Visual walkthrough — Transformations — translation, reflection, rotation, enlargement (basic)

The parent note handed you this formula like a gift already wrapped:

$x^{\prime }=x\cos \theta -y\sin \theta ,y^{\prime }=x\sin \theta +y\cos \theta$

But where does it come from? Why cosine and sine? Why the minus sign in only one place? This page builds that whole formula from a single dot spinning on a circle — no matrices assumed, no trig identities dropped from the sky. Everything you need is earned on the way.

We are going deeper than the parent topic. We will lean on Trigonometry, Coordinate Geometry, Vectors and the Distance Formula — but only after we build each idea in pictures.

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Step 1 — What "rotation" even means

WHAT. We take a point $P$ and turn it about the origin by an angle we call $\theta$ (the Greek letter "theta" — just a name for "how much we turn"). Turning anticlockwise (the same way a clock's hands do not go) is our positive direction.

WHY. Before any formula, we must nail down what stays fixed and what moves. In a rotation the distance from the origin never changes — that single fact is the seed the whole formula grows from.

PICTURE. The red dot swings along the dashed circle. Its distance $r$ from the origin (the radius of that circle) is frozen; only its angle changes.

Here:

• $r$ = the length of the arrow from origin to $P$ — the radius, unchanged by rotation.
• $\theta$ = the angle we sweep, measured anticlockwise from where $P$ started.

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Step 2 — Describe the point by an ANGLE, not just $x$ and $y$

WHAT. We rewrite the point $P=(x,y)$ using two new numbers: its distance $r$ from the origin and the angle $\varphi$ (Greek "phi") it currently makes with the positive $x$-axis. This pair $(r,\varphi )$ is called polar coordinates.

WHY. Here is the key move, and the reason for everything: rotation is awkward in $(x,y)$ because it mixes both coordinates together. But in $(r,\varphi )$ rotation is trivial — it just adds $\theta$ to the angle and leaves $r$ alone. We switch languages to the one where the problem is easy, solve it, then switch back. That is why trigonometry enters: sine and cosine are the exact dictionary between "angle" and "$x,y$".

PICTURE. The point sits on a right triangle: the horizontal leg is $x$, the vertical leg is $y$, and the slanted side (the hypotenuse) is $r$.

Rearranging each — multiply both sides by $r$ — gives the point's coordinates in terms of its angle:

$x=r\cos \varphi ,y=r\sin \varphi$

• $r\cos \varphi$ : take the full length $r$, keep only the fraction $\cos \varphi$ that points sideways → that's $x$.
• $r\sin \varphi$ : take the length $r$, keep only the fraction $\sin \varphi$ that points upward → that's $y$.

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Step 3 — Turning is just "$\varphi$ becomes $\varphi +\theta$"

WHAT. Rotate $P$ anticlockwise by $\theta$. In polar language the new point $P^{\prime }$ has the same $r$ but angle $\varphi +\theta$.

WHY. From Step 1, distance is frozen ($r$ stays). From Step 2, the only thing describing "which direction" is the angle. Turning by $\theta$ adds $\theta$ to that direction. Nothing else can change.

PICTURE. Same circle, same radius. The old arrow at angle $\varphi$; the new red arrow at angle $\varphi +\theta$. The little wedge between them is exactly $\theta$.

So by the same dictionary from Step 2, the new coordinates are:

$x^{\prime }=r\cos (\varphi +\theta ),y^{\prime }=r\sin (\varphi +\theta )$

• $x^{\prime }$ : the sideways part of the new arrow, which sits at angle $\varphi +\theta$.
• $y^{\prime }$ : the upward part of that same new arrow.

We are basically done — except $x^{\prime }$ and $y^{\prime }$ still mention $\varphi$, and we started only knowing $x,y$. We must eliminate $\varphi$. That is the whole job of the next step.

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Step 4 — Split the turned angle with the addition formulas

WHAT. We expand $\cos (\varphi +\theta )$ and $\sin (\varphi +\theta )$ using the angle-addition identities:

$\cos (\varphi +\theta )=\cos \varphi \cos \theta -\sin \varphi \sin \theta$ $\sin (\varphi +\theta )=\sin \varphi \cos \theta +\cos \varphi \sin \theta$

WHY. These identities are the tool that separates "where the point was" ($\varphi$, hidden inside $\cos \varphi ,\sin \varphi$) from "how far we turned" ($\theta$). We need that separation because we know $x,y$ (which carry $\varphi$) but not $\varphi$ itself. Splitting lets $\theta$ stand alone as the knob we control.

PICTURE. A geometric proof of the cosine rule for adding angles: a unit arm swept first by $\varphi$, then further by $\theta$, with the horizontal reach broken into the $\cos \varphi \cos \theta$ piece minus the $\sin \varphi \sin \theta$ piece.

Reading the picture: the total horizontal reach after both turns is the long $\cos \varphi \cos \theta$ segment minus the overshoot $\sin \varphi \sin \theta$ — that minus sign is the only minus in the whole final formula, and now you can see where it lives.

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Step 5 — Substitute and watch $\varphi$ vanish

WHAT. Put the expansions from Step 4 into the $x^{\prime },y^{\prime }$ from Step 3, then multiply the $r$ through.

