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Successive 2D rotations are additive

Short answer

Multiply matrices: R(θ1)·R(θ2) yields entries cosθ1cosθ2−sinθ1sinθ2 and sinθ1cosθ2+cosθ1sinθ2, which equal cos(θ1+θ2) and sin(θ1+θ2). Hence R(θ1)·R(θ2)=R(θ1+θ2), proving additivity.

Justify with necessary matrix operations that two successive rotations in 2-D is additive.

Claim: R(θ1) · R(θ2) = R(θ1 + θ2), so two successive rotations by θ1 and θ2 are the same as one rotation by θ1 + θ2.

2D rotation about the origin (homogeneous form):

R(θ) = | cosθ  -sinθ  0 |
       | sinθ   cosθ  0 |
       | 0      0     1 |

Rotating first by θ2 and then by θ1 gives the product R(θ1)·R(θ2):

R(θ1)·R(θ2) =
| cosθ1cosθ2 - sinθ1sinθ2   -(cosθ1sinθ2 + sinθ1cosθ2)   0 |
| sinθ1cosθ2 + cosθ1sinθ2    cosθ1cosθ2 - sinθ1sinθ2    0 |
| 0                          0                          1 |

Using the trigonometric identities

  • cos(θ1 + θ2) = cosθ1cosθ2 − sinθ1sinθ2
  • sin(θ1 + θ2) = sinθ1cosθ2 + cosθ1sinθ2
R(θ1)·R(θ2) = | cos(θ1+θ2)  -sin(θ1+θ2)  0 |
              | sin(θ1+θ2)   cos(θ1+θ2)  0 | = R(θ1 + θ2)
              | 0            0           1 |

Hence successive rotations are additive. Since θ1 + θ2 = θ2 + θ1, rotations about the same pivot point are also commutative.

✓ Answer checked by Claude Opus 5.5 (high effort)

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