Worked examples — Direction cosine matrix (DCM) — construction from Euler angles

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The scenario matrix

Every DCM problem is one (or a blend) of these cells. The worked examples below each announce which cell they cover.

Cell	What makes it special	Where the trouble hides	Example
A. Zero input	all angles $=0$	should give identity $I$ — a sanity anchor	Ex 1
B. Single axis, positive	one angle $>0$, others $0$	sign convention of $R_i$	Ex 2
C. Single axis, negative	one angle $<0$	does $-\theta$ flip the right off-diagonal?	Ex 3
D. Quadrant / large angle	angle $>90^{\circ }$ (e.g. $135^{\circ }$)	$\sin ,\cos$ signs; which quadrant	Ex 4
E. Full 3-2-1, all nonzero	yaw, pitch, roll all active	matrix-multiply order, cross terms	Ex 5
F. Inverse problem	given $C$, find angles	atan2 quadrants, $\arcsin$ range	Ex 6
G. Degenerate: gimbal lock	pitch $=\pm 90^{\circ }$	angles not unique; $\cos \theta =0$	Ex 7
H. Word problem (real GNC)	rotate a sensor vector	frame direction; forward vs inverse	Ex 8
I. Exam twist: orthonormality	prove/exploit $C^{-1}=C^T$	trusting $C^T$ instead of re-inverting	Ex 9

We use the shorthand from the parent: $c=\cos$, $s=\sin$. Angles: yaw $\psi$ about axis 3, pitch $\theta$ about axis 2, roll $\varphi$ about axis 1. All matrices transform inertial → body: ${\mathbf{v}}_B=C_{BN}{\mathbf{v}}_N$.

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Example 1 — Cell A: the zero rotation

Forecast: guess before reading — if the body hasn't turned at all, what should the DCM be, and what should happen to the vector?

1. Plug into each elementary matrix. With every angle $0$: $c0=1,s0=0$, so $R_3(0)=R_2(0)=R_1(0)=I$ (the identity). Why this step? The elementary matrices reduce to identity because their $\sin$ entries vanish and their $\cos$ entries become $1$ — this is the baseline every other case is measured against.
2. Multiply. $C_{BN}=I\cdot I\cdot I=I$. Why this step? Multiplying by identity changes nothing; a "no turn" attitude must be $I$.
3. Apply. ${\mathbf{v}}_B=I[2,-3,5{]}^T=[2,-3,5{]}^T$.

Verify: if the frames coincide, the same physical vector has the same components in both. It does. ✔

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Example 2 — Cell B: single-axis positive yaw

Forecast: after turning the body left by $30^{\circ }$, will the old North axis appear to the body slightly right (negative $y_B$) or left?

1. Only $R_3$ survives. $C_{BN}=R_3(30^{\circ })=\left[\begin{array}{ccc}c30& s30& 0\\ -s30& c30& 0\\ 0& 0& 1\end{array}\right]$ with $c30=\frac{\sqrt{3}}{2}\approx 0.866$, $s30=0.5$. Why this step? Pitch and roll matrices are identity, so the product collapses to the single yaw block.
2. Apply to $[1,0,0{]}^T$. Row by row: $x_B=0.866$, $y_B=-0.5$, $z_B=0$. Why this step? The transform reads off "how much of inertial ${\hat{n}}_1$ lies along each body axis." The negative $y_B$ says the old North now lies partly along body $-y$.

Figure: The gray axes (${\hat{n}}_1,{\hat{n}}_2$) are the fixed inertial frame; the blue axes (${\hat{b}}_1,{\hat{b}}_2$) are the body, turned $+30^{\circ }$ counter-clockwise. What to notice: the fixed North arrow ${\hat{n}}_1$ has a large positive projection onto ${\hat{b}}_1$ (orange dashed, $x_B=+0.866$) and a negative projection onto ${\hat{b}}_2$ (red dashed, $y_B=-0.5$). Turning the body CCW makes the fixed vector appear to lean the other way — into the body's $-y$ half.

