Question

Let $f(x) = \sin x \sin(x + \frac{\pi}{3}), x \in \mathbb{R}$. Then the minimum value of $f$ is equal to

• A. $\frac{1}{4}$
• B. $\frac{-1}{4}$
• C. $\frac{3}{4}$
• D. $\frac{-3}{4}$
• E. $\frac{3}{2}$

Hint:

Any function of the form \( a + b \cos \theta \) has a minimum value of \( a - |b| \). Transforming products into sums makes this property immediately applicable.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
To find the minimum value of a product of sine functions, it is best to use product-to-sum formulas to convert it into a single trigonometric expression with a constant offset.
Key Formula or Approach:
Use the identity: \( 2\sin A \sin B = \cos(A-B) - \cos(A+B) \).

Step 2: Detailed Explanation:
Multiply and divide the function by 2:
\[ f(x) = \frac{1}{2} \left[ 2 \sin x \sin(x + \frac{\pi}{3}) \right] \]
Apply the identity with \( A = x + \frac{\pi}{3} \) and \( B = x \):
\[ f(x) = \frac{1}{2} \left[ \cos(x + \frac{\pi}{3} - x) - \cos(x + \frac{\pi}{3} + x) \right] \]
\[ f(x) = \frac{1}{2} \left[ \cos(\frac{\pi}{3}) - \cos(2x + \frac{\pi}{3}) \right] \]
Substitute \( \cos(\frac{\pi}{3}) = \frac{1}{2} \):
\[ f(x) = \frac{1}{4} - \frac{1}{2} \cos(2x + \frac{\pi}{3}) \]
To minimize $f(x)$, we need to subtract the largest possible value of the cosine term.
The maximum value of \( \cos(2x + \frac{\pi}{3}) \) is 1.
\[ \text{Minimum } f = \frac{1}{4} - \frac{1}{2}(1) = \frac{1}{4} - \frac{2}{4} = -\frac{1}{4} \]

Step 3: Final Answer:
The minimum value of the function is $-\frac{1}{4}$.