Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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:


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.


share|cite|improve this question
For your second question: "rotate by $-\theta$ anticlockwise" is just another way of saying "rotate by $\theta$ clockwise". Note that the cosine is even and the sine is odd. – J. M. Apr 26 '11 at 16:40
To evaluate $\sin(45^\circ)$ and $\cos(45^\circ)$, notice that $45^\circ$ is half of a right angle, and so is the angle in an isosceles right triangle. Draw such a triangle, apply the Pythagorean theorem and use the fact that the smaller sides have the same length to conclude that $\cos(45^\circ)=\sin(45^\circ)=\frac{1}{\sqrt 2}$. As for your evaluations, your calculator was in radian mode with degree inputs. – Jonas Meyer Apr 26 '11 at 16:46
The cheater's way of motivating the rotation formulae: convert your Cartesian coordinates to polar coordinates, add the rotation angle to the polar angle (corresponding to an anticlockwise rotation), and convert back, making use of the sum-of-angles formulae for trigonometric functions. – J. M. Apr 26 '11 at 16:48
"your calculator was in radian mode with degree inputs." - it's a very common mistake to make. Be very careful with your manipulations if you insist on having to use degrees. – J. M. Apr 26 '11 at 16:50
Thanks for your answers. Need a bit of time to absorb what you're saying. I'll post back once I'm done. – Joe Apr 26 '11 at 17:38
up vote 3 down vote accepted

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.

share|cite|improve this answer
Thanks for your answer. It's going to take me a while to work through it. I'll post back once I'm done. :) – Joe Apr 26 '11 at 17:37
Great Answer. I drew out the unit vector stuff and saw it working. Need to practice, practice, practice. – Joe Apr 26 '11 at 18:06

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.