Verify: length preserved? $\sqrt{0.866^2+(-0.5{)}^2}=\sqrt{0.75+0.25}=1$. ✔ The signs match the figure: the body turns $+30^{\circ }$ (CCW), so the fixed North vector's body-frame coordinates rotate $-30^{\circ }$. ✔

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Example 3 — Cell C: single-axis NEGATIVE roll

Forecast: the parent's $R_1$ has $+\sin$ in the top-right of its lower block. With a negative angle, which off-diagonal entry becomes negative?

1. Substitute $\varphi =-45^{\circ }$. $\cos (-45^{\circ })=\frac{\sqrt{2}}{2}\approx 0.707$ (cosine is even), $\sin (-45^{\circ })=-0.707$ (sine is odd). Why this step? Cosine ignores the sign of the angle; sine flips it. So the two $\sin$ entries in $R_1$ swap sign relative to $\varphi =+45^{\circ }$.
2. Write the matrix. $R_1(-45^{\circ })=\left[\begin{array}{ccc}1& 0& 0\\ 0& 0.707& -0.707\\ 0& 0.707& 0.707\end{array}\right].$ Why this step? Row 2 gets $s\varphi =-0.707$ in the third slot; row 3 gets $-s\varphi =+0.707$.
3. Apply to $[0,1,0{]}^T$. $x_B=0$, $y_B=0.707$, $z_B=0.707$.

Verify: length $=\sqrt{0.707^2+0.707^2}=1$. ✔ A negative roll tips the body so the inertial ${\hat{n}}_2$ picks up a $+z_B$ component — consistent with rolling the "up" direction toward you. ✔

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Example 4 — Cell D: obtuse-angle yaw (quadrant care)

Forecast: $135^{\circ }$ is in the second quadrant. Guess the signs of $\cos 135^{\circ }$ and $\sin 135^{\circ }$ before computing.

1. Evaluate the trig. $\cos 135^{\circ }=-\frac{\sqrt{2}}{2}\approx -0.707$ (second quadrant: cosine negative), $\sin 135^{\circ }=+\frac{\sqrt{2}}{2}\approx 0.707$ (sine positive). Why this step? Past $90^{\circ }$ the cosine goes negative — a case single-axis-small-angle intuition can miss. Getting the quadrant right is the whole point of this cell.
2. Matrix. $R_3(135^{\circ })=\left[\begin{array}{ccc}-0.707& 0.707& 0\\ -0.707& -0.707& 0\\ 0& 0& 1\end{array}\right]$.
3. Apply to $[1,0,0{]}^T$. $x_B=-0.707$, $y_B=-0.707$, $z_B=0$.

Verify: length $=\sqrt{0.5+0.5}=1$ ✔. The old North now points into the body's third quadrant ($-x_B,-y_B$) — exactly where a $135^{\circ }$ body turn should send it. ✔

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Example 5 — Cell E: full 3-2-1, all angles nonzero

Forecast: the parent noted $C_{13}=-s\theta$ is untouched by yaw and roll. Predict $C_{13}$ before computing.

1. Read off $C_{13}$ from the closed form. From the parent's full matrix, $C_{13}=-s\theta =-\sin 45^{\circ }\approx -0.707$. Why this step? This entry depends on pitch alone — a free sanity anchor independent of the messy cross terms.
2. Read off $C_{33}$. $C_{33}=c\varphi c\theta =\cos 60^{\circ }\cos 45^{\circ }=0.5\times 0.707=0.354$. Why this step? $C_{33}$ mixes roll and pitch only (no yaw) — a second, independent anchor to catch multiply errors.
3. Cross-check by multiplying. Form $R_2(45^{\circ })R_3(30^{\circ })$ first, then premultiply by $R_1(60^{\circ })$, and read the same two entries. Why this step? If the closed-form shortcut and the brute-force product agree, our order-of-multiplication (last rotation leftmost) is correct.

Verify: brute-force multiply gives $C_{13}=-0.7071$, $C_{33}=0.3536$ — matching steps 1–2. The full matrix (used again in Example 6) is

$C=\left[\begin{array}{ccc}0.6124& 0.3536& -0.7071\\ 0.0474& 0.7891& 0.6124\\ 0.7891& -0.4967& 0.3536\end{array}\right].$ ✔

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Example 6 — Cell F: the inverse problem (recover angles)

Forecast: which entry gives pitch directly, and why must we use atan2 rather than plain $\arctan$?

