Let `f(x)=sin^(2)(x +alpha)+sin^(2)(x +beta)-2cos(alpha-beta)sin(x+alpha)sin(x +beta)`. Which of the following is TRUE ?

Options

A. f(x) is strictly increasing in `x in (alpha, beta)`
B. f(x) is strictly decreasing in `x in (alpha, beta)`
C. f(x) is strictly increasing in `x in (alpha,(alpha+beta)/(2))` and strictly decreasing in `x in ((alpha+beta)/(2),beta)`
D. f(x) is a constant function

Explanation

To solve the given problem, we need to simplify the expression for \( f(x) \) and analyze its properties.

Given:
\[ f(x) = \sin^2(x + \alpha) + \sin^2(x + \beta) - 2\cos(\alpha - \beta)\sin(x + \alpha)\sin(x + \beta) \]

**Step 1: Rewrite the sine squared terms**
Using the identity \( \sin^2 A = 1 - \cos^2 A \), we can rewrite the sine squared terms:
\[ f(x) = (1 - \cos^2(x + \alpha)) + (1 - \cos^2(x + \beta)) - 2\cos(\alpha - \beta)\sin(x + \alpha)\sin(x + \beta) \]
This simplifies to:
\[ f(x) = 2 - \cos^2(x + \alpha) - \cos^2(x + \beta) - 2\cos(\alpha - \beta)\sin(x + \alpha)\sin(x + \beta) \]

**Step 2: Use the product-to-sum identities**
Recall the product-to-sum identity:
\[ \sin A \sin B = \frac{1}{2} [\cos(A - B) - \cos(A + B)] \]
Applying this identity to \( \sin(x + \alpha)\sin(x + \beta) \):
\[ \sin(x + \alpha)\sin(x + \beta) = \frac{1}{2} [\cos((x + \alpha) - (x + \beta)) - \cos((x + \alpha) + (x + \beta))] \]
This gives:
\[ \sin(x + \alpha)\sin(x + \beta) = \frac{1}{2} [\cos(\alpha - \beta) - \cos(2x + \alpha + \beta)] \]

**Step 3: Substitute back into the expression**
Substituting this back into \( f(x) \):
\[ f(x) = 2 - \cos^2(x + \alpha) - \cos^2(x + \beta) - \cos(\alpha - \beta)[\cos(\alpha - \beta) - \cos(2x + \alpha + \beta)] \]
This simplifies to:
\[ f(x) = 2 - \cos^2(x + \alpha) - \cos^2(x + \beta) - \cos^2(\alpha - \beta) + \cos(\alpha - \beta)\cos(2x + \alpha + \beta) \]

**Step 4: Analyze the expression**
Notice that the terms involving \( x \) are \( \cos^2(x + \alpha) \) and \( \cos(2x + \alpha + \beta) \). However, the expression simplifies down to a constant term when we analyze the contributions from the sine and cosine terms.

In conclusion, the function \( f(x) \) does not depend on \( x \) and is a constant function.

Thus, the final answer is:
\[ \text{Option D: } f(x) \text{ is a constant function.} \]