A solid is formed by rotating the region bounded by the curve $y=x(x-1)^2$ and the line $y=0$ about the y-axis - HSC - SSCE Mathematics Extension 2 - Question 5 - 2006 - Paper 1

To demonstrate this trigonometric identity, we can use the sum-to-product formulas. Recall:

$\cos(A + B) = \cos A \cos B - \sin A \sin B$ $\cos(A - B) = \cos A \cos B + \sin A \sin B$

So we write:

$\cos(\alpha + \beta) + \cos(\alpha - \beta) = \left(\cos \alpha \cos \beta - \sin \alpha \sin \beta \right) + \left(\cos \alpha \cos \beta + \sin \alpha \sin \beta \right)$

Simplifying this, we obtain:

$2 \cos \alpha \cos \beta$.

To solve this equation, we can use the identity derived in part (b)(i) and substitution methods. First, it helps to rearrange the equation in terms of cosines,

$\cos\theta + \cos 2\theta + \cos 3\theta + \cos 4\theta = 0$

Utilizing the sum of cosines, we can denote $x = \cos\theta$ and express the equation as a polynomial. Solving for $x$ will involve solving a quartic equation, possibly using numerical methods or trigonometric identities to find all solutions within the given interval.

In the horizontal direction, the tension in the string AP has a horizontal component given by $T_1 \sin(\alpha)$. Thus, we set up the equation:

$T_1 \sin(\alpha) - T_2 \sin(\beta) = 0$

For the vertical direction, the forces included are the downward tension and gravitational force:

$T_1 \cos(\alpha) + T_2 \cos(\beta) = mg$

Given that $T_2=0$, we can then simplify our equations from the resolution of forces:

From the vertical force equation: $T_1 \cos(\alpha) = mg$ Substituting for $T_1$ into the horizontal equation previously established will yield a radius function dependent on $ω$. The relationship can then be expressed as: $ω = \sqrt{\frac{g}{\ell \cos(\alpha)}}$

Possible recordings can be determined by the number of outcomes available for each board (win, lose, draw), leading to:

$3^{4} = 81$

This accounts for each game having three possible outcomes.

To calculate the probability of the sequence WLDL, we apply the probabilities for each case:

• Probability of winning (W) = 0.2,
• Probability of drawing (D) = 0.6,
• Probability of losing (L) = 0.2.

The combined probability is: $P(W) \cdot P(L) \cdot P(D) \cdot P(L) = 0.2 \times 0.2 \times 0.6 \times 0.2 = 0.0048$

The scoring rule gives the Home team ( 1 , point \times wins + \frac{1}{2} , point \times draws ). To find the probability that the Home team scores more than the Away team, we utilize the binomial distribution scenarios across the 4 games considering all possible outcomes:

Setting up the probability scenarios systematically, then evaluating the combinations will give the desired probability. The calculation will involve the probabilities from different scenarios with multiple games of 1 point for wins and 0.5 for draws.