# How to graph trigonometric functions

I am trying to complete some homework for my physics course and I have come to realise that I do not understand how parameters inside a trigonometric function affect the function and therefore a graph of the function.

I have the following question where I need to graph some functions on a disturbance vs time (t) graph where T is the period of the function. For an example the function $y=\sin{\left(\frac{2\pi t}{T}\right)}$, is one that I have to put on this graph. My first idea was as the sine function has a period of $2\pi$ and $T$ is the given period of $y$ that I could cancel the two periods leaving just $\sin{(t)}$ but after more thought I do not really think this is correct as it is not given that $2\pi=T$. After this I am not really sure how to further work the problem out.

I would like to know how to visualise or workout how parameters inside a trig function affect the behaviour of the function.

EDIT:

I do not understand how to calculate the period of the function given above, I am confused because I understand the period of the function to be given by the coefficient of in this case $t$ as $t$ is multiplied by $2\pi$ and $\frac{1}{T}$ I am not sure which one I would use to calculate the period or both of them?

I am mostly confused that the function presented above seems to include a variable for its period as a parameter which would then be used to calculate its period and I don't see how this can happen. I hope that makes sense, at the moment basically when I look at the function and try to calculate its period I just see an infinite loop scenario, I hope that is clear and if it isn't let me know so I can clarify further. Thanks!

EDIT 2:

Along with my last edit, the only real stab I could make at this would be the following the period $T$ is given by $T=\frac{2\pi}{2\pi}=1$, so in this case the function $y=\sin\left(\frac{2\pi t}{T}\right)$ would just have a period $T=1$ and therefore be a normal sine wave?

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The period of the trigonometric function $sin(x)$ is as you said $2\pi$.

The period of $sin(ax)$ is $T=\frac{2\pi}{a}$.

You can then use $0, \frac{T}{4},2\frac{T}{4},3\frac{T}{4}$ and $T$. You'll see that nice angles will appear and computations will be easy.

take a look at the following images, I think they 'll help you clear out things a bit

I used a freeware program to draw those (geogebra easy to use, you can download it if you like and experiment a bit on your own).

If the constant is negative you'll have something like that (since $sin(-x)=-sin(x)$

(of course functions $sin(2x)$ is actually $sin(2x)+2$ etc but I left that out. It's easier I believe to visualize it like this)

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Thanks, this is a nice answer but on reflection I am hoping you can explain one thing further to me, about calculating the period of the function. In the example presented in my initial question the function has as one of its parameters $T$ which is given to be its period. I am confused as to how a function can have a period which is dependent upon its period..if that makes sense? – Aesir Jan 12 '13 at 15:53
I have added an additional part to the question and would be grateful if your answer could be extended to cover that part. – Aesir Jan 12 '13 at 16:00
@Aesir I believe constant $a$ in this case is $a=\frac{2\pi}{T}$ so the period of the function is (let's say T') $T'=\frac{2\pi}{\frac{2\pi}{T}}=\frac{2\pi T}{2\pi}=T$. – epsilon Jan 12 '13 at 16:24