Adding $2\pi$ not getting same value out of trig function

Question

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

Answer 1

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

Answer 2

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 $12\pi$.

Answer 3

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

Answer 4

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 $\sin.$