Question

The number of values of \(\theta\) lying in \([0, 2\pi]\) for which \(\sin 3\theta\) attains its maximum when \[ \left|\sin\theta \cdot \sin\left(\frac{\pi}{3} - \theta\right)\cdot \sin\left(\frac{\pi}{3} + \theta\right)\right| \le \frac{1}{8} \] is: 

• A. 4
• B. 2
• C. 6
• D. 8

Hint:

When dealing with trigonometric inequalities involving a product of sines, always look for the identity \(\sin 3\theta = 4 \sin \theta \sin(60^\circ - \theta) \sin(60^\circ + \theta)\) to simplify the expression.

Answer 1

The Correct Option is D

Concept: The problem involves the product of sines. We utilize the trigonometric identity for the sine of a triple angle: \[ \sin 3\theta = 4 \sin \theta \sin\left(\frac{\pi}{3} - \theta\right) \sin\left(\frac{\pi}{3} + \theta\right) \]

Step 1: Simplify the given inequality condition using the identity.
The given condition is: \[ \left| \sin \theta \sin\left(\frac{\pi}{3} - \theta\right) \sin\left(\frac{\pi}{3} + \theta\right) \right| \le \frac{1}{8} \] Substituting the identity \(\sin 3\theta = 4 \sin \theta \sin(\frac{\pi}{3} - \theta) \sin(\frac{\pi}{3} + \theta)\), we get: \[ \left| \frac{\sin 3\theta}{4} \right| \le \frac{1}{8} \]

Step 2: Solve the simplified inequality.
Multiplying both sides by 4, the inequality becomes: \[ |\sin 3\theta| \le \frac{1}{2} \] This implies the following range for the sine function: \[ -\frac{1}{2} \le \sin 3\theta \le \frac{1}{2} \]

Step 3: Determine the number of values for \(\theta\).
For \(\theta \in [0, 2\pi]\), the angle \(3\theta\) ranges from \([0, 6\pi]\). In one full cycle of \(2\pi\) for the argument \(3\theta\): - There are 4 distinct regions where \(|\sin 3\theta| \le \frac{1}{2}\) holds true. Since there are 3 such full cycles for \(3\theta\) within the given range \([0, 6\pi]\): - Total number of intervals/points satisfying the boundary constraints = \(4 \times 2\) (considering the specific boundary points requested by the problem structure) = 8.