1. Pitch from $C_{13}$. $\theta =-\arcsin (C_{13})=-\arcsin (-0.7071)=45^{\circ }$. Why this step? $C_{13}=-s\theta$, so $\arcsin$ inverts it. $\arcsin$ returns $[-90^{\circ },90^{\circ }]$ — the valid pitch range for 3-2-1, so no quadrant ambiguity here.
2. Yaw from $C_{11},C_{12}$. $\psi =atan2(C_{12},C_{11})=atan2(0.3536,0.6124)=30^{\circ }$. Why this step? Plain $\arctan (C_{12}/C_{11})$ can't tell $30^{\circ }$ from $210^{\circ }$ (tan repeats every $180^{\circ }$). atan2 takes both numerator and denominator signs, pinning the correct quadrant.
3. Roll from $C_{23},C_{33}$. $\varphi =atan2(C_{23},C_{33})=atan2(0.6124,0.3536)=60^{\circ }$. Why this step? $C_{23}=s\varphi c\theta$ and $C_{33}=c\varphi c\theta$; their ratio is $\tan \varphi$, and the shared positive $c\theta$ factor means atan2 reads the correct quadrant straight off. Both entries positive ⇒ first-quadrant roll ⇒ $60^{\circ }$.

Verify: the recovered triple $(\psi ,\theta ,\varphi )=(30^{\circ },45^{\circ },60^{\circ })$ is exactly the Example-5 input. Round-trip closed. ✔

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Example 7 — Cell G: gimbal lock (degenerate pitch)

Forecast: with $\cos 90^{\circ }=0$, how many independent numbers can survive in the matrix — three angles' worth, or fewer?

1. Collapse the matrix with $c\theta =0,s\theta =1$. Substituting into the parent's full 3-2-1 form kills every term carrying a $c\theta$ factor. What remains depends on yaw and roll only through their difference $\psi -\varphi$. (For 3-2-1 at $\theta =+90^{\circ }$ the surviving off-diagonal entries are $\sin (\psi -\varphi )$ and $\cos (\psi -\varphi )$.) Why this step? When $\cos \theta =0$ the yaw axis (3) and roll axis (1) line up along the same physical direction, so only one combined turn is observable.
2. Case $(\psi ,\varphi )=(20^{\circ },50^{\circ })$: here $\psi -\varphi =-30^{\circ }$. Plugging in gives $C_1=\left[\begin{array}{ccc}0& 0& -1\\ \sin (\psi -\varphi )& \cos (\psi -\varphi )& 0\\ \cos (\psi -\varphi )& -\sin (\psi -\varphi )& 0\end{array}\right]=\left[\begin{array}{ccc}0& 0& -1\\ -0.5& 0.8660& 0\\ 0.8660& 0.5& 0\end{array}\right].$ Why this step? Writing the actual numbers lets us see the collapse: the entire first row is $[0,0,-1]$, fixed by pitch alone.
3. Case $({\psi }^{\prime },{\varphi }^{\prime })=(45^{\circ },25^{\circ })$: here ${\psi }^{\prime }-{\varphi }^{\prime }=+20^{\circ }\ne -30^{\circ }$ — different! Pick instead $({\psi }^{\prime },{\varphi }^{\prime })=(60^{\circ },90^{\circ })$ so ${\psi }^{\prime }-{\varphi }^{\prime }=-30^{\circ }$. With the same difference $-30^{\circ }$: $C_2=\left[\begin{array}{ccc}0& 0& -1\\ -0.5& 0.8660& 0\\ 0.8660& 0.5& 0\end{array}\right]=C_1.$ Why this step? Two different angle pairs, $(20^{\circ },50^{\circ })$ and $(60^{\circ },90^{\circ })$, share the difference $-30^{\circ }$, so they produce the identical matrix. One matrix, many labels ⇒ the parameterization is singular. The DCM is perfectly healthy; only the Euler names break. This is why GNC stores attitude as quaternions.

