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:
\[ \boxed{130}. \]
If \( z \) is a complex number and \( k \in \mathbb{R} \), such that \( |z| = 1 \), \[ \frac{2 + k^2 z}{k + \overline{z}} = kz, \] then the maximum distance from \( k + i k^2 \) to the circle \( |z - (1 + 2i)| = 1 \) is:
The largest $ n \in \mathbb{N} $ such that $ 3^n $ divides 50! is:
Two light beams fall on a transparent material block at point 1 and 2 with angle \( \theta_1 \) and \( \theta_2 \), respectively, as shown in the figure. After refraction, the beams intersect at point 3 which is exactly on the interface at the other end of the block. Given: the distance between 1 and 2, \( d = 4/3 \) cm and \( \theta_1 = \theta_2 = \cos^{-1} \frac{n_2}{2n_1} \), where \( n_2 \) is the refractive index of the block and \( n_1 \) is the refractive index of the outside medium, then the thickness of the block is cm. 
A bob of mass \(m\) is suspended at a point \(O\) by a light string of length \(l\) and left to perform vertical motion (circular) as shown in the figure. Initially, by applying horizontal velocity \(v_0\) at the point ‘A’, the string becomes slack when the bob reaches at the point ‘D’. The ratio of the kinetic energy of the bob at the points B and C is: 