# How do you prove Euler's angle formula?

Euler's rotation theorem states that any rotation in $\mathbb{R}^3$ can be described by $3$ parameters.

Theorem Any rotation of the $xyz$-space is the composition of a rotation around the $z$ axis, followed by a rotation about the $x$ axis, followed by a rotation around the $z$ axis.

While this theorem is not too hard to prove I am looking for different arguments implying this result whether elementary (for instance using classical geometry, or linear algebra) or advanced (say using Lie theory).

(Another title for this post could be: what is your favorite proof of Euler angle formula/Euler rotation theorem.)

PS The reason for asking this question is that many of the (online) resources that I have consulted omit the proof and the other ones are not very enlightening for me.

• Artin's Algebra discusses this in some detail. Feb 22, 2015 at 0:45
• @rogerl Thanks! Do you know in which chapter of Artin's book this is discussed?
– EPS
Feb 22, 2015 at 1:01
• @rogerl I'm afraid you might have confused this theorem with something else: I was not able to find it in Artin's book.
– EPS
Feb 22, 2015 at 1:54
• Any arguments that you've seen before and wouldn't need to see again? Feb 22, 2015 at 6:54
• @Muphrid The geometric solution that I knew was very similar to what David has posted below; it is however good to see his solution because I still learned something from it. The Lie theoretic solution that I have in mind is proving a similar result for $SU(2)$, where the algebraic proof might be easier, and then mapping it to $SO(3)$. There is another argument in Sepanski's book on compact Lie groups (as an exercise) that I have seen but I am not sure if I can follow the hints of the author. I will sketch it below.
– EPS
Feb 22, 2015 at 18:01

Consider an arbitrary rotation $R$ in $\mathbb R^3.$ Let the images of the $x$-, $y$-, and $z$-axes under $R$ be labeled the $x'$-, $y'$-, and $z'$-axes respectively.

There is a rotation by some angle $\alpha$ around the $z$-axis that takes the $z'$-axis to the $yz$-plane. Let the $z''$-axis be the image of the $z'$-axis under this rotation.

There is then a rotation by some angle $\beta$ around the $x$-axis that takes the positive $z''$-axis to the positive $z$-axis.

Let the $x'''$-axis be the image of the $x'$-axis under the previous two rotations. A rotation by some angle $\gamma$ around the $z$-axis then takes the positive $x'''$-axis to the positive $x$-axis.

These three rotations, performed in this sequence, take the $x'$-, $y'$-, and $z'$-axes back to the the $x$-, $y$-, and $z$-axes in their proper orientation. To obtain the desired rotation $R$, therefore, simply reverse these three rotations. That is, the desired $zxz$-convention Euler angles are $(\phi, \theta, \psi) = (-\gamma, -\beta, -\alpha)$.

Regarding the dreaded "gimbal lock" problem, consider a rotation $R$ that consists of a very-small-angle rotation about the $y$-axis. That is, $R$ takes the positive $y$-axis to itself, but displaces the $x$- and $z$-axes slightly. To reverse $R$ via $zxz$ Euler angles, we must first rotate through an angle $\frac\pi2$ (measured in radians) about the $z$-axis in order to take points from the $xz$-plane to the $yz$-plane. We then make a very small rotation about the $x$-axis followed by another rotation of $\frac\pi2$ about the $z$-axis. If we want the rotation $R$ to be performed on a reference body mounted on an axle within a ring-shaped gimbal which in turn is mounted on an axle within another gimbal which itself is mounted on an axle fixed to a non-rotating frame, then the $zxz$ Euler angles correspond to a starting position in which the axle between two two gimbals lies along the $x$-axis and the other two axles lie along the $z$-axis. That is, the starting position for this Euler convention is already in gimbal lock. It is impossible to produce a continuous rotation about the $y$-axis by changing the $(\phi, \theta, \psi)$ Euler angles continuously. This makes this an unsuitable configuration of the gimbals of a spacecraft's attitude indicator, which must be able to rotate freely within its outer frame.

• +1 I had not heard of "gimbal lock" until quite recently. Very interesting! "This makes this an unsuitable configuration of the gimbals of a spacecraft's attitude indicator." Is it possible for you to explain how this difficulty is overcome in practice?
– EPS
Feb 22, 2015 at 17:56
• I'm not an expert on spacecraft attitude control, but I can point to Gimbal Angles, Gimbal Lock, and a Fourth Gimbal for Christmas. To summarize, the IMU (device that showed attitude) of the Apollo lander had three gimbal axles; astronauts were trained to avoid putting the lander in an attitude that would make two axles of the gimbals too nearly parallel. A diagram of the IMU shows the axles pairwise orthogonal. A rotation from that configuration would correspond to $yzx$-convention Euler angles. Feb 22, 2015 at 18:26
• Also see math.stackexchange.com/questions/8980/… regarding gimbal lock. Feb 22, 2015 at 18:26
• This is great David. Many thanks!
– EPS
Feb 22, 2015 at 20:40

Since e.g. this Wikipedia article gives conversions between Euler angles and rotation matrices, all you have to do is check whether these apply to all possible inputs, and really have the desired properties. Sure, you'll need some special considerations for the gimbal-locked case, but that shouldn't be too hard either.