Notes
Sine and cosine
- Both have period $2\pi$ and range $[-1, 1]$. Sine starts at 0; cosine starts at its maximum 1.
- For $y = A\sin(Bx - C) + D$: amplitude $|A|$, period $\displaystyle \frac{2\pi}{|B|}$, phase shift $\displaystyle \frac{C}{B}$ (right if positive), midline $y = D$.
- Range: $[D - |A|, D + |A|]$. Negative $A$ reflects the graph over the midline.
Modeling
- Midline $\displaystyle = \frac{\max + \min}{2}$, amplitude $\displaystyle = \frac{\max - \min}{2}$, $\displaystyle B = \frac{2\pi}{\text{period}}$.
Tangent and reciprocals
- $y = \tan x$ has period $\pi$, zeros at $n\pi$, and asymptotes at $\displaystyle \frac{\pi}{2} + n\pi$. For $\tan(Bx)$ the period is $\displaystyle \frac{\pi}{|B|}$.
- $\csc x$ has asymptotes where $\sin x = 0$; $\sec x$ where $\cos x = 0$. Both have $|y| \ge 1$.
Topics in this unit
Each topic has its own short notes and practice.
Practice
Amplitude is $|A| = 2$ and period is $\displaystyle \frac{2\pi}{B} = \frac{2\pi}{3}$.
Shifting cosine right $\displaystyle \frac{\pi}{2}$ puts its maximum at $\displaystyle x = \frac{\pi}{2}$, which is the sine curve.
Midline 1 and amplitude 2 give $[1 - 2, 1 + 2]$. The negative sign does not change the range.
Period $\displaystyle = \frac{2\pi}{\pi} = 2$.
Period $\displaystyle = 2\pi \div \frac{1}{2} = 4\pi$.
$\tan x = 0$ when $\sin x = 0$. Tangent has range all reals, period $\pi$, and no amplitude.
Free response
Write your answer first, then open the scoring guide.
Free-response question
The water depth at a harbor, in feet, is $d(t) = 5\cos\left(\frac{\pi t}{6}\right) + 12$, where $t$ is the number of hours after high tide.
(a) State the amplitude, the period and the midline.
Show the answer and scoring guide
Amplitude 5. Period $\displaystyle \frac{2\pi}{\pi/6} = 12$ hours. Midline $d = 12$.
Check off the points you earned:
(b) Find the maximum and minimum depths.
Show the answer and scoring guide
Maximum $12 + 5 = 17$ ft. Minimum $12 - 5 = 7$ ft.
Check off the points you earned:
(c) Find the depth 2 hours after high tide, and the time of the first low tide.
Show the answer and scoring guide
$\displaystyle d(2) = 5\cos\frac{\pi}{3} + 12 = 5 \cdot \frac{1}{2} + 12 = 14.5$ ft. Low tide happens when $\displaystyle \cos\frac{\pi t}{6} = -1$, so $\displaystyle \frac{\pi t}{6} = \pi$ and $t = 6$ hours.
Check off the points you earned: