Trigonometric Graphs (1 Viewer)

coyazayo · Jan 23, 2016

Hi i was just graphing trigonometric functions when I came across a cosine function with a period of 3pi. I also came across another cosine function with a period of 2pi/3. So my question is: how do you exactly graph theses? Functions with period of 2pi, pi etc... are real easy but these I really need help with. Thanks in advance!!

DatAtarLyfe · Jan 23, 2016

If you look at a regular cosine function that fluctuates in a 2pi period, you can observe that it is broken up into 4 equal pieces. It cuts the axis in the first quarter of the period (pi/2), creates a minimum stat point at pi, cuts again at 3pi/2 and then a maximum stat point at 2pi
Similarly, you can apply the same principle to different periods. If you break the designated period into 4 parts, you can replicate it the same way as the regular cosine.
Say for example your 3pi function. Breaking it into quarters we get 3pi/4, 3pi/2 , 9pi/4 and 3pi. So it cuts at 3pi/4, min stat point at 3pi/2, cuts again at 9pi/4 and max stat point at 3pi.

Of course, other factors such as the frequency, amplitude and initial phase also define the graph and hence alter the way you break it up

InteGrand · Jan 23, 2016

coyazayo said:
Hi i was just graphing trigonometric functions when I came across a cosine function with a period of 3pi. I also came across another cosine function with a period of 2pi/3. So my question is: how do you exactly graph theses? Functions with period of 2pi, pi etc... are real easy but these I really need help with. Thanks in advance!!

$\noindent One very simple way is to first simply sketch the shape of the cosine graph like usual, without labelling the $x$-axis values yet (just draw in the first cycle of it). Then once we've drawn the shape in, we focus on getting the $x$-axis right.$

$\noindent Since the period is $3\pi$ (say), we'd label the right endpoint of this cycle as having $x=3\pi$. (The left endpoint is $x=0$, since we drew the first period of the graph, by which I mean starting from $x=0$). Then we know that the cosine graph hits the $x$-axis first when it is one-quarter of the way through its cycle, so make this value $x=\frac{3\pi}{4}$. It hits its first minimum of $y=-1$ $\Big{(}$ I'm assuming the equation is $y=\cos \left(\frac{2}{3}x\right)$$\Big{)}$ halfway through the cycle, so make this $x=\frac{3\pi}{2}$. It hits the $x$-axis next three-quarters of the way through its cycle, so make this $x=\frac{3}{4}\cdot 3\pi = \frac{9\pi}{4}$. So now we've labelled the key $x$-values. The $y$-values are clearly easy to do, just make the maximums at $y=1$ and the minimums at $y=-1$.$

$\noindent For a general period of $T$, do a similar process but instead of $3\pi$, use $T$, whatever $T$ is.$

DatAtarLyfe · Jan 23, 2016

Ha, beat IG to it

InteGrand · Jan 23, 2016

DatAtarLyfe said:
Ha, beat IG to it

$\noindent Haha yeah :-). But you wrote $\frac{5\pi}{4}$ instead of $\frac{9\pi}{4}$. :$\mathcal{P}$$

DatAtarLyfe · Jan 23, 2016

Whoops, thanks

Trigonometric Graphs (1 Viewer)

coyazayo

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DatAtarLyfe

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InteGrand

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InteGrand

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DatAtarLyfe

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