Start with $x^{\prime }$:

$x^{\prime }=r\cos (\varphi +\theta )=r(\cos \varphi \cos \theta -\sin \varphi \sin \theta )$

Distribute the $r$ into each term:

$x^{\prime }=\underset{=x}{\underbrace{(r\cos \varphi )}}\cos \theta -\underset{=y}{\underbrace{(r\sin \varphi )}}\sin \theta$

WHY this is the magic moment. Look at the braces: $r\cos \varphi$ is exactly $x$ and $r\sin \varphi$ is exactly $y$ (straight from Step 2). The hidden angle $\varphi$ disappears, swallowed back into the $x$ and $y$ we already knew. We are left with only $x$, $y$, and our chosen turn $\theta$:

$\boxed{x^{\prime }=x\cos \theta -y\sin \theta }$

• $x\cos \theta$ : how much of the old sideways position survives as sideways after turning.
• $-y\sin \theta$ : the old upward part now leans sideways — and it leans in the negative $x$ direction for an anticlockwise turn, hence the minus.

Do the identical trick for $y^{\prime }$:

$y^{\prime }=r\sin (\varphi +\theta )=r(\sin \varphi \cos \theta +\cos \varphi \sin \theta )=\underset{=y}{\underbrace{(r\sin \varphi )}}\cos \theta +\underset{=x}{\underbrace{(r\cos \varphi )}}\sin \theta$

$\boxed{y^{\prime }=x\sin \theta +y\cos \theta }$

• $x\sin \theta$ : the old sideways part now tips upward — positive, so $+$.
• $y\cos \theta$ : the surviving upward part.

That is exactly the parent's formula, now fully earned.

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Step 6 — The matrix is just these two lines stacked

WHAT. Write both results together. Stacking $x^{\prime }$ and $y^{\prime }$ and pulling out the pattern gives the rotation matrix:

$\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)=\left(\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)$

WHY. A matrix is nothing mystical here — it is a compact box that says "row 1 makes $x^{\prime }$, row 2 makes $y^{\prime }$." Read row 1 across: $\cos \theta \cdot x+(-\sin \theta )\cdot y$ — that is $x^{\prime }$. Read row 2: $\sin \theta \cdot x+\cos \theta \cdot y$ = $y^{\prime }$. Nothing new; only tidier.

PICTURE. The four grid arrows: where the point $(1,0)$ lands (first column) and where $(0,1)$ lands (second column). The columns of the matrix are literally "where the two basic direction-arrows go."

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Step 7 — Every special case falls out by plugging in

WHAT. Test the formula at the famous angles. No new work — just substitute known values of $\cos \theta ,\sin \theta$.

$\theta$	$\cos \theta$	$\sin \theta$	$(x,y)\to$
$90^{\circ }$	$0$	$1$	$(-y,x)$
$180^{\circ }$	$-1$	$0$	$(-x,-y)$
$270^{\circ }$	$0$	$-1$	$(y,-x)$
$0^{\circ }$ (do nothing)	$1$	$0$	$(x,y)$

WHY show all of them. Each is a quadrant-check. A point in quadrant I (right-up) must land in quadrant II (left-up) after a $90^{\circ }$ anticlockwise turn — and $(-y,x)$ does exactly that when $x,y>0$. The $0^{\circ }$ row is the degenerate case: turning by nothing leaves the point untouched, which the formula respects ($x^{\prime }=x,y^{\prime }=y$).

PICTURE. One dot in quadrant I, and its three images after $90^{\circ }$, $180^{\circ }$, $270^{\circ }$, marching anticlockwise through quadrants II, III, IV.

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The one-picture summary

The whole derivation in a single frame: point at angle $\varphi$, radius $r$ frozen, swept by $\theta$ to angle $\varphi +\theta$; the coordinates read off both arrows via cosine (sideways) and sine (upward).

Recall Retell it like Feynman

A rotation is just swinging a point on a string tied to the origin — the string length $r$ never changes, only the direction. Directions are best described by an angle, and sine/cosine are the translator between an angle and the point's $x,y$. Turning by $\theta$ simply adds $\theta$ to the angle — dead easy in angle-language. Then I use the addition formula to split "old angle plus turn" into pieces; the "old angle" pieces are secretly just my original $x$ and $y$ (because $x=r\cos \varphi$, $y=r\sin \varphi$), so they collapse back and vanish. What's left is $x^{\prime }=x\cos \theta -y\sin \theta$ and $y^{\prime }=x\sin \theta +y\cos \theta$. The single minus sign comes from an anticlockwise turn pulling the "up" part toward the left. Stack those two lines and you've written the rotation matrix; plug in $90^{\circ }$, $180^{\circ }$, $270^{\circ }$, $0^{\circ }$ to get every classic shortcut, and plug in the origin to see the one point that never budges.

The formula that undoes a rotation — turning back by $-\theta$ — is its inverse; swap the sign of $\theta$ and every $\sin \theta$ flips sign, sending you the other way around the circle.

Recall Quick self-test

Why does $r$ stay constant under rotation? ::: Rotation swings the point on a fixed-length string from the origin; only direction changes, not distance. Where does the minus sign in $x^{\prime }=x\cos \theta -y\sin \theta$ come from? ::: The angle-addition formula $\cos (\varphi +\theta )=\cos \varphi \cos \theta -\sin \varphi \sin \theta$; geometrically, the old "up" part leans into negative $x$ for an anticlockwise turn. What are the columns of the rotation matrix? ::: Where the arrows $(1,0)$ and $(0,1)$ land after the turn. Image of $(x,y)$ under $90^{\circ }$ anticlockwise? ::: $(-y,x)$.