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 sines, we can use the product-to-sum trigonometric identity to transform it into a linear expression of cosine.
Step 2: Key Formula or Approach:
Use the identity: \(\sin A \sin B = \frac{1}{2}[\cos(A - B) - \cos(A + B)]\).
Identify the maximum value of the resulting variable cosine term to minimize the overall function.
Step 3: Detailed Explanation:
Let \(A = x\) and \(B = x + \frac{\pi}{3}\).
\[ f(x) = \frac{1}{2}\left[\cos\left(x - \left(x + \frac{\pi}{3}\right)\right) - \cos\left(x + x + \frac{\pi}{3}\right)\right] \] \[ f(x) = \frac{1}{2}\left[\cos\left(-\frac{\pi}{3}\right) - \cos\left(2x + \frac{\pi}{3}\right)\right] \] Since cosine is an even function, \(\cos(-\theta) = \cos(\theta)\):
\[ \cos\left(-\frac{\pi}{3}\right) = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \] Substitute this back:
\[ f(x) = \frac{1}{2}\left[\frac{1}{2} - \cos\left(2x + \frac{\pi}{3}\right)\right] \] \[ f(x) = \frac{1}{4} - \frac{1}{2}\cos\left(2x + \frac{\pi}{3}\right) \] To find the minimum value of \(f(x)\), we need the subtracted term to be as large as possible.
The maximum value of \(\cos(2x + \frac{\pi}{3})\) is \(1\).
Substitute 1 for the cosine term:
\[ f_{\text{min}} = \frac{1}{4} - \frac{1}{2}(1) = \frac{1}{4} - \frac{2}{4} = -\frac{1}{4} \] Step 4: Final Answer:
The minimum value is \(-\frac{1}{4}\).