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My notes on Transformations has the following formula for rotating a point counter-clockwise about the origin:

$\begin{align*}x^\prime&=x\cos\theta - y\sin\theta\\y^\prime&=x\sin\theta + y\cos\theta\end{align*}$

Why? And why does changing the direction mean that we change the signs before the sin values so that rotating a point around the origin clockwise can be found by:

$\begin{align*}x^\prime&=x\cos\theta + y\sin\theta\\y^\prime&=-x\sin\theta + y\cos\theta\end{align*}$

I am trying to work through an example, to rotate a point at (20, 0) $45^o$ clockwise, but I don't get the answer supplied by the notes.

My Workings

$\begin{align*}x^\prime&=20\cos(45) + 0\sin(45) = 10.51\\y^\prime&=-20\sin(45) + 0\cos(45) = -17.01\end{align*}$

but my notes give the answer as being:

$\begin{align*}x^\prime&=14.14\\y^\prime&=-14.14\end{align*}$

I don't want to just follow the formula (especially as I seem to be getting the wrong answer!). I really want to understand how the new position relates to the the addition and / or subtraction of sin and cos, but I don't have any intuitions about it.

Be grateful for any help.

Regards

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    Working above example with polar coordinates as suggested by @J.M. - Original polar coords = (20, $0^o$) subtract $45^o$ of rotation to give coords after rotation of (20, $315^o$). We know hypotenuse is 20, so x is 20 x cos(315) = 14.14, and y is 20 x sin(315) = -14.14. Good to think about it that way too. Many thanks for the explanation J.M.2011-04-26

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If you accept that counterclockwise rotation by $\theta$ is a linear transformation, then the transformation is determined by its effect on the standard basis vectors $(1,0)$ and $(0,1)$. So, let's look at where those vectors are sent.

The formulas come from the trigonometry going on, and I encourage you to draw this out at least once. The vector $(1,0)$ points straight right along the $x$-axis, and after applying the transformation (rotating it by $\theta$), we get a new vector, still on the unit circle, with angle $\theta$ from the $x$-axis. Essentially by definition of cosine and sine, this means that the new $x$ coordinate is $\cos(\theta)$, and the new $y$ coordinate is $\sin(\theta)$

Similarly, when we look at what happens to $(0,1)$ on that picture, we see that it gets sent to a vector with $x$-coordinate $-\sin(\theta)$ and $y$-coordinate $\cos(\theta)$.

That means that the transformation by $\theta$ degrees can be enacted by the matrix $\begin{pmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{pmatrix}$

Consequently, we can figure out what this transformation does to an arbitrary vector $(x,y)^T$ by left-multiplying it by this matrix to get (assuming I did this right) $\begin{pmatrix} x \cdot \cos(\theta) - y \cdot \sin(\theta) \\ x \cdot \sin(\theta) + y \cdot \cos(\theta) \end{pmatrix}$

You could also check this in a less linear-algebra-ish way by just thinking about the rotation of the arbitrary vector to start with, but I think if you understand one way you'll be close to understanding the other.

Also, you can come up with the matrix for clockwise rotation in a similar way: it sends $(1,0)$ to $(\cos(\theta), -\sin(\theta))$ and sends $(0,1)$ to $(\sin(\theta),\cos(\theta))$, so the corresponding matrix for the transformation is $\begin{pmatrix} \cos(\theta) & \sin(\theta) \\ -\sin(\theta) & \cos(\theta) \end{pmatrix}$

and the result of applying this matrix to an arbitrary vector $(x,y)^T$ is $\begin{pmatrix} x \cdot \cos(\theta) + y \cdot \sin(\theta) \\ -x \cdot \sin(\theta) + y \cdot \cos(\theta) \end{pmatrix}$

As far as why you're getting the wrong answer: I strongly suspect that you're using a calculator that expects radian inputs to the trig functions. You're probably plugging in 45 thinking that's 45 degrees, but it's interpreting it as 45 radians and giving you a weird answer. Either try again using $\pi / 4$ radians, or change your calculator to work with degrees.

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    Great Answer. I drew out the unit vector stuff and saw it working. Need to practice, practice, practice.2011-04-26