# Find the rotation matrix for +15° out of the rotation matrix +60° without using trigonometric functions

Find the rotation matrix for +15° out of the rotation matrix +60° without using trigonometric functions. And as much as I would love to tell you what I did so far. I don't even know where to start with this exercise..

The matrix for +60° is this. $$\left[ \begin{matrix} 1/2 & -\sqrt 3/2\\ \sqrt 3/2 & 1/2 \\ \end{matrix} \right]$$

How do I even start?

-
It is preferable to have the question self-contained, so I have incorporated the title. Please take a look and see if you agree. – Ross Millikan Jun 29 '11 at 16:39

## 6 Answers

There is a far simpler way of doing this, assuming that you are allowed to use the transformation that reflects across the line $y = x$ (trivially obtained, no trig involved) and the line $y = -x$.

Let's be exact. The reflection matrices across $y = x$, denoted by $R_+$, and across $y = -x$, denoted by $R_-$, are given by $$R_+ =\left(\begin{matrix} 0 & 1\\ 1 & 0 \\ \end{matrix}\right) \qquad \qquad R_- =\left(\begin{matrix} 0 & -1\\ -1 & 0 \\ \end{matrix}\right)$$ But then $R_+$ applied to the first column of the 60 degree rotation gives the first column of the 15 degree rotation, and $R_-$ applied to the second column of the 60 degree rotation gives the second column of the 15 degree rotation.

So it can be done using very little math, no challenging multiplication, and best of all, no roots of matrices. I admit, when I first looked at it, I thought only of the way that Ross wrote of.

-
Good on you! (and some more characters) – Ross Millikan Jun 29 '11 at 16:34
+1! I'd hastily dismissed the possibility of getting to anything involving 15 using only 60 (or its multiples). The idea of finding the columns individually is very clever! – Zev Chonoles Jun 29 '11 at 16:34
that sounds nice, but it is not allowed, as the exercise clearly states that we need to find the result explicitly with square roots. – nino Jun 29 '11 at 16:50
@nino: Well I wish that were included in the prompt! What about this: multiply your matrix by the matrix $\left(\begin{matrix} \frac{1}{\sqrt2} & \frac{1}{\sqrt2}\\ \frac{-1}{\sqrt2} & \frac{1}{\sqrt2} \\ \end{matrix}\right)$. This involves square roots, and is still not so bad. – mixedmath Jun 29 '11 at 16:56
That could work and would be doable in a somehow realistic time frame. I think I have to ask my teacher if this would be a realistic approach which doesn't count as using trigonometric functions. Thanks for your help! – nino Jun 29 '11 at 17:04

Does the following use trigonometric functions? (The matrix for rotation by $60^\circ$ certainly does.)

Follow the rotation by $60^\circ$ by a rotation through $-45^\circ$, the matrix for which is easily written down. So we are multiplying the given matrix by another easily written down matrix.

My guess given the context is that this approach is the intended one, since a simple matrix operation is involved.

-

The only approach I can see is to take the fourth root of the matrix, that is, find a $2 \times 2$ matrix $A$, such that $A^4=\left(\begin{matrix} 1/2 & -\sqrt 3/2\\ \sqrt 3/2 & 1/2 \\ \end{matrix}\right)$. If you expect that it is of the form $\left(\begin{matrix} a & -b\\ b & a \\ \end{matrix}\right)$ with $a^2+b^2=1$, it should go through.

Added: If you first take a square root, you get $a^2-b^2=1/2, -2ab=\sqrt{3}/2$, which gives $b^2=\frac{3}{16a^2}, a^2-\frac{3}{16a^2}=\frac{1}{2}, a=\frac{\sqrt{3}}{2}, b=\frac{1}{2}$ (taking the positive root). This give the $30^{\circ}$ matrix as $A^2=\left(\begin{matrix} \sqrt 3/2 & -1/2\\ 1/2 & \sqrt 3/2 \\ \end{matrix}\right)$.

