Let `f(x)=(sinx+3sin3x+5sin5x+3sin7x)/(sin2x+2sin4x+3sin6x)`, wherever defined. If `x_(1)+x_(2)=(pi)/(2)`, where `f(x)` is defined at `x_(1) and x_(2)`, then `f^(2)(x_(1))+f^(2)(x_(2))` is

Options

A. `cos^(2)x`
B. `sin^(2)x`
C. 4
D. 1

Answer

To solve the problem, we need to find \( f^2(x_1) + f^2(x_2) \) where \( f(x) = \frac{\sin x + 3\sin 3x + 5\sin 5x + 3\sin 7x}{\sin 2x + 2\sin 4x + 3\sin 6x} \) and \( x_1 + x_2 = \frac{\pi}{2} \).

### Step 1: Simplify \( f(x) \)

First, we will simplify the function \( f(x) \).

**Numerator:**
\[ \sin x + 3\sin 3x + 5\sin 5x + 3\sin 7x \]
We can group the terms:
\[ = \sin x + 3(\sin 3x + \sin 7x) + 5\sin 5x \]
Using the sine addition formula, we can express \( \sin 3x + \sin 7x \):
\[ \sin 3x + \sin 7x = 2\sin(5x)\cos(2x) \]
Thus, the numerator becomes:
\[ \sin x + 6\sin 5x \cos 2x + 5\sin 5x \]
This simplifies to:
\[ \sin x + 5\sin 5x + 6\sin 5x \cos 2x \]

**Denominator:**
\[ \sin 2x + 2\sin 4x + 3\sin 6x \]
Using the sine addition formula again, we can express \( \sin 4x + \sin 6x \):
\[ \sin 4x + \sin 6x = 2\sin(5x)\cos(x) \]
Thus, the denominator becomes:
\[ \sin 2x + 2(2\sin 5x \cos x) \]
This simplifies to:
\[ \sin 2x + 4\sin 5x \cos x \]

### Step 2: Final Simplification of \( f(x) \)

Now, we can express \( f(x) \):
\[ f(x) = \frac{\sin x + 6\sin 5x \cos 2x + 5\sin 5x}{\sin 2x + 4\sin 5x \cos x} \]

### Step 3: Evaluate \( f(x_1) \) and \( f(x_2) \)

Given \( x_1 + x_2 = \frac{\pi}{2} \), we can use the identity:
\[ \cos\left(\frac{\pi}{2} - x\right) = \sin x \]
Thus,
\[ f(x_2) = f\left(\frac{\pi}{2} - x_1\right) = 2\sin x_1 \]

### Step 4: Calculate \( f^2(x_1) + f^2(x_2) \)

Now we can calculate \( f^2(x_1) + f^2(x_2) \):
\[ f(x_1) = 2\cos x_1 \]
\[ f(x_2) = 2\sin x_1 \]
Thus,
\[ f^2(x_1) + f^2(x_2) = (2\cos x_1)^2 + (2\sin x_1)^2 \]
\[ = 4\cos^2 x_1 + 4\sin^2 x_1 \]
Using the Pythagorean identity \( \sin^2 x + \cos^2 x = 1 \):
\[ = 4(\cos^2 x_1 + \sin^2 x_1) = 4 \cdot 1 = 4 \]

### Final Answer
\[ f^2(x_1) + f^2(x_2) = 4 \]