The expression provided can be simplified using Euler's formula and basic trigonometric identities.

Problem:

$\frac{(\cos 20^\circ + i \sin 20^\circ)^6}{(\cos 50^\circ + i \sin 50^\circ)^2} \cdot (\cos \theta - i \sin \theta)^2$

Solution Steps:

1. Represent terms using Euler's formula: Euler's formula states: $\cos x + i \sin x = e^{ix}$ So, the expression becomes: $\frac{(e^{i \cdot 20^\circ})^6}{(e^{i \cdot 50^\circ})^2} \cdot (e^{-i \cdot \theta})^2$

2. Simplify powers: Using the property $(e^{i x})^n = e^{i n x}$, we have: $(e^{i \cdot 20^\circ})^6 = e^{i \cdot 120^\circ}, \quad (e^{i \cdot 50^\circ})^2 = e^{i \cdot 100^\circ}, \quad (e^{-i \cdot \theta})^2 = e^{-i \cdot 2\theta}$

   Substituting these into the expression: $\frac{e^{i \cdot 120^\circ}}{e^{i \cdot 100^\circ}} \cdot e^{-i \cdot 2\theta}$

3. Combine exponents: Using the property $\frac{e^{ix}}{e^{iy}} = e^{i(x-y)}$, we get: $e^{i \cdot (120^\circ - 100^\circ)} \cdot e^{-i \cdot 2\theta}$ Simplify $120^\circ - 100^\circ$: $e^{i \cdot 20^\circ} \cdot e^{-i \cdot 2\theta}$

   Combine further: $e^{i \cdot (20^\circ - 2\theta)}$

4. Return to trigonometric form: Using Euler's formula backward: $e^{i \cdot (20^\circ - 2\theta)} = \cos(20^\circ - 2\theta) + i \sin(20^\circ - 2\theta)$

Final Simplified Expression:

$\cos(20^\circ - 2\theta) + i \sin(20^\circ - 2\theta)$

Let me know if you have questions or want me to explain specific steps further.

Related Questions:

1. How does Euler's formula simplify trigonometric expressions?
2. Can you prove the exponentiation properties used in this problem?
3. What are the real and imaginary parts of the result?
4. How would this change if $\theta = 10^\circ$?
5. Can you visualize this result on the complex plane?

Tip:

When working with complex numbers in exponential form, always leverage the rules of exponents and Euler's formula to simplify calculations efficiently.