A particle is undergoing simple harmonic motion with period \( \frac{\pi}{3} \) - HSC - SSCE Mathematics Extension 2 - Question 13 - 2020 - Paper 1

To find the point of intersection, equate the components of the two lines:

1. For x-component:
   (3 + 2h_1 = -6 + 3h_2 \Rightarrow (1))

2. For y-component:
   (-1 + h_1 = 2 + 2h_2 \Rightarrow (2))

3. For z-component:
   (7 + h_1 = -2 - 3h_2 \Rightarrow (3))

From equation (1):
(2h_2 - 2h_1 = 9 \Rightarrow h_2 = 1 + h_1)

Substituting (h_2) into (2):
(-1 + h_1 = 2 + 2(1 + h_1) \Rightarrow -1 + h_1 = 4 + 2h_1)

Solving gives:
( -1 - 4 = 2h_1 - h_1 \Rightarrow h_1 = -5 ) and substituting back into equation gives (h_2 = -4.

Finally, substituting (h_1) and (h_2) back into the line equation gives the point of intersection as:

$\mathbf{r} = \begin{pmatrix} -2 \ \frac{3}{2} \ -5 \end{pmatrix}$

By Pythagoras, we have:

$(a + b)^2 = x^2 + (a - b)^2 \Rightarrow x^2 = (a + b)^2 - (a - b)^2$

The properties of the triangle imply: (x \leq a + b) Thus, it can be shown that: ( \frac{a + b}{2} \geq \sqrt{ab} ) using the inequality of means (AM-GM).

Let (p = a) and (q = b), then:

By following the derivation steps, we arrive at:

$p^2 + 4q^2 - 4pq = (p - 2q)^2 \geq 0,$ which confirms that the inequality holds.

Using Euler's formula, we have:

$e^{i\theta} = \cos(\theta) + i\sin(\theta) \text{ and } e^{-i\theta} = \cos(\theta) - i\sin(\theta).$

Thus,

$e^{i\theta} + e^{-i\theta} = 2\cos(\theta).$

Starting with:

$(e^{i\theta} + e^{-i\theta})^2 = 2\cos(\theta)\cdot 2\cos(\theta) = 4\cos^2(\theta),$

and using the cosine of double angle:

$\cos(2\theta) = 2\cos^2(\theta) - 1$

we derive the required expression through expansions.

Using the identity derived:

$\int_{0}^{\frac{\pi}{2}} \cos^2(\theta) \, d\theta = \int_{0}^{\frac{\pi}{2}} \frac{1}{8}(\cos(4\theta) + 4\cos(2\theta) + 3) \, d\theta$

Calculating each integral gives:

$= \left[ \frac{3\pi}{16} \right] - 0 + \frac{3\pi}{16}$.

Finally, the answer results in ( \frac{3\pi}{16} ).