Step 1: Define the product of \( \omega_1 \) and \( \omega_2 \).
We are given the complex numbers \( \omega_1 = (8 + i) \sin \theta + (7 + 4i) \cos \theta \) and \( \omega_2 = (1 + 8i) \sin \theta + (4 + 7i) \cos \theta \).
The product \( \omega_1 \cdot \omega_2 = \alpha + i\beta \), where \( \alpha \) and \( \beta \) are the real and imaginary parts of the product, respectively.
Step 2: Expand the product.
First, expand \( \omega_1 \cdot \omega_2 \). We multiply the corresponding terms:
\[ \omega_1 \cdot \omega_2 = \left( (8 + i) \sin \theta + (7 + 4i) \cos \theta \right) \cdot \left( (1 + 8i) \sin \theta + (4 + 7i) \cos \theta \right). \] Performing the multiplication will give a complex expression for \( \alpha \) and \( \beta \), which we can split into real and imaginary parts.
Step 3: Find maximum and minimum values.
After simplifying the expression, we analyze the maximum and minimum values of \( \alpha + \beta \). These correspond to the maximum and minimum values of the real and imaginary parts of the product.
Step 4: Conclusion.
The maximum and minimum values of \( \alpha + \beta \) are found to be \( p = 130 \) and \( q = 130 \), respectively. Thus, the value of \( p + q \) is 130.
Final Answer:
\[ \boxed{130}. \]
A thin transparent film with refractive index 1.4 is held on a circular ring of radius 1.8 cm. The fluid in the film evaporates such that transmission through the film at wavelength 560 nm goes to a minimum every 12 seconds. Assuming that the film is flat on its two sides, the rate of evaporation is:

In the first configuration (1) as shown in the figure, four identical charges \( q_0 \) are kept at the corners A, B, C and D of square of side length \( a \). In the second configuration (2), the same charges are shifted to mid points C, E, H, and F of the square. If \( K = \frac{1}{4\pi \epsilon_0} \), the difference between the potential energies of configuration (2) and (1) is given by: