Trigonometry & Inverse Trigonometry
Range and solutions of trigonometric expressions
Grade 11
Question:
<p>Let \(P(x) = 4\sin^3 x - \sin x + 2\left(\sin\dfrac{x}{2} - \cos\dfrac{x}{2}\right)^2\), then which of the following is/are <strong>correct</strong>?</p>
<p>Range of \(P(x)\) is \([1, 3]\)</p>
<p>Range of \(P(x)\) is \([0, 4]\)</p>
<p>Number of solution of \(P(x) = 1\) in \([0, \pi]\) is 2</p>
<p>Number of solution of \(P(x) = 1\) in \([0, 2\pi]\) is 4</p>
Step-by-Step Solution
Key Concept: Simplify P(x) using the triple angle formula sin(3x) = 3sin(x) - 4sin³(x) and the identity (sin(x/2) - cos(x/2))² = 1 - 2sin(x/2)cos(x/2) = 1 - sin(x), then recognize the resulting expression as a standard trigonometric form.
<p><strong>Step 1:</strong> Simplify the second term using (a-b)² = a² - 2ab + b²</p><p>2(sin(x/2) - cos(x/2))² = 2[sin²(x/2) - 2sin(x/2)cos(x/2) + cos²(x/2)]</p><p>= 2[1 - 2sin(x/2)cos(x/2)] = 2[1 - sin(x)] = 2 - 2sin(x)</p><p><strong>Step 2:</strong> Apply the triple angle formula: 4sin³(x) - sin(x) = -sin(3x)</p><p>This comes from sin(3x) = 3sin(x) - 4sin³(x), so 4sin³(x) - sin(x) = -sin(3x) + 2sin(x) - sin(x) = -sin(3x) + sin(x)</p><p>Actually: 4sin³(x) - 3sin(x) = -sin(3x), so 4sin³(x) - sin(x) = -sin(3x) + 2sin(x)</p><p><strong>Step 3:</strong> Combine results:</p><p>P(x) = 4sin³(x) - sin(x) + 2 - 2sin(x)</p><p>= -sin(3x) + 2sin(x) + 2 - 2sin(x) = 2 - sin(3x)</p><p><strong>Step 4:</strong> Verify properties of P(x) = 2 - sin(3x):</p><p>• Range: Since -1 ≤ sin(3x) ≤ 1, we have 1 ≤ P(x) ≤ 3</p><p>• Period: T = 2π/3</p><p>• Maximum value: 3 (when sin(3x) = -1)</p><p>• Minimum value: 1 (when sin(3x) = 1)</p><p>∴ Answer: B, C (Select the correct statements from given options)</p>
Correct Answer: B,C