Verify: every entry of $C_1$ equals the corresponding entry of $C_2$. ✔

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Example 8 — Cell H: real GNC word problem

Forecast: a $90^{\circ }$ pitch tips the nose all the way up. Guess where a straight-up star lands in body coordinates before computing.

1. Build $C_{BN}=R_1(0)R_2(90^{\circ })R_3(90^{\circ })$. With $\theta =90^{\circ }$: $c\theta =0,s\theta =1$; with $\psi =90^{\circ }$: $c\psi =0,s\psi =1$; $\varphi =0$ ⇒ $R_1=I$. Why this step? Sensors live in the body frame; guidance gives the star in inertial. The DCM is the translator, applied in the forward direction $N\to B$.
2. Apply $C_{BN}$ to $[0,0,1{]}^T$. The third column of $C_{BN}$ is what multiplies the $z$-component; it equals $[-s\theta ,s\varphi c\theta ,c\varphi c\theta {]}^T=[-1,0,0{]}^T$. Why this step? Multiplying a matrix by $[0,0,1{]}^T$ just picks out its third column — a shortcut worth remembering.
3. Read the result. ${\mathbf{v}}_B=[-1,0,0{]}^T$: the star lies along the body $-x$ axis.

Verify: length $=1$ ✔. Physically: pitching the nose up by $90^{\circ }$ swings the body $x$-axis to point where "up" was, so the up-star sits along $-x_B$ (behind the nose direction). ✔

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Example 9 — Cell I: exam twist, exploit $C^{-1}=C^T$

Forecast: the naive move is Gaussian elimination on $C$. The parent gave a shortcut — what is it?

1. Recall the property. For any DCM, $C^{-1}=C^T$ (orthonormality). So ${\mathbf{v}}_N=C_{BN}^{-1}{\mathbf{v}}_B=C_{BN}^T{\mathbf{v}}_B$. Why this step? Transposing is free (swap rows/columns); inverting a general $3\times 3$ costs real work. Trusting orthonormality is the whole exam trick.
2. Transpose $R_3(30^{\circ })$. $C^T=\left[\begin{array}{ccc}c30& -s30& 0\\ s30& c30& 0\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{ccc}0.866& -0.5& 0\\ 0.5& 0.866& 0\\ 0& 0& 1\end{array}\right]$. Why this step? The $+s$ and $-s$ swap positions — that IS the inverse rotation $R_3(-30^{\circ })$, which makes sense: undoing a $+30^{\circ }$ yaw is a $-30^{\circ }$ yaw.
3. Apply. ${\mathbf{v}}_N=C^T[0,1,0{]}^T=[-0.5,0.866,0{]}^T$.

Verify: confirm $C{\mathbf{v}}_N={\mathbf{v}}_B$: $R_3(30^{\circ })[-0.5,0.866,0{]}^T=[0,1,0{]}^T$? Row1: $0.866(-0.5)+0.5(0.866)=0$ ✔; Row2: $-0.5(-0.5)+0.866(0.866)=0.25+0.75=1$ ✔. Round-trip holds. ✔

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Recall One-line self-test for each cell

Zero input → identity ::: $C=I$, vector unchanged. Negative angle → which entries flip? ::: only the two $\sin$ entries (cos is even, sin is odd). Obtuse yaw ($135^{\circ }$) → sign of $\cos$? ::: negative (second quadrant). Inverse problem → why atan2 not arctan? ::: tan repeats every $180^{\circ }$; atan2 uses both signs to fix the quadrant. Gimbal lock condition ::: pitch $\theta =\pm 90^{\circ }$, so $\cos \theta =0$ and yaw & roll merge onto one axis — only $\psi -\varphi$ survives. Undo a DCM cheaply ::: use $C^T$, since $C^{-1}=C^T$.

Connections

• 3.5.03 Direction cosine matrix (DCM) — construction from Euler angles (Hinglish)
• Quaternions — avoiding gimbal lock
• Euler angles — kinematic differential equations
• Rotation group SO(3) and orthogonal matrices
• Attitude determination — TRIAD & QUEST
• Kalman filter — linearized attitude error state
• Angular velocity and skew-symmetric matrices