Spin a logo by exactly 90^\circ, tilt a steering wheel by 33.7^\circ, sweep a clock hand smoothly through every angle in between — there's a matrix for that, and it's built directly out of sine and cosine. Every rotating icon, every turning clock hand, every spinning wheel you've ever seen on a screen is running the same short formula underneath, over and over, many times a second.

To rotate the whole plane about the origin by an angle \theta, we only need to know where \hat{\imath} and \hat{\jmath} go — and trigonometry tells us exactly, straight from the unit circle. The point \hat{\imath}=(1,0) sits at angle 0^\circ on the unit circle; rotating it by \theta just moves it round to angle \theta, landing at (\cos\theta, \sin\theta) — that's the very definition of sine and cosine. The point \hat{\jmath}=(0,1) sits at angle 90^\circ; rotating it by \theta moves it to angle 90^\circ+\theta, which the angle-sum identities turn into (-\sin\theta, \cos\theta). Stacking those two landing spots as columns gives the rotation matrix:

Turn the dial

Spin the angle below. The grid rotates rigidly — distances and angles are perfectly preserved, nothing stretches or skews. Watch the readout: it is exactly the two columns we just derived, (\cos\theta,\sin\theta) and (-\sin\theta,\cos\theta), updating live as \theta sweeps round.

Worked example: the 90° quarter-turn

Set \theta = 90^\circ. Since \cos 90^\circ = 0 and \sin 90^\circ = 1, the formula collapses to something beautifully simple:

Check it against \hat{\imath}=(1,0) directly: R(90^\circ)\begin{bmatrix}1\\0\end{bmatrix} = \begin{bmatrix}0\cdot 1 + (-1)\cdot 0\\ 1\cdot 1 + 0\cdot 0\end{bmatrix} = \begin{bmatrix}0\\1\end{bmatrix} — exactly where the first column says it should land. Try \hat{\jmath}=(0,1) too: R(90^\circ)\begin{bmatrix}0\\1\end{bmatrix} = \begin{bmatrix}-1\\0\end{bmatrix}, the second column. No surprises — the matrix multiplication just hands back the columns we already knew, because that's all a matrix ever does.

Worked example: rotating a point by 45° and 60°

Some angles give clean, memorable trig values, which makes them perfect for practising by hand. At 45^\circ, \cos 45^\circ=\sin 45^\circ=\tfrac{\sqrt2}{2}, so:

— the point (1,0) swings up to sit exactly on the diagonal line y=x, as you'd expect from splitting a right angle clean in half. At 60^\circ, \cos 60^\circ=\tfrac12 and \sin 60^\circ=\tfrac{\sqrt3}{2}, so rotating the vector (2,0) gives:

Notice the length is unchanged either way: (1,0) and (2,0) had lengths 1 and 2, and so do their rotated images — a rotation never stretches anything, it only turns it. Check the second one: the rotated vector is (1,\sqrt3), with length \sqrt{1^2+(\sqrt3)^2}=\sqrt{1+3}=\sqrt4=2 — exactly the length we started with.

Worked example: composing two rotations

What happens if you rotate by 30^\circ and then, on the result, rotate again by 60^\circ? Intuitively the two turns should just add up to one 90^\circ turn — and matrix multiplication confirms it exactly, because applying R(30^\circ) and then R(60^\circ) is the single matrix R(60^\circ)R(30^\circ):

The messy square roots all cancel out, leaving exactly the clean 90^\circ matrix from the first worked example. That is not a coincidence of these particular numbers — the top-left entry of the product works out to \cos60^\circ\cos30^\circ - \sin60^\circ\sin30^\circ, which is precisely the angle-sum identity for \cos(60^\circ+30^\circ). Multiplying rotation matrices and adding angles are, quietly, the exact same fact — angles of rotation simply add when you compose them, R(a)R(b)=R(a+b), however ugly the intermediate matrix multiplication looks. This is a first, gentle taste of a much bigger idea for later: composing two transformations — doing one, then another — is always the same as multiplying their matrices together, the subject of composing transformations.

Rotations that don't distort

A rotation is a rigid motion: it preserves every length and every angle, so it never changes area — its determinant is exactly 1. Its columns are perpendicular unit vectors, which makes it an orthogonal matrix, and undoing a rotation is simply rotating back: R(\theta)^{-1} = R(-\theta). Rotations are the workhorses of computer graphics, robotics and any place a thing needs to turn without warping.

You can check the "unit vector" claim yourself: the first column has length \sqrt{\cos^2\theta + \sin^2\theta}, which the Pythagorean identity \cos^2\theta+\sin^2\theta=1 collapses straight to 1, for every angle \theta — no exceptions, no special cases. And the inverse rule is worth a quick check too: rotating (1,0) by 30^\circ and then by -30^\circ should land you exactly back where you started, since \cos(-30^\circ)=\cos(30^\circ) and \sin(-30^\circ)=-\sin(30^\circ) — the second rotation is the mirror image of the first, undoing it perfectly.

See it explained