expression (typically y = 1 or a definite value)</p>
Correct Answer: ABCD