Starting with a direction cosine matrix (DCM), we need to determine the three Euler angles. The connection is made by comparing the DCM elements with the combined Euler 213 sequence. It is important to note that the 12 combinations of Euler angles for a given sequence can be found from a given DCM. The DCM matrix is
![$\displaystyle DCM = \left[ \begin{array}{ccc} A_{11} & A_{12} & A_{13} \ A_{21} & A_{22} & A_{23} \ A_{31} & A_{32} & A_{33} \end{array} \right]$ $\displaystyle DCM = \left[ \begin{array}{ccc} A_{11} & A_{12} & A_{13} \ A_{21} & A_{22} & A_{23} \ A_{31} & A_{32} & A_{33} \end{array} \right]$](http://images.physicslibrary.org/cache/objects/68/l2h/img1.png) |
(1) |
The Euler 213 sequence is
![$\displaystyle R_3(\psi)R_1(\theta)R_2(\phi) = \left[ \begin{array}{ccc} c_{\psi... ...\ c_{\theta} s_{\phi} & -s_{\theta} & c_{\theta} c_{\phi} \end{array} \right]$ $\displaystyle R_3(\psi)R_1(\theta)R_2(\phi) = \left[ \begin{array}{ccc} c_{\psi... ...\ c_{\theta} s_{\phi} & -s_{\theta} & c_{\theta} c_{\phi} \end{array} \right]$](http://images.physicslibrary.org/cache/objects/68/l2h/img2.png) |
(2) |
If we examine the element in coloum 2 row 3, then by inspection

Solving for yields
 |
(3) |
Care must now be taken when evaluating the inverse sine. It is a multivalued function, which will have values of and
. Analytically, the convention is to choose the principle value such that

If a numerical program is used, a function asin() usually does this for us. The next step is to analyze the ratio
. Using these values from the Euler sequence we get

Rearranging the minus sign and using the tangent yields

Solving the quadrant ambiquity caused by the inverse tangent is done by examining the signs of the numerator and denominator. Denoting y as the numerator and x as the denominator, then the quadrant is chosen by:
quadrant 1
![$\left[ \begin{array}{cc} y & x \ + & + \ \end{array} \right] $ $\left[ \begin{array}{cc} y & x \ + & + \ \end{array} \right] $](http://images.physicslibrary.org/cache/objects/68/l2h/img12.png)
so if x and y are both positive, then
. Similarily for the other quadrant possibilites
quadrant 2
![$\left[ \begin{array}{cc} y & x \ + & - \ \end{array} \right] $ $\left[ \begin{array}{cc} y & x \ + & - \ \end{array} \right] $](http://images.physicslibrary.org/cache/objects/68/l2h/img14.png)
quadrant 3
![$\left[ \begin{array}{cc} y & x \ - & - \ \end{array} \right] $ $\left[ \begin{array}{cc} y & x \ - & - \ \end{array} \right] $](http://images.physicslibrary.org/cache/objects/68/l2h/img15.png)
quadrant 4
![$\left[ \begin{array}{cc} y & x \ - & + \ \end{array} \right] $ $\left[ \begin{array}{cc} y & x \ - & + \ \end{array} \right] $](http://images.physicslibrary.org/cache/objects/68/l2h/img16.png)
Ofcourse, it is much simplier to use a calculator or numerical program that uses the atan2() function which will choose the principle value. In a similar fashion we can find the final Euler angle by looking at the ratio
which gives the relation

Canceling terms and rearrangeing gives us

which uses the same method to resolove quadrant ambiguity as above. To summarize, we will give the formulas for the conversion from a direction cosine matrix to the Euler 213 angles in Matlab syntax. Be careful how you implement this in other numerical programs. In Matlab it goes atan2(y,x) and in Mathematica it is ArcTan[x, y].
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