What are you talking about specifically? Are you talking about just $\displaystyle \sin{(x)}$ or are you talking about $\displaystyle \frac{25}{4}\sin\frac{2\pi(x-82)}{365}+\frac{49}{4}$ ? If you are referring to $\displaystyle \sin{(x)}$ , this function has a domain of all real numbers. This means I can take any real number to be the x value, plug it in, and I will get some number y that is between -1 and 1, inclusive. (That would be the range).

If we take a look at $\displaystyle \frac{2\pi(x-82)}{365}$ we can see that this expression results in a real number. This is so because the domain of this expression is all real numbers. Therefore, $\displaystyle \sin\frac{2\pi(x-82)}{365}$ is defined for all values of x that are real numbers. If that is true, then by the properties of sine functions and function transformations, we can determine that although the function is defined for all x such that x is a real number, it's amplitude is still 1 but it's period changed.

I can restrict your domain and tell you that $\displaystyle \frac{25}{4}\sin\frac{2\pi(x-82)}{365}+\frac{49}{4}$ is only defined on the interval $\displaystyle 1 \leq x \leq 365$ . Still, 1 is a real number, and so is 365 so I'm going to get real number results. The sine portion of the function can take on any number from its domain, but we aren't talking about any number; we're talking in the context of the problem, so we have to focus on a restricted domain, namely $\displaystyle 1 \leq x \leq 365$ . Does this restricted domain change what the sine function does? Of course not. Sine still does it's thing. I'm just limiting you on your

*choices* of x values to plug in.

To sum up, you have something*sin(stuff) + something, but the sin(stuff) still does its sine function thing, with a few alterations. Is the amplitude affected in just the sin(stuff) portion? No. Is its period affected? Yes. Does it still take on 1 and -1 if we limit the interval on which it is defined? No ... use a calculator and you'll see that we are at about -0.98 when x=1 and when x=365 . But careful!

**That's just the sin(stuff) portion! **
Here's the graph of $\displaystyle \sin \frac{2\pi(x-82)}{365}$ on the interval $\displaystyle 1 \leq x \leq 365$ :

Attachment 27825
Correct. To be clear why, let's look at $\displaystyle \sin x$ by itself. Again, the amplitude is 1. Oh, wait a second, I don't want you looking at the whole entire graph, so I'm going to restrict your domain to $\displaystyle \pi \leq x \leq 2 \pi$ . What happened? We're just looking at a portion of the graph of the function since we are only plotting points on the given interval. Does $\displaystyle \sin x $ on the interval $\displaystyle \pi \leq x \leq 2 \pi$ ever reach its normal maximum point of 1? No, of course not. That's because we're only looking at a specific set of x values defined by that interval.

I hope that helps. (Nod)