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The image provides the following mathematical problem:

"If the complex number

\[
z = \frac{\cos \frac{x}{2} + i \sin \frac{x}{2}}{\cos x + i \sin \frac{x}{2}}, \quad x \in \left(0, \frac{\pi}{2}\right)
\]

is purely real, then the value of \( \cos^2 x \) is?"

### Solution:

1. The given complex number is:

\[
z = \frac{\cos \frac{x}{2} + i \sin \frac{x}{2}}{\cos x + i \sin \frac{x}{2}}
\]

2. We are told that \( z \) is purely real. This means that the imaginary part of \( z \) must be zero.

To proceed, let's simplify the expression for \( z \):

The numerator is \( \cos \frac{x}{2} + i \sin \frac{x}{2} \), which can be written in exponential form using Euler's formula as:

\[
\cos \frac{x}{2} + i \sin \frac{x}{2} = e^{i \frac{x}{2}}
\]

Similarly, the denominator is \( \cos x + i \sin \frac{x}{2} \), but we cannot directly express this as a single exponential form, so we will work with the real and imaginary parts separately.

3. To find when \( z \) is purely real, multiply the numerator and the denominator by the conjugate of the denominator to eliminate the imaginary part:

\[
z = \frac{\left( \cos \frac{x}{2} + i \sin \frac{x}{2} \right) \left( \cos x - i \sin \frac{x}{2} \right)}{\left( \cos x + i \sin \frac{x}{2} \right) \left( \cos x - i \sin \frac{x}{2} \right)}
\]

The denominator simplifies using the identity \( (a + bi)(a - bi) = a^2 + b^2 \):

\[
\text{Denominator} = \cos^2 x + \sin^2 \frac{x}{2}
\]

The numerator expands as:

\[
\text{Numerator} = \left( \cos \frac{x}{2} + i \sin \frac{x}{2} \right) \left( \cos x - i \sin \frac{x}{2} \right)
\]
\[
= \cos \frac{x}{2} \cos x - \cos \frac{x}{2} i \sin \frac{x}{2} + i \sin \frac{x}{2} \cos x + \sin^2 \frac{x}{2}
\]
\[
= \cos \frac{x}{2} \cos x + i \left( \sin \frac{x}{2} \cos x - \cos \frac{x}{2} \sin \frac{x}{2} \right) + \sin^2 \frac{x}{2}
\]

Since \( z \) is purely real, the imaginary part must be zero. Therefore, we set the imaginary part equal to zero:

\[
\sin \frac{x}{2} \cos x - \cos \frac{x}{2} \sin \frac{x}{2} = 0
\]

This simplifies to:

\[
\sin \frac{x}{2} \left( \cos x - \cos \frac{x}{2} \right) = 0
\]

For non-zero \( \sin \frac{x}{2} \), we must have:

\[
\cos x = \cos \frac{x}{2}
\]

This equation implies that:

\[
x = \frac{\pi}{3}
\]

4. Now, we can find \( \cos^2 x \):

\[
\cos^2 \frac{\pi}{3} = \cos^2 \left( \frac{\pi}{3} \right) = \left( \frac{1}{2} \right)^2 = \frac{1}{4}
\]

Thus, the value of \( \cos^2 x \) is \( \frac{1}{4} \).

---

Would you like more details or have any questions? Here are five related questions to explore:

1. How does Euler's formula help simplify complex numbers?
2. What are the steps to find when a complex number is purely real?
3. How can you find the conjugate of a complex number and why is it useful?
4. Why does setting the imaginary part to zero solve this problem?
5. Can this approach be applied to other intervals of \( x \)?

**Tip:** Always remember that complex numbers are real when the imaginary part vanishes, and this often involves multiplying by the conjugate to simplify.

If the complex number (cos(x/2) + i sin(x/2)) / (cos(x) + i sin(x/2)), x ∈ (0, π/2), i = √(-1), is purely real, then the value of cos²(x) is?

Learn how to determine the value of cos²(x) when given a complex number of the form (cos(x/2) + i sin(x/2)) / (cos(x) + i sin(x/2)) that is purely real. This problem involves key mathematical concepts including Euler's Formula, complex conjugates, and trigonometric identities. Suitable for students in Grades 11-12.

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