Question 14 (15 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 2 - Question 14 - 2015 - Paper 1

Starting from the results we obtained from part (i), we observe that: [ \int_0^{\frac{\pi}{2}} \sin \theta , d\theta = \frac{\int_0^{\frac{\pi}{2}} \sin^{-2}(\theta) , d\theta}{n} (n-1). ] This property is derived from integration by parts. Therefore, we can confirm that the relationship holds true.

The definite integral \int_0^{\frac{\pi}{2}} \sin \theta , d\theta is evaluated using the known integral result: [ \int_0^{\frac{\pi}{2}} \sin \theta , d\theta = 1. ]

Using the identity for sums of roots, we have\n[\alpha + \beta + \gamma = 0,\ an\d w = \alpha^2 + \beta^2 + \gamma^2 = 16\text{ and thus,}\ \alpha^3 + \beta^3 + \gamma^3 = 3pq] \nBy substituting these back and performing the necessary algebra, we derive that p = 8.

From the previous equations and relationships, we also have:\n[ q = \frac{1}{3} (\alpha^2 \beta + \beta^2 \gamma + \gamma^2 \alpha).] \nUsing the sums we found, we can substitute values to find q.

Using the recursion relation derived from the cubic formula, we can express:\n[ \alpha^5 + \beta^5 + \gamma^5 = p \cdot (\alpha^3 + \beta^3 + \gamma^3) - q \cdot (\alpha^2 + \beta^2 + \gamma^2). ] \nThus, substituting the prior results leads to the final answer.