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

To prove that $\triangle ASP \parallel \triangle ASBP$, we can use the properties of alternate angles formed by the tangents PS and PT. Since PS and PT are tangents to the circles $\mathcal{C}_1$ and $\mathcal{C}_2$, we can state that the angles $\angle ASP$ and $\angle ABP$ are equal. Therefore, by the Alternate Interior Angles Theorem, the triangles are parallel.

Using the previous result, we can apply the tangent-secant theorem (or power of a point theorem) which states that the square of the length of the tangent $SP$ from point S to circle $\mathcal{C}_1$ is equal to the product of the lengths of the segments created by the secant line through point P. This gives us:

$SP^2 = AP \times BP.$

For the second part, we already deduced the relationship, which gives:

$PT = PS.$

We know that DT is drawn from S, meeting the bisector of angle ZSP at D. By the properties of angles and tangents from point S, and the fact that D lies on the angle bisector, we can conclude that triangle DSP is congruent to triangle DTP. This congruence leads to the conclusion that DT must pass through the center of circle $\mathcal{C}_2$, confirming our statement.

Using the definition of $P_n$, we get:

$P_n = \cos(\frac{\alpha}{2})P_{n-1}.$

Thus:

$P_n \sin(\frac{\alpha}{2}) = \cos(\frac{\alpha}{2})P_{n-1} \sin(\frac{\alpha}{2}) = \frac{1}{2}P_{n-1} \sin(\frac{\alpha}{2^n}).$

From the previous result, by iterating this relationship, we can express $P_n$ in terms of $\sin(\alpha)$, leading us to the conclusion:

$P_n = \frac{\sin(\alpha)}{2^n \sin(\frac{\alpha}{2^n})}.$

To show that $\sin(\alpha) < \alpha$ for $x > 0$, we will use calculus or the properties of sine function. We note that within the interval $(0, \pi)$, the sine function is always less than the angle in radians, leading us to:

$\sin(\theta) < \alpha$ for $x > 0$.