# Basic trigonometry

Given an angle $\theta$ (in radians), write $\theta = 2\pi t + \phi$, where $\phi$ is between $0$ and $2\pi$, $0\leq \phi\lt 2\pi$, and $t$ is an integer. Call $k=2\pi t$ the number of "full circles" or "full turns" corresponding to the angle.

If $\theta = \frac{13\pi}{4}$, am I right that k=$2\pi$?

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@Gjorgji: this is very poorly written; is $k$ multiplying the $4$? What does "angle = 5pi/4" mean? Could you please use mark-up and write it coherently? I honestly cannot fathom just what it is you are trying to say or ask. – Arturo Magidin Nov 30 '10 at 21:24
How can i edit it so it looks like an equation? – Gjorgji Nov 30 '10 at 21:32
@Gjorgji: But what does your equation about $k$ mean? It makes absolutely no sense. If $k=1$, then $k$ is not $2\pi$. If $k$ is "1 angle", then what does "1 angle" mean? Does "angle" mean "full turn"? What do you mean, "$k=2\pi$"? Sorry, but this is still gibberish. – Arturo Magidin Nov 30 '10 at 21:44
k is the number of full circles and the angle is the remaining one lower than $2\pi$ (360), i got confused but yeah 360=$2\pi$ so 1 turn=$2\pi$ – Gjorgji Nov 30 '10 at 21:51
@Gjorgji: And we are supposed to guess all that? Sigh. I've completely rewritten the question. Is this what you meant to ask? But if $k$ is the "number of full circles", then the answer should not be $2\pi$, because you do not do $2\pi$ full circles, you only do $1$. – Arturo Magidin Nov 30 '10 at 21:53

I'm guessing that the homework problem is something like "find an angle between 0 and $2\pi$ that's equal to $13\pi/4$". That is, for what $k$ such that $k$ is a multiple of $2\pi$ do we have $${13\pi\over 4} = k + {5\pi\over 4}$$
The answer is indeed $k=2\pi$. The first equation in the original question should just have "$k$" in place of "($2\pi$)".