# Coefficients in Trigonometric Functions

So, previously in my math classes I have been taught that The phase shift of $$\sin(ax+c)$$ is c.

But in my new class they say that $$\sin(ax+c) = \sin(a(x+c))$$ which I believe to be wrong. But I have an exam tommorrow and I do not wish to be marked down for having the wrong phase shift.

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The phase shift is $c$. The equation $\sin(ax+c) = \sin(a(x+c))$ is false in general (it's true if $a=1$ of course). The $a$ is the frequency parameter, and doesn't affect the phase shift. You can think of phase shift as the angle you get at time zero, or in this case when $x=0$. – Alan Guo Oct 22 '12 at 4:59
So I was correct? And my professor is teaching it wrong? In the textbook it says: $$y=\csc(2x-\pi/4) = \csc 2(x-\pi/8)$$ – Matt Jens Jensen Oct 22 '12 at 5:03
The equation you cite from your textbook is very different from the one you display in the body of the question. – Gerry Myerson Oct 22 '12 at 5:08
Not in theory. It is essentially the same idea. Isn't it? – Matt Jens Jensen Oct 22 '12 at 5:09
No, no, no. $ax+c$ is not the same as $a(x+c)$, but $2x-(\pi/4)$ is the same as $2(x-(\pi/8))$, in theory, in practice, and in every country on this or any other planet. – Gerry Myerson Oct 22 '12 at 5:13

There seem to be different conventions about phase shift. See this Regents link, where it says that if $$y=\sin(2x-(\pi/2))=\sin(2(x-(\pi/4)))$$ then physicists and engineers say the phase shift is $\pi/2$ but some math textbooks say it's $\pi/4$. Find out for certain which convention your class is using, and use it. If it's too late for that, then declare which convention you are using, and stick with it.