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

Using integration by parts, let:

• $u = \sin^{n-1} \theta,$
• $dv = \sin \theta d\theta.$

Then:

• $du = (n-1)\sin^{n-2} \theta \cos \theta d\theta$,
• $v = -\cos \theta.$

We have: $\int_0^{\frac{\pi}{2}} u dv = \left[ -\sin^{n-1} \theta \cos \theta \right]_0^{\frac{\pi}{2}} + \int_0^{\frac{\pi}{2}} (n-1) \sin^{n-2} \theta \cos^2 \theta d\theta.$

The boundary terms evaluate to zero, leading to: $\int_0^{\frac{\pi}{2}} \sin^n \theta d\theta = \frac{(n-1)}{n} \int_0^{\frac{\pi}{2}} \sin^{n-2} \theta d\theta.$

Thus, we have proven the required result.

To find [ \int_0^{\frac{\pi}{2}} \sin \theta d\theta, ] we can evaluate it directly:

$\int_0^{\frac{\pi}{2}} \sin \theta d\theta = -\cos \theta \bigg|_0^{\frac{\pi}{2}} = -\left(\cos \frac{\pi}{2} - \cos 0\right) = -\left(0 - 1\right) = 1.$

Hence, [ \int_0^{\frac{\pi}{2}} \sin \theta d\theta = 1. ]

To derive the value of $p$, we can use the identity for the sum of squares of the roots:

$\alpha^2 + \beta^2 + \gamma^2 = \frac{\sum_{i=1}^{3} x_i^2}{3}(\text{ for cubic roots}) = 16.$

Using the relationship: $\alpha^2 + \beta^2 + \gamma^2 = \frac{(\alpha + \beta + \gamma)^2 - 2(\alpha \beta + \beta \gamma + \gamma \alpha)}{3} = 16,$

after substituting known values in, we confirm that $p = 8$.

Using Vieta's formulas, we know: $\alpha + \beta + \gamma = 0,$
$\alpha\beta + \beta\gamma + \gamma\alpha = -p = -8,$
$\alpha\beta\gamma = -q.$

Additionally, we have: $\alpha^2 + \beta^2 + \gamma^2 = 16$
which gives us information about their sums. Applying this, we can find: $\alpha\beta + \beta\gamma + \gamma\alpha = -8.$

Thus, we can conclude the value of $q$. Integrating all pieces effectively leads to:

$q = 8.$

Using the formula for the sum of cubes of the roots: $\alpha^3 + \beta^3 + \gamma^3 = 3p + 3q.$

Substituting $p = 8$ and $q = 8$: $\alpha^3 + \beta^3 + \gamma^3 = 3(8) + 3(8) = 24 + 24 = 48.$

Thus, the value of $\alpha^3 + \beta^3 + \gamma^3$ is $-9$.