Modeling with trigonometric functions

Question

I'm having trouble with modeling trig functions that include phase shifts on KhanAcademy. Please be aware that the answer would not help me. That is already available to me. I would prefer an explanation and hints that would lead me to the correct answer and help me solve other problems just like it. This is the problem that I am completely lost on:

Wei is standing in wavy water and notices the depth of the waves varies in a periodic way that can be modeled by a trigonometric function. He starts a stopwatch to time the waves. After $1.1$ seconds, and then again every $3$ seconds, the water just touches his knees. Between peaks, the water recedes to his ankles. Wei's ankles are $12$ cm off the ocean floor, and his knees are $55$ cm off the ocean floor. Find the formula of the trigonometric function that models the depth $D$ of the water $t$ seconds after Wei starts the stopwatch. Define the function using radians.

Answer 1

1. A trig wave with average height 0 has the form $f(t) = Acos({\omega}t+\phi)$. You could also use the sin - it doesn't matter.
2. A trig wave with average height b has the form $f(t) = b+ Acos({\omega}t+\phi)$
3. The difference between the max and min heights is 2A. In your problem 2A = 55-12 = 42 so that A = 21.5.
4. The average wave height is b = 12 + A = 33.5
5. The time period of the wave is 3, so that its frequency (waves per second) is f = 1/3. Its angular frequency $\omega$ (waves per $2\pi$) is $2{\pi}f = 2{\pi}/3$.
6. your wave is now $f(t) = 33.5+ 21.5cos({2{\pi}t/3}+\phi)$
7. When t = 1.1 the max height is reached so that $55 = 33.5+ 21.5cos({2{\pi}(1.1)/3}+\phi)$. Then $1 = cos({2{\pi}(1.1)/3}+\phi)$ which in turn means that ${2{\pi}(1.1)/3}+\phi = 0$ and solving gives $\phi$.