the rotation of a sinusoid curve

**ines** · December 17th 2010, 12:11 PM

Hello,
I have a sinusoid curve in an euclidian orthonormal plane [OX,OY].
the equation of the curve is y=a sin x.

Now, I want to get the equation of the curve when I rotate it by an $\alpha$ angle with respect to the (ox) axis like it is shown in the figure submitted.

To do that,I'll define tehe vectorial euclidian rotation :in an oriented vectorial euclidian plan, a vectorial rotation is simply defined by its angle $\alpha$ . Its matrix in an orthonormal direct basis is :
$ $\begin{pmatrix} \cos\alpha & -\sin \alpha\\ \sin \alpha & \cos\alpha \end{pmatrix}$$

In another way, a vector $\vec U$ (x,y) has as image the vector $\vec V$ (x',y') that can be calculated like that :

$ $\begin{pmatrix} x'\\y'\end{pmatrix} =\begin{pmatrix}\cos\alpha&-\sin\alpha \\ \sin\alpha&\cos\alpha\end{pmat rix}\begin{pmatrix}x\\y\end{pm atrix},$ $

then we have :
$x' = x \cos \alpha - y \sin \alpha\,$
and
$y' = x \sin \alpha + y \cos \alpha\,$

as we have

$ $y=a \sin x$$ , then we have

$y' = x \sin \alpha + a \sin x \cos \alpha\,$

the problem is that I must write $y'$ in function of $x'$ .

Thank you for reading me. I'll be vey glad if you can help me

**snowtea** · December 17th 2010, 03:12 PM

Were you given restrictions on $\alpha$ ? Because for $\alpha > 45^o$ , y' cannot be expressed as a function of x' (since it fails the vertical line test).

**ines** · December 19th 2010, 03:05 AM

Hello Snowtea,

thank you for answering but I didn't understand why I must give restrictions to

?
I tried to solve the problem like that but It seems to me complicated:

As we have y=a sin x.

and if we rotate the vector v(x',y') in an angle -

, we have then u(x,y) and the matrix of rotation is:

$$\begin{pmatrix} x\\y\end{pmatrix}=\begin{pmatrix}-\cos\alpha&-\sin\alpha \\ \sin\alpha&-\cos\alpha\end{pmatrix}\begin{pmatrix}x'\\y'\end{p matrix},$ $

then we have

$ $x = -x' \cos \alpha - y' \sin \alpha\,$ $
and
$y = x' \sin \alpha - y' \cos \alpha\,$

then we replace x and y in the equation y=a sin x by their expressions in funtion of x' and y'.

we then have:
$x' \sin \alpha - y' \cos \alpha\ = a sin(-x' \cos \alpha - y' \sin \alpha)$

The problm now is how to write the equation in a simple form y'=f(x').

Could you help me.
Thank you

**snowtea** · December 19th 2010, 06:51 AM

You can only write y' = f(x') when for every x', you have a unique y' that satisfies the equation.

If $\alpha > \pi/4$ , just consider the simple case when $\alpha = \pi/2$ .
Then for $x'=0$ , $y'\in\{0, \pi, -\pi, ...\}$ are all solutions, so y' cannot be written as a function of x'.

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