Now the same approach gives $a^2-b^2=\sqrt 3/2, -2ab=1/2$ and you can get a quadratic in $a^2$ again.

-
there is an easier way! – mixedmath Jun 29 '11 at 16:26
Ok, that sounds reasonable, but I am not able to solve it, as I get a^4-b^4-4a^2b^2=1/2 and -4^3b+4ab^3=-sqrt(3)/2 I have no idea how i would go from here. – nino Jun 29 '11 at 16:41
@mixedmath and that would be? – nino Jun 29 '11 at 16:41
@nino: see my answer above ;p – mixedmath Jun 29 '11 at 16:41
For a rotation though known $\theta$ like $60^\circ$, square root, then fourth, is easy. Let the rotation take $(1,0)$ to $(a,b)$. Add these two vectors. Normalize the result. Repeat. – André Nicolas Jun 29 '11 at 18:43

Or you can use twice the half-angle formulas: http://oakroadsystems.com/twt/double.htm#SineHalf

Or maybe easier (avoiding square roots), to use the angle sum with $60^o$ and $-45^o$

-
he says that he wants to do this without using trig, and I think that using such formulas is of the same nature of using trig functions. Don't you? – mixedmath Jun 29 '11 at 16:34
@mixedmath: you're right; I repeatedly jump in and do not take time to read the question. My bad, I will try to change this. – gary Jun 29 '11 at 16:37

Since the idea is quite different, this is not tacked on to my previous answer. There may be a connection to the idea of mixedmath. The motivation was actually Ross Millikan's sketch of a calculation of the fourth root of a rotation matrix.

Suppose that we are explicitly given a rotation matrix, where for simplicity the rotation angle $\theta$ is between $0$ and $\pi$. We want to find the matrix for a rotation by $\theta/2$, $\theta/4$, or more generally $\theta/2^k$.

Compute what the rotation does to $(1,0)$. Say the result is $(a,b)$.

Find the sum $(1,0)+(a,b)=(1+a,b)$. Divide by the norm. This tells us where the point (1,0) is taken by rotation through $\theta/2$. Now we can immediately write down the appropriate rotation matrix.

For the rotation through $\theta/4$, repeat. For rotation through $\theta/2^n$, repeat more often.

We could think of this process as using the half-angle formulas for $\sin$ and $\cos$. That's certainly not the way I got to it, the idea was entirely primitive geometric. I prefer to think of the half-angle formulas as consequences of the above-described process.

-

I am answering in order to present the explicit result and comment on another answer. If the rotation matrix by the angle $\theta$ is denoted by $rot(\theta)$ then

$$rot(15^\circ) = rot(-45^\circ) rot(60^\circ) = rot(60^\circ) rot(-45^\circ)$$

as already stated by @user6312. So,

$$rot(15^\circ) = \left(\begin{matrix} \frac{1}{\sqrt2} & \frac{1}{\sqrt2}\\ \frac{-1}{\sqrt2} & \frac{1}{\sqrt2} \\ \end{matrix}\right) \left(\begin{matrix} \frac{1}{2} & \frac{-\sqrt3}{2}\\ \frac{\sqrt3}{2} & \frac{1}{2} \\ \end{matrix}\right) = \left(\begin{matrix} \frac{\sqrt6 + \sqrt2}{4} & -\frac{\sqrt6 - \sqrt2}{4}\\ \frac{\sqrt6 - \sqrt2}{4} & \frac{\sqrt6 + \sqrt2}{4} \\ \end{matrix}\right)$$ This is in fact a nice derivation of the explicit formulae for $cos(15^\circ)$ (matrix element 1,1) and $sin(15^\circ)$ (matrix element 2,1). Geometric derivations of these formulae can be found e.g. in R. Knott.

The suggestion by @mixedmath is not correct. Applying $R_+$ and $R_+$ to the columns of $rot(60^\circ)$ only gives values of the form 1/2 or $\sqrt3/2$, and never the values of $rot(15^\circ)$ (or am I missing something?).

-