Even though the wave of $\sin(2x-4)$ looks like this and it starts from position 2 out of curiosity I would like to know whether we could start it from position 4 like the wave of $\sin(2x)$ shown here.
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You can shift the origin to $x=a$ by replacing the argument $x$ by $x-a$:
$$\sin(2(x-a))\tag{1}$$
Since for $\sin(2x-4)=\sin(2(x-2))$ you have $a=2$, the origin is shifted to $x=2$. If you want to shift it to $x=4$, you get from Eq. $(1)$
$$\sin(2(x-4))=\sin(2x-8)\tag{2}$$
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$\begingroup$ :Yeah that's really great.The last line was what I needed. $\endgroup$ – justin Jan 19 '16 at 8:32
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I am not sure whether I understand the question completely.
First of all do you mean this?
Here ONE zero crossing of the sin-wave is at the point (4,0).
In general the sinus is has no real starting point since its periodic, therefor you get an infinite amount of zero-crossings.
Then you might want to look at the equation this way:
$$sin(2*pi*Frequency*x+Phase) \tag{1}$$
Here Phase represents the shift to the left.
In your case with the nested frequency and phase it looks like this:
$$sin(2*(x-2))$$ or $$sin(2*(x-4))$$
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$\begingroup$ Please use Latex formatting to make your equations more readable. Please also correct the argument of your first sine function; the frequency shouldn't be in the denominator. $\endgroup$ – Matt L. Jan 18 '16 at 11:58
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$\begingroup$ Thank you for the corrections, I noticed the Frequency error, but was not able to edit it earlier. $\endgroup$ – NeinDochOah Jan 18 '16 at 22:39
