# Adding $2\pi$ not getting same value out of trig function [closed]

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

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

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

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 $$\sin(\frac16 x + 2\pi)$$.

• Good point (+1) Nov 2, 2018 at 12:30
• @Natalie Johnson I added a final line on how to make it work as expected.
– RGS
Nov 2, 2018 at 12:40

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 $$\sin.$$

• That is not quite what happened with the OP
– RGS
Nov 2, 2018 at 12:28
• 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 Nov 2, 2018 at 12:35
• @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. Nov 2, 2018 at 12:46

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 $$12\pi$$.

• this is true but doesn't address the issue of the OP
– RGS
Nov 2, 2018 at 12:27
• @RGS: You must have a different impression about what the OP's issue is than I (and two other answerers) have. Nov 2, 2018 at 12:28
• 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.
– RGS
Nov 2, 2018 at 12:30
• @RGS whats the solution then? Nov 2, 2018 at 12:37

You aren't adding $$2\pi$$ to the argument, but to $$x$$: $$\sin\left(\frac{x+2\pi}{6}\right) \neq \sin\left(\frac{x}{6}+2\pi\right)$$

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