# A question on transformation

I am doing a transformation problem of getting the graph of $$\sin (2x – \pi/6)$$ by applying transformations to $$F(x) = \sin x$$

In the process, I let $$f(x) = F(2x) = \sin 2x$$.

Next, I then let $$g(x) = f(x – \pi/12) = \sin 2[x – \pi/12] = \sin (2x – \pi/6)$$. The graphs are plotted as shown.

From f(x) to g(x), the above equations clearly shows there is a phase-shift of $$\pi/12$$ and this agrees with the red and blue lines

However, if I just comparing the functions $$\sin (2x)$$ and $$\sin (2x – \pi/6)$$ directly, shouldn’t there be just a right shift of $$\pi/6$$?.

Here the graph is plotted as functions of x, taking values of x in the horizontal axis. So there is a shift of $$\pi/12$$.

If we plot it as a function of 2x, then we would have a shift of $$2.\frac{\pi}{12} = \frac{\pi}{6}$$

• Right. I forgot I have to plot it as function of 2x. – Mick Apr 29 at 17:01

Based on your reasoning, I presume you want to use the fact that the transformation $$f(x)\to f(x+a)$$ corresponds to a horizontal shift of the function by $$|a|$$ (where the direction depends on the sign of $$a$$)

Nonetheless, when the function is of the form $$\sin(2x)$$ in order to obtain the function $$\sin(2x-\pi/6)$$ you need to find $$a$$ such that $$\sin({\color{red}2}(x+a))=\sin(2x+\color{red}{2}a)=\sin(2x-\pi/6)$$ and thus the needed shift is $$a=\frac{\pi/6}{2}=\frac{\pi}{12}$$.

But this does not contradict the property of the transformation: if you want to shift the function $$f(x)=\sin(2x)$$ by $$\pi/6$$ to the right then you do apply the transfomration as usual $$f(x-\pi/6)=\sin(\color{red}{2}(x-\pi/6))=\sin(2x-\pi/3)$$ but as you see the actual shift is hidden by the factor 2.

In general, when constructing an harmonic oscillation of the form $$\sin(\omega x+\varphi)$$ you need to translate/shift the function horizontally by $$\varphi/\omega$$