Question

Find minimum value of \( \sin x \sin \left( x + \frac{\pi}{3} \right) \)

• A. \( \frac{1}{2} \)
• B. 0
• C. 1
• D. \( \frac{\sqrt{3}}{2} \)

Hint:

Using trigonometric identities like the product-to-sum identity helps simplify expressions and find the minimum or maximum values more easily.

Answer 1

The Correct Option is B

Step 1: Use trigonometric identities.
We start with the product-to-sum identity for sine: \[ \sin A \sin B = \frac{1}{2} \left[ \cos(A - B) - \cos(A + B) \right] \] For the given expression, \( A = x \) and \( B = x + \frac{\pi}{3} \), so applying the identity: \[ \sin x \sin \left( x + \frac{\pi}{3} \right) = \frac{1}{2} \left[ \cos \left( x - \left( x + \frac{\pi}{3} \right) \right) - \cos \left( x + \left( x + \frac{\pi}{3} \right) \right) \right] \] This simplifies to: \[ \sin x \sin \left( x + \frac{\pi}{3} \right) = \frac{1}{2} \left[ \cos \left( - \frac{\pi}{3} \right) - \cos \left( 2x + \frac{\pi}{3} \right) \right] \]
Step 2: Find the minimum value.
We know that the cosine function has a range of [-1, 1], so the minimum value of \( \cos \left( 2x + \frac{\pi}{3} \right) \) is -1. Therefore, the minimum value of the given expression is: \[ \frac{1}{2} \left[ \frac{1}{2} - (-1) \right] = 0 \]
Final Answer: 0.