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I am having a moment of weakness and lost the plot, why is this wrong

Image of sine plots

I assumed adding $2\pi$ would give the same result, clearly not the case on the graph (using desmos graphing tool) $\sin (x + 2\pi)$ does not equal $\sin (x)$ in this plot

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  • $\begingroup$ You plotted a sine and two vertical lines... What is the question about? $\endgroup$
    – Yuriy S
    Nov 2, 2018 at 12:41
  • $\begingroup$ Are you working in degrees or radians? Units are important... $\endgroup$
    – Xander Henderson
    Nov 2, 2018 at 18:58

4 Answers 4

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Note that desmos is plotting two different lines, the black vertical line

$$x = \frac16 \pi$$

and the red vertical line

$$x = \frac16 \pi + 2\pi$$

The purple graph is a graph of

$$\sin(\frac16 x)$$

which will look exactly the same if you write

$$\sin(\frac16x + 2\pi)$$

and is completely uncorrelated with the other two plots you made. In other words, when you wrote $x = \frac16 \pi$ and $x = \frac16\pi + 2\pi$ you didn't interact with the command $\sin \frac16 x$ you already had.

For it to work as expected you need to add the $2\pi$ inside the first command you have on desmos, or write a completely new one with $\sin(\frac16 x + 2\pi)$.

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    $\begingroup$ Good point (+1) $\endgroup$ Nov 2, 2018 at 12:30
  • $\begingroup$ @Natalie Johnson I added a final line on how to make it work as expected. $\endgroup$
    – RGS
    Nov 2, 2018 at 12:40
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You don't have $\sin x$, but $\sin(x/6),$ so when you add $2\pi$ to $x,$ this is what happens:

$$\sin\left(\frac{1}{6}(x+2\pi)\right) = \sin\left(\frac{1}{6}x+\frac{1}{6}2\pi\right) =\sin\left(\frac{1}{6}x+\frac{\pi}{3}\right).$$

So you've really added only $\pi/3$ to the argument of $\sin.$

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  • $\begingroup$ That is not quite what happened with the OP $\endgroup$
    – RGS
    Nov 2, 2018 at 12:28
  • $\begingroup$ But hang on a minute. The unit circle is 2 *pi so a time period is always 2*pi... but sin(1/2 x) , sin(x) and sin (2x) all have different periods and not 2*pi $\endgroup$ Nov 2, 2018 at 12:35
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    $\begingroup$ @NatalieJohnson No. You seem to realize that these functions have different periods, but yet assert the "time period is always $2\pi$. You might be confused because you're putting $\frac{1}{6}\pi$ and $\frac{1}{6}\pi + 2\pi$ in for $x$, where it gets multiplied by $\frac{1}{6}$ again. $\endgroup$
    – B. Goddard
    Nov 2, 2018 at 12:46
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When you add $2\pi$ to $x$, $\frac16x$ increases only by $\frac13\pi$, which you shouldn't expect to preserve the sine.

The period of $\sin(\frac16x)$ is $6$ times $2\pi$, or $12\pi$.

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  • $\begingroup$ this is true but doesn't address the issue of the OP $\endgroup$
    – RGS
    Nov 2, 2018 at 12:27
  • $\begingroup$ @RGS: You must have a different impression about what the OP's issue is than I (and two other answerers) have. $\endgroup$ Nov 2, 2018 at 12:28
  • $\begingroup$ From my pov, the OP thinks he did something that he did not do. Notice he did not add $2\pi$ to any argument of any trig function. $\endgroup$
    – RGS
    Nov 2, 2018 at 12:30
  • $\begingroup$ @RGS whats the solution then? $\endgroup$ Nov 2, 2018 at 12:37
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You aren't adding $2\pi$ to the argument, but to $x$: $$\sin\left(\frac{x+2\pi}{6}\right) \neq \sin\left(\frac{x}{6}+2\pi\right)$$

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  • $\begingroup$ That was not quite what happened $\endgroup$
    – RGS
    Nov 2, 2018 at 12:28

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