algebra precalculus - Confusion with modeling a trigonometric function

I am studying trigonometry on Khan Academy and came across this problem:

The daily low temperature in Guangzhou, China, varies over time in a periodic way that can be modeled by a trigonometric function.
The period of change is exactly $1$ year. The temperature peaks around July 26 at $78°F$, and has its minimum half a year later at $49°F$. Assuming a year is exactly 365 days, July 26 is $\frac { 206 }{ 365 } $

​​ of a year after January 1.
Find the formula of the trigonometric function that models the daily low temperature $T$ in Guangzhou t years after January 1, 2015. Define the function using radians.

$T(t)=$

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So the steps I took are:

1) Finding the amplitude:

$$\frac { 78-49 }{ 2 } =14.5$$

2) Finding the midline:
$$\frac { 78+49 }{ 2 } =63.5$$

3) Figuring out whether to use cosine or sine:

I figured that I can treat July 26th as the beginning of the year and then shift the function to make it the 206th day of the year. So I used cosine. Since at $0$, a cosine function is at its max value.

4) The period:

$1$ year is a period so it must be $$\frac { 2\pi }{ 365 } $$

5) The function without the shift is now:
$$14.5cos(\frac { 2\pi }{ 365 } u)+63.5$$

6) Now I must find the value of $u$ in order to properly shift the function. I imagine that this must be $t-206$ since it is $206$ days after January 1.

I feel like I must be missing something here or got one of the steps wrong. Please guide me in the right direction.

Answer

It looks like you mostly have the right idea. The thing you forgot is that $t$ is the number of years, not days. The answer you should get is

$$
14.5 \cos \left(2 \pi \left(t - \frac{206}{365} \right)\right) + 63.5
$$