Question

The maximum value of \[ E = 16\sin\frac{x}{2}\cos^3\frac{x}{2} \] where \(x\in[0,\pi]\), is

• A. \(3\)
• B. \(3\sqrt{3}\)
• C. \(6\sqrt{3}\)
• D. \(6\)

Hint:

When trigonometric powers appear, reduce them using identities before differentiating.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
To find the maximum value of a trigonometric expression, we can use the derivative method or transform the expression into a form where the Arithmetic Mean-Geometric Mean (AM-GM) inequality or standard identities apply.
Step 2: Key Formula or Approach:
Let \( \theta = x/2 \). Then \( E = 16 \sin \theta \cos^3 \theta \). Since \( x \in [0, \pi] \), \( \theta \in [0, \pi/2] \), so \( \sin \theta \) and \( \cos \theta \) are non-negative.
Step 3: Detailed Explanation:
1. To use AM-GM on \( \sin^2 \theta \cdot \cos^6 \theta \), we split \( \cos^6 \theta \) into three equal parts: \[ E^2 = 16^2 \cdot \sin^2 \theta \cdot \cos^2 \theta \cdot \cos^2 \theta \cdot \cos^2 \theta \] To match the sum \( \sin^2 \theta + \cos^2 \theta = 1 \), we write: \[ E^2 = 16^2 \cdot 3^3 \cdot \left[ \sin^2 \theta \cdot \frac{\cos^2 \theta}{3} \cdot \frac{\cos^2 \theta}{3} \cdot \frac{\cos^2 \theta}{3} \right] \] 2. Apply AM-GM: \[ \left( \frac{\sin^2 \theta + \frac{\cos^2 \theta}{3} + \frac{\cos^2 \theta}{3} + \frac{\cos^2 \theta}{3}}{4} \right)^4 \geq \sin^2 \theta \cdot \left(\frac{\cos^2 \theta}{3}\right)^3 \] \[ \left( \frac{\sin^2 \theta + \cos^2 \theta}{4} \right)^4 \geq \frac{E^2}{16^2 \cdot 27} \implies \left(\frac{1}{4}\right)^4 \geq \frac{E^2}{256 \cdot 27} \] \[ \frac{1}{256} \geq \frac{E^2}{256 \cdot 27} \implies E^2 \leq 27 \] 3. Maximum value \( E = \sqrt{27} = 3\sqrt{3} \).
Step 4: Final Answer:
The maximum value of E is \( 3\sqrt{3} \).