(1 + sin theta + i * cos theta) ^ n + (1 + sin theta - i * cos theta) ^ n = 2 ^ (n + 1) * cos^n (pi/4 - theta/2) * cos((n*pi)/4 - (n*theta)/2)

We are tasked with proving the following equation:

\[
(1 + \sin \theta + i \cos \theta)^n + (1 + \sin \theta - i \cos \theta)^n = 2^{n+1} \cos^n \left( \frac{\pi}{4} - \frac{\theta}{2} \right) \cos \left( \frac{n\pi}{4} - \frac{n\theta}{2} \right)
\]

### Step 1: Complex Form Simplification
The terms on the left-hand side involve complex numbers of the form \( z_1 = 1 + \sin \theta + i \cos \theta \) and \( z_2 = 1 + \sin \theta - i \cos \theta \). These can be written as:

\[
z_1 = (1 + \sin \theta) + i \cos \theta
\]
\[
z_2 = (1 + \sin \theta) - i \cos \theta
\]

### Step 2: Converting to Exponential Form
Recognize that each complex number can be rewritten in polar form. For \( z_1 \) and \( z_2 \), use the fact that \( 1 + \sin \theta \) is the real part and \( \cos \theta \) is the imaginary part.

\[
z_1 = \sqrt{(1 + \sin \theta)^2 + \cos^2 \theta} e^{i \phi}
\]

Where the magnitude \( |z_1| = \sqrt{(1 + \sin \theta)^2 + \cos^2 \theta} = \sqrt{1 + 2\sin \theta + \sin^2 \theta + \cos^2 \theta} = \sqrt{2 + 2\sin \theta} = 2 \cos \left( \frac{\pi}{4} - \frac{\theta}{2} \right) \).

Thus,

\[
z_1 = 2 \cos \left( \frac{\pi}{4} - \frac{\theta}{2} \right) e^{i \left( \frac{\pi}{4} - \frac{\theta}{2} \right)}
\]
and
\[
z_2 = 2 \cos \left( \frac{\pi}{4} - \frac{\theta}{2} \right) e^{-i \left( \frac{\pi}{4} - \frac{\theta}{2} \right)}.
\]

### Step 3: Raising to Power \(n\)
Now, raise \( z_1 \) and \( z_2 \) to the power \( n \):

\[
z_1^n = \left[ 2 \cos \left( \frac{\pi}{4} - \frac{\theta}{2} \right) \right]^n e^{in \left( \frac{\pi}{4} - \frac{\theta}{2} \right)}
\]
and
\[
z_2^n = \left[ 2 \cos \left( \frac{\pi}{4} - \frac{\theta}{2} \right) \right]^n e^{-in \left( \frac{\pi}{4} - \frac{\theta}{2} \right)}.
\]

### Step 4: Summing \( z_1^n \) and \( z_2^n \)
Now, we sum \( z_1^n \) and \( z_2^n \):

\[
z_1^n + z_2^n = 2 \left[ 2 \cos \left( \frac{\pi}{4} - \frac{\theta}{2} \right) \right]^n \cos \left( n \left( \frac{\pi}{4} - \frac{\theta}{2} \right) \right)
\]

\[
= 2^{n+1} \cos^n \left( \frac{\pi}{4} - \frac{\theta}{2} \right) \cos \left( \frac{n\pi}{4} - \frac{n\theta}{2} \right)
\]

This matches the right-hand side of the equation we set out to prove.

Thus, the equation is verified.

---

Would you like more details or clarification on any step?

Here are 5 questions to consider based on this explanation:
1. What is the significance of using polar form for complex numbers in this proof?
2. How would the proof change if \( \theta \) were negative?
3. Can this technique of summing powers of complex numbers be applied to other trigonometric identities?
4. How can the relationship between the magnitude and argument of complex numbers help in solving other trigonometric equations?
5. What role does Euler's formula play in simplifying trigonometric expressions involving complex numbers?

**Tip**: Understanding polar and exponential forms of complex numbers is key to simplifying expressions involving powers and products of complex numbers.

This problem involves proving a complex trigonometric identity by leveraging Euler's formula and polar form of complex numbers. It covers topics like polar coordinates, exponentiation, and the application of De Moivre's theorem. Suitable for advanced high school or college-level students.

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