Question8. A given sinusoidal function has an amplitude of 8 , an axis at $y=12$, a period of $9 \pi$, and a maximum at $x=3 \pi$. Determine an equation for the function, and find all intersections of the function and $y=16$ between $x=-5 \pi$ and $x=12 \pi$. [ 8 marks]

Studdy Solution
Solve for intersections with $y = 16$:
Set the equation equal to 16:
$$8 \cos\left(\frac{2}{9}(x - 3\pi)\right) + 12 = 16$$
Subtract 12 from both sides:
$$8 \cos\left(\frac{2}{9}(x - 3\pi)\right) = 4$$
Divide by 8:
$$\cos\left(\frac{2}{9}(x - 3\pi)\right) = \frac{1}{2}$$
The general solutions for $\cos(\theta) = \frac{1}{2}$ are:
$$\theta = \frac{\pi}{3} + 2k\pi \quad \text{or} \quad \theta = -\frac{\pi}{3} + 2k\pi$$
Substitute back for $\theta$:
$$\frac{2}{9}(x - 3\pi) = \frac{\pi}{3} + 2k\pi \quad \text{or} \quad \frac{2}{9}(x - 3\pi) = -\frac{\pi}{3} + 2k\pi$$
Solve for $x$:
1. $x - 3\pi = \frac{9\pi}{6} + 9k\pi$ $$x = \frac{9\pi}{6} + 3\pi + 9k\pi$$ $$x = \frac{11\pi}{2} + 9k\pi$$
2. $x - 3\pi = -\frac{9\pi}{6} + 9k\pi$ $$x = -\frac{9\pi}{6} + 3\pi + 9k\pi$$ $$x = \frac{3\pi}{2} + 9k\pi$$
Find $x$ within $[-5\pi, 12\pi]$:
For $x = \frac{11\pi}{2} + 9k\pi$: - $k = -1$: $x = \frac{11\pi}{2} - 9\pi = -\frac{7\pi}{2}$ - $k = 0$: $x = \frac{11\pi}{2}$
For $x = \frac{3\pi}{2} + 9k\pi$: - $k = 0$: $x = \frac{3\pi}{2}$ - $k = 1$: $x = \frac{3\pi}{2} + 9\pi = \frac{21\pi}{2}$ (out of range)
The intersections are at $x = -\frac{7\pi}{2}, \frac{3\pi}{2}, \frac{11\pi}{2}$.
The equation of the function is:
$$y = 8 \cos\left(\frac{2}{9}(x - 3\pi)\right) + 12$$
Intersections with $y = 16$ are at:
$$x = -\frac{7\pi}{2}, \frac{3\pi}{2}, \frac{11\pi}{2}$$