# How to calculate Euler angles of inverse matrix?

Given three Euler angles $\alpha$, $\beta$ and $\gamma$, I can construct rotation matrix $R$ using proper Euler angles

$$R\left( \alpha ,\beta ,\gamma \right) = R_i\left( \alpha \right) R_j\left( \beta \right) R_k\left( \gamma \right), \qquad \left\langle i, j, k \right\rangle \in \pi\left( x, y, z \right),$$

where $\pi\left( x, y, z \right)$ is a set of all possible permutations of triplet $\left\langle x, y, z \right\rangle$.

Also Tait–Bryan angles can be used

$$R\left( \alpha ,\beta ,\gamma \right) = R_i\left( \alpha \right) R_j\left( \beta \right) R_i\left( \gamma \right), \qquad i, j \in \left\{ x, y, z \right\}, i \neq j.$$

Matrices are defined as folllows

$$R_x\left( \alpha \right) = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos{\alpha} & - \sin{\alpha} \\ 0 & \sin{\alpha} & \cos{\alpha} \end{bmatrix}$$

$$R_y\left( \beta \right) = \begin{bmatrix} \cos{\beta} & 0 & \sin{\beta} \\ 0 & 1 & 0 \\ - \sin{\beta} & 0 & \cos{\beta} \end{bmatrix}$$

$$R_z\left( \gamma \right) = \begin{bmatrix} \cos{\gamma} & - \sin{\gamma} & 0 \\ \sin{\gamma} & \cos{\gamma} & 0 \\ 0 & 0 & 1 \end{bmatrix}$$

Is there a formula that calculates Euler angles for inverse of matrix that was generated with known Euler angles? I mean such function $f$ that $$R\left( f\left( \alpha, \beta, \gamma \right) \right) = R^T\left( \alpha, \beta, \gamma \right)$$

I can simply generate a matrix by $\alpha$, $\beta$ and $\gamma$ angles, transpose it and decompose, but is there a faster way to do this?

• What about Tait-Bryan ? Here the order is the same... – Widawensen Nov 14 '17 at 18:28
• @Widawensen oh, thanks, I didn't notice. I should try it. – Charlie Nov 14 '17 at 18:29

$A$ is an inverse matrix for $R$ when $RA=I$.
Certainly composition of operations $$R_i\left( \alpha \right) R_j\left( \beta \right) R_k\left( \gamma \right) \ \ (R_k\left( -\gamma \right) R_j\left( -\beta \right) R_i\left( -\alpha \right)) =I$$ however the order of rotations for $A$ is inverse here and I don't know whether this is acceptable for the solution of your problem.