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 show this identity:

1. We can use the cosine angle addition formula: $\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$ $\cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta$
2. By adding the two equations: $\cos(\alpha + \beta) + \cos(\alpha - \beta) = 2\cos \alpha \cos \beta$ This proves the identity as required.

To solve the equation:

1. From part (i), we know how to relate angles using cosines. We first apply the identities:
   • Rewrite $\cos 2\theta$, $\cos 3\theta$, and $\cos 4\theta$ using double angle formulas.
2. Substitute:
   • The equation simplifies to a polynomial which can be factored or solved numerically.
3. The roots can be found in the given interval by testing values or using numerical methods.

To resolve the forces on particle P:

1. In the horizontal direction, the tension $T_1$ provides a component: $T_1 \sin \alpha$ This is the centripetal force due to the motion, equal to $m \omega^2 r$.
2. In the vertical direction, we balance the forces: $T_1 \cos \alpha + T_2 = mg$ where $T_2$ is the tension in the second string.

To find $\omega$ under the condition that $T_2 = 0$:

1. We start from the vertical force equation: $T_1 \cos \alpha = mg$ Thus, we express $T_1$: $T_1 = \frac{mg}{\cos \alpha}$
2. Using the horizontal component: $T_1 \sin \alpha = m \omega^2 l$ Substituting for $T_1$: $\frac{mg \sin \alpha}{\cos \alpha} = m \omega^2 l$
3. Simplifying and solving for $\omega$, we get: $\omega = \sqrt{\frac{g \tan \alpha}{l}}$

To find the number of different recordings:

1. For each of the four games, there are three potential outcomes (Win, Draw, Loss). Thus, $3^4 = 81$ different outcomes are possible.

To calculate the probability of the result WDLD:

1. The probability of each outcome is:
   • Win (W): 0.2
   • Draw (D): 0.6
   • Loss (L): 0.2
2. Thus, the probability of WDLD is: $P(WDLD) = P(W) \cdot P(D) \cdot P(L) \cdot P(D) = 0.2 \cdot 0.6 \cdot 0.2 \cdot 0.6 = 0.0144$.

To find this probability:

1. The scoring system gives:
   • 1 point for a win, \frac{1}{2} point for a draw, and 0 points for a loss.
2. Considering all configurations of outcomes, we can calculate the expected scores and compare.
3. This leads us to a binomial probability distribution, where we can compute individual probabilities and sum accordingly to find the desired probability.