Understanding the Trigonometric Expression

The problem asks for the maximum value of the function:

$$ f(x) = 4\sin^2 x + 3\cos^2 x + \sin \frac x 2 + \cos \frac x 2 $$

To find the maximum value, we will break down the function into parts and analyze each part separately.

Simplifying the \( \sin^2 x \) and \( \cos^2 x \) Component

Consider the first part of the expression: \( 4\sin^2 x + 3\cos^2 x \).

We can rewrite this using the trigonometric identity \( \cos^2 x = 1 - \sin^2 x \):

$$ 4\sin^2 x + 3\cos^2 x = 4\sin^2 x + 3(1 - \sin^2 x) $$

Distributing the 3:

$$ = 4\sin^2 x + 3 - 3\sin^2 x $$

Combining like terms:

$$ = (4\sin^2 x - 3\sin^2 x) + 3 $$ $$ = \sin^2 x + 3 $$

The value of \( \sin x \) ranges from -1 to 1. Therefore, the value of \( \sin^2 x \) ranges from 0 to 1.

The maximum value of \( \sin^2 x \) is 1.

Consequently, the maximum value of \( 3 + \sin^2 x \) is \( 3 + 1 = 4 \).

Finding the Maximum of \( \sin \frac x 2 + \cos \frac x 2 \)

Now, let's analyze the second part of the expression: \( \sin \frac x 2 + \cos \frac x 2 \).

This is in the form \( a\sin\theta + b\cos\theta \), where \( a=1 \), \( b=1 \), and \( \theta = \frac x 2 \). This form can be converted to \( R\sin(\theta + \alpha) \) or \( R\cos(\theta - \alpha) \).

We use the formula \( a\sin\theta + b\cos\theta = R\cos(\theta - \alpha) \), where \( R = \sqrt{a^2 + b^2} \).

Calculate \( R \):

$$ R = \sqrt{1^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2} $$

Now, rewrite the expression:

$$ \sin \frac x 2 + \cos \frac x 2 = \sqrt{2} \left( \frac{1}{\sqrt{2}} \sin \frac x 2 + \frac{1}{\sqrt{2}} \cos \frac x 2 \right) $$

Since \( \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}} \) and \( \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}} \), we have:

$$ = \sqrt{2} \left( \sin \frac x 2 \cos \frac{\pi}{4} + \cos \frac x 2 \sin \frac{\pi}{4} \right) $$

Using the sine addition identity \( \sin(A+B) = \sin A \cos B + \cos A \sin B \):

$$ = \sqrt{2} \sin \left( \frac x 2 + \frac{\pi}{4} \right) $$

The maximum value of the sine function is 1.

Therefore, the maximum value of \( \sqrt{2} \sin \left( \frac x 2 + \frac{\pi}{4} \right) \) is \( \sqrt{2} \times 1 = \sqrt{2} \).

Determining the Overall Maximum Value

The original function is the sum of the two parts we analyzed: \( f(x) = (3 + \sin^2 x) + (\sin \frac x 2 + \cos \frac x 2) \).

We found that the maximum value of the first part (\( 3 + \sin^2 x \)) is 4, and the maximum value of the second part (\( \sin \frac x 2 + \cos \frac x 2 \)) is \( \sqrt{2} \).

To find the maximum value of the sum, we need to see if these maximum values can occur simultaneously for the same value of \( x \).

The maximum value of \( 3 + \sin^2 x \) occurs when \( \sin^2 x = 1 \). This happens when \( x = \frac{\pi}{2} + k\pi \) for any integer \( k \).

The maximum value of \( \sin \frac x 2 + \cos \frac x 2 \) occurs when \( \sin \left( \frac x 2 + \frac{\pi}{4} \right) = 1 \). This requires:

$$ \frac x 2 + \frac{\pi}{4} = \frac{\pi}{2} + 2m\pi \quad \text{(where } m \text{ is an integer)} $$

Solving for \( x \):

$$ \frac x 2 = \frac{\pi}{2} - \frac{\pi}{4} + 2m\pi $$ $$ \frac x 2 = \frac{\pi}{4} + 2m\pi $$ $$ x = \frac{\pi}{2} + 4m\pi $$

Consider the value \( x = \frac{\pi}{2} \). This value satisfies both conditions:

• When \( x = \frac{\pi}{2} \), \( \sin^2 x = \sin^2(\frac{\pi}{2}) = 1^2 = 1 \). The first part is \( 3 + 1 = 4 \).
• When \( x = \frac{\pi}{2} \), \( \sin \frac x 2 + \cos \frac x 2 = \sin(\frac{\pi}{4}) + \cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \). The second part is \( \sqrt{2} \).

Since both maximums can be achieved at \( x = \frac{\pi}{2} \), the maximum value of the entire function is the sum of the individual maximums.

$$ \text{Maximum Value} = (\text{Max of } 3 + \sin^2 x) + (\text{Max of } \sin \frac x 2 + \cos \frac x 2) $$ $$ \text{Maximum Value} = 4 + \sqrt{2} $$