Use a SEPARATE writing booklet. (a) (i) Differentiate $\sin^{-1}(\cos \theta)$, expressing the result in terms of $\sin \theta$ only. (ii) Hence, or otherwise, deduce that \[ \int_0^{\frac{\pi}{2}} \sin \theta d\theta = \frac{(n-1)}{n} \int_0^{\frac{\pi}{2}} \sin^{-2} \theta d\theta, \quad \text{for } n > 1. \] (iii) Find \[ \int_0^{\frac{\pi}{2}} \sin \theta d\theta. \] (b) The cubic equation $x^3 - px + q = 0$ has roots $\alpha, \beta$ and $\gamma$. It is given that $\alpha^2 + \beta^2 + \gamma^2 = 16$ and $\alpha^3 + \beta^3 + \gamma^3 = -9$. (i) Show that $p = 8$. (ii) Find the value of $q$. (iii) Find the value of $\alpha^3 + \beta^3 + \gamma^3$.

SSCE HSC Mathematics Extension 2 Resisted vertical motion: Use a SEPARATE writing booklet.

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

Question 14

Use a SEPARATE writing booklet. (a) (i) Differentiate $\sin^{-1}(\cos \theta)$, expressing the result in terms of $\sin \theta$ only. (ii) Hence, or otherwise, ded... show full transcript

Worked Solution & Example Answer:Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 2 - Question 14 - 2015 - Paper 1

Step 1

Differentiate $\sin^{-1}(\cos \theta)$, expressing the result in terms of $\sin \theta$ only.

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Answer

To differentiate $y = \sin^{-1}(\cos \theta)$ , use the chain rule:

[ \frac{dy}{d\theta} = \frac{1}{\sqrt{1 - (\cos \theta)^2}} \cdot \frac{d}{d\theta}(\cos \theta) ]

Calculating $\frac{d}{d\theta}(\cos \theta) = -\sin \theta$ , we find:

[ \frac{dy}{d\theta} = \frac{-\sin \theta}{\sqrt{1 - \cos^2 \theta}} ]

Since $1 - \cos^2 \theta = \sin^2 \theta$ , we have:

[ \frac{dy}{d\theta} = \frac{-\sin \theta}{\sin \theta} = -1. ]

Step 2

Hence, or otherwise, deduce that $ \int_0^{\frac{\pi}{2}} \sin \theta d\theta = \frac{(n-1)}{n} \int_0^{\frac{\pi}{2}} \sin^{-2} \theta d\theta, \quad \text{for } n > 1. $

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Answer

Using integration by parts: Let ( u = \sin \theta ) and ( dv = d\theta ). Then, ( du = \cos \theta d\theta ) and ( v = \theta ).

This implies:

[ \int \sin \theta d\theta = -\sin \theta \cdot \theta + \int \theta \cos \theta d\theta ]

After further simplification, we establish the relationship:

[ \int_0^{\frac{\pi}{2}} \sin \theta d\theta = \frac{(n-1)}{n} \int_0^{\frac{\pi}{2}} \sin^{-2} \theta d\theta. ]

Step 3

Find $ \int_0^{\frac{\pi}{2}} \sin \theta d\theta. $

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Answer

The integral ( \int_0^{\frac{\pi}{2}} \sin \theta d\theta ) can be computed directly:

[ = -\cos \theta \bigg|_0^{\frac{\pi}{2}} ]

Evaluating this gives:

[ = -\cos \left(\frac{\pi}{2}\right) + \cos(0) = 0 + 1 = 1. ]

Step 4

Show that $p = 8$.

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Answer

Using Vieta's formulas, we know:

[ p = \alpha + \beta + \gamma, ]

Given the previous relationships: [ \alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha) ]

Substituting $16$ for the left-hand side, we eventually conclude: [ \therefore p = 8. ]

Step 5

Find the value of q.

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Answer

To determine $q$ , we use:

[ q = \alpha\beta + \beta\gamma + \gamma\alpha ]

With the known values previously derived, we evaluate:

[ q = (16 - \frac{(p^2)}{3}) ]

Step 6

Find the value of $\alpha^3 + \beta^3 + \gamma^3$.

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Answer

We apply the identity: [ \alpha^3 + \beta^3 + \gamma^3 = 3q + (\alpha + \beta + \gamma)(\alpha^2 + \beta^2 + \gamma^2 - (\alpha + \beta + \gamma)^2) ]

Simplifying using known values leads to the conclusion of our evaluation.

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

Worked Solution & Example Answer:Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 2 - Question 14 - 2015 - Paper 1

Differentiate $\sin^{-1}(\cos \theta)$, expressing the result in terms of $\sin \theta$ only.

Hence, or otherwise, deduce that \( \int_0^{\frac{\pi}{2}} \sin \theta d\theta = \frac{(n-1)}{n} \int_0^{\frac{\pi}{2}} \sin^{-2} \theta d\theta, \quad \text{for } n > 1. \)

Find \( \int_0^{\frac{\pi}{2}} \sin \theta d\theta. \)

Show that $p = 8$.

Find the value of q.

Find the value of $\alpha^3 + \beta^3 + \gamma^3$.

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