7. (a) The region bounded by $0 \leq x \leq \sqrt{3}$, $0 \leq y \leq 3(\sqrt{3} - x^2)$ is rotated about the y-axis to form a solid - HSC - SSCE Mathematics Extension 2 - Question 7 - 2003 - Paper 1

To prove that the angles $\angle ASP$ and $\angle SBP$ are parallel:

1. Consider the tangents PS and PT from point P to the circles $\mathcal{C}_1$ and $\mathcal{C}_2$. By the property of tangents, the angles formed where the tangents meet the segments are equal: $\angle APS = \angle PTS$ and $\angle BPS = \angle PTB.$
2. Since PS and PT are tangents and S is on the circumference, it follows that $\angle ASP$ is equal to $\angle BPS$.
3. By Alternate Interior Angles Theorem, if a transversal intersects two lines such that the pair of alternate interior angles are equal, this implies the lines are parallel. Thus, $\angle ASP \parallel \angle SBP$.

Using the previously established angles:

1. From the properties of tangents and secants, we can use the tangent-secant theorem: $SP^2 = AP \cdot BP$.
2. The proof follows by noting that triangle $ASP$ is similar to triangle $BSP$ due to the corresponding angles being equal.
3. Hence, $SP^2 = AP \cdot BP$.
4. Now, since $PT$ and $PS$ are both tangents from point P to circles, we know $PT = PS$. Therefore the deduced relationship holds true.

To prove that line DT passes through the center of $\mathcal{C}_2$:

1. Note that line segment SD is drawn perpendicular to SP.
2. The angle bisector theorem states that the bisector of an angle divides the opposite side in the ratio of the other two sides.
3. Thus, from triangle $DSP$ and triangle $DTP$, since angles $DPS$ and $DPT$ are congruent (as angle bisectors) we see that the segments created are proportional.
4. Therefore, line DT must intersect the center of $\mathcal{C}_2$, confirming that DT passes through the center.

To show the relationship, we can utilize the product relation of the cosines:

1. Writing $P_n$ in terms of the cosine products: $P_n = \cos(\frac{\alpha}{2}) \cos(\frac{\alpha}{4}) \cos(\frac{\alpha}{8}) \cdots \cos(\frac{\alpha}{2^n}).$
2. Using the trigonometric identity $\sin x = 2 \sin(\frac{x}{2}) \cos(\frac{x}{2})$, we arrive at: $\sin(\frac{\alpha}{2^n}) = 2 \sin(\frac{\alpha}{2^{n+1}}) \cos(\frac{\alpha}{2^{n+1}}).$
3. Substituting back, we can rearrange this to demonstrate that: $P_n \cdot \sin(\frac{\alpha}{2^n}) = \frac{1}{2} P_{n-1} \sin(\frac{\alpha}{2^{n-1}}).$

To deduce this relationship:

1. Starting from the previous part where we established that: $P_n \sin(\frac{\alpha}{2^n}) = \frac{1}{2} P_{n-1} \sin(\frac{\alpha}{2^{n-1}}).$
2. Iterating for value $n=1$, we find: $P_1 \sin(\frac{\alpha}{2}) = \frac{1}{2} P_0 \sin(\alpha).$
3. Recognizing that $P_0 = 1$ leads us to find: $P_n = \frac{2^{n-1} \sin(\frac{\alpha}{2})}{(2^n \cdot \sin(\frac{\alpha}{2^n}))} = \frac{\sin \alpha}{2^n \cdot \sin(\frac{\alpha}{2^n})}$ confirming the relationship.

To show this compound inequality:

1. We know from the property of the sine function that for $0 < \alpha < \pi$, it holds that: $\sin(\alpha) < \alpha$ which satisfies the left part of our inequality.
2. For $P_n \sin(\frac{\alpha}{2^n})$, since $P_n$ is a product of cosines, and recognizing that each cosine is less than 1, it follows that: $P_n < 1.$ Thus: $P_n \sin(\frac{\alpha}{2^n}) < \frac{\alpha}{2^n} < P_n \sin(\frac{\alpha}{2^n}).$
3. Therefore, this confirms the complete inequality $\sin \alpha < \alpha < P_n \sin(\frac{\alpha}{2^n})$.