Math / Trigonometry

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

STEP 1

1. The sinusoidal function is either a sine or cosine function.
2. The amplitude of the function is $8$ .
3. The vertical shift (axis) of the function is at $y = 12$ .
4. The period of the function is $9\pi$ .
5. The function reaches its maximum at $x = 3\pi$ .

STEP 2

1. Determine the form of the sinusoidal function.
2. Calculate the frequency of the function.
3. Write the equation of the sinusoidal function.
4. Solve for intersections with $y = 16$ within the specified interval.

STEP 3

Determine the form of the sinusoidal function:
Given the maximum at $x = 3\pi$ , the function is likely a cosine function, as cosine reaches its maximum at $x = 0$ in its standard form. Therefore, the function can be expressed as:
$y = A \cos(B(x - C)) + D$
where: - $A$ is the amplitude, - $B$ affects the period, - $C$ is the horizontal shift, - $D$ is the vertical shift.

STEP 4

Calculate the frequency of the function:
The period $T$ is given by $T = \frac{2\pi}{B}$ . We know $T = 9\pi$ , so:
$9\pi = \frac{2\pi}{B}$
Solving for $B$ :
$B = \frac{2\pi}{9\pi} = \frac{2}{9}$

STEP 5

Write the equation of the sinusoidal function:
The amplitude $A = 8$ , the vertical shift $D = 12$ , and the maximum occurs at $x = 3\pi$ , so the horizontal shift $C = 3\pi$ . The equation is:
$y = 8 \cos\left(\frac{2}{9}(x - 3\pi)\right) + 12$

STEP 6

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:
$x = -\frac{7\pi}{2}, \frac{3\pi}{2}, \frac{11\pi}{2}$

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