Visual walkthrough — Sum and difference formulas — sin(A±B), cos(A±B), tan(A±B) — proofs

Before step 1, we agree on the one object everything lives on.

---

Step 1 — Place two points at angles $A$ and $B$

WHAT. Mark two points on the circle: $P$ reached by turning through angle $A$, and $Q$ reached by turning through angle $B$.

WHY. We want a relationship about the combined angle. The only difference between $P$ and $Q$ is the angle between them, and that angle is $A-B$. So $A-B$ is baked into this picture geometrically — we haven't written a formula yet, but it's already there as the opening between the two arms.

PICTURE.

Reading the coordinates straight off the definition: $P=(\underset{\text{x of }P}{\underbrace{\cos A}},\underset{\text{y of }P}{\underbrace{\sin A}}),Q=(\underset{\text{x of }Q}{\underbrace{\cos B}},\underset{\text{y of }Q}{\underbrace{\sin B}})$

The grey dashed line joining $P$ and $Q$ is the chord. Our whole strategy: measure that one chord in two different ways.

---

Step 2 — Measure the chord using coordinates

WHAT. Use the distance formula (Pythagoras on the horizontal and vertical gaps) to get the squared length $P{Q}^2$.

WHY THIS TOOL — why distance, why squared? We need a number that both descriptions of the chord must agree on. Length is that number. We keep it squared to avoid a square root — squaring keeps everything as polynomials in $\cos$ and $\sin$, which we can expand and cancel cleanly.

PICTURE.

The horizontal gap is $\cos A-\cos B$ (orange), the vertical gap is $\sin A-\sin B$ (violet). Pythagoras on that little right triangle: $P{Q}^2=\underset{{\text{horizontal leg}}^2}{\underbrace{(\cos A-\cos B{)}^2}}+\underset{{\text{vertical leg}}^2}{\underbrace{(\sin A-\sin B{)}^2}}$

Multiply each bracket out: $P{Q}^2=\underset{\text{from horizontal}}{\underbrace{{\cos }^{2}A-2\cos A\cos B+{\cos }^{2}B}}+\underset{\text{from vertical}}{\underbrace{{\sin }^{2}A-2\sin A\sin B+{\sin }^{2}B}}$

---

Step 3 — Collapse it with the Pythagorean identity

WHAT. Group the squared terms and use ${\cos }^2\theta +{\sin }^2\theta =1$ twice.

WHY. Four of the six terms are "just the circle" — they must sum to something clean, because $P$ and $Q$ sit on a radius-$1$ circle. The Pythagorean Identity is exactly the statement "the tip's distance from the centre is $1$", so it turns those four terms into the constant $2$, leaving only the interesting cross-terms.

PICTURE.

$P{Q}^2=\underset{=1}{\underbrace{({\cos }^{2}A+{\sin }^{2}A)}}+\underset{=1}{\underbrace{({\cos }^{2}B+{\sin }^{2}B)}}-2(\cos A\cos B+\sin A\sin B)$

$\boxed{P{Q}^2=2-2\underset{\text{this is our target expression!}}{\underbrace{(\cos A\cos B+\sin A\sin B)}}}$

Notice: the exact combination $\cos A\cos B+\sin A\sin B$ has appeared on its own. Our job is now just to show it equals $\cos (A-B)$ — which the next step does with a single rotation.

---

Step 4 — Rotate the picture so the angle sits at the axis

WHAT. Spin the entire figure clockwise by angle $B$. Now $Q$ lands on the positive $x$-axis at $(1,0)$, and $P$ lands at angle $A-B$.

WHY THIS TOOL — why rotate? A rotation is a rigid motion: it slides the whole picture without stretching, so the chord length does not change. This is the key that lets us measure the same chord a second, simpler way — one where only the single angle $A-B$ is left in the picture.

PICTURE.

After the spin:

• $Q^{\prime }=(\underset{\cos 0}{\underbrace{1}},\underset{\sin 0}{\underbrace{0}})$ — sitting on the axis,
• $P^{\prime }=(\underset{x}{\underbrace{\cos (A-B)}},\underset{y}{\underbrace{\sin (A-B)}})$ — at the leftover angle $A-B$.

The red arc is the angle $A-B$, unchanged by the spin (both arms turned by the same $B$, so the opening between them is untouched).

---

Step 5 — Measure the same chord again, now the easy way

WHAT. Distance formula on the rotated points $P^{\prime }$ and $Q^{\prime }$.

WHY. Because $Q^{\prime }$ is at $(1,0)$, the algebra is far lighter — and it produces $\cos (A-B)$ directly.

PICTURE.

$P{Q}^2=\underset{{\text{horizontal gap}}^2}{\underbrace{(\cos (A-B)-1{)}^2}}+\underset{{\text{vertical gap}}^2}{\underbrace{(\sin (A-B)-0{)}^2}}$

Expand and use the Pythagorean identity one more time: $=\underset{\text{first bracket}}{\underbrace{{\cos }^2(A-B)-2\cos (A-B)+1}}+\underset{\text{second bracket}}{\underbrace{{\sin }^2(A-B)}}$ $=\underset{=1}{\underbrace{[{\cos }^2(A-B)+{\sin }^2(A-B)]}}-2\cos (A-B)+1=2-2\cos (A-B)$

---

Step 6 — Set the two measurements equal

WHAT. The chord is one physical thing, so its two squared lengths (Step 3 and Step 5) must match.

WHY. This is the whole point of measuring twice — equality forces the identity.

$\underset{\text{Step 5}}{\underbrace{2-2\cos (A-B)}}=\underset{\text{Step 3}}{\underbrace{2-2(\cos A\cos B+\sin A\sin B)}}$

Subtract $2$ from both sides, divide by $-2$: $\boxed{\cos (A-B)=\cos A\cos B+\sin A\sin B}✓$

---

Step 7 — The degenerate cases (nothing left uncovered)

A proof you can trust must survive the boring inputs. Test them on the final formula.

PICTURE.

• $B=0$ (points coincide, chord has length $0$): formula gives $\cos (A-0)=\cos A\cos 0+\sin A\sin 0=\cos A\cdot 1+\sin A\cdot 0=\cos A$. ✓ Consistent.
• $A=B$ ($P$ and $Q$ are the same point): $\cos (0)={\cos }^{2}A+{\sin }^{2}A=1$. ✓ This is literally the Pythagorean Identity falling out as a special case.
• $A-B=90°$ (arms perpendicular): $\cos 90°=0$, so the formula forces $\cos A\cos B+\sin A\sin B=0$ — the "perpendicular ⇒ dot product zero" fact, hidden inside.
• Negative / reflex angles: nothing in Steps 1–6 assumed $A,B$ were small or positive. Coordinates $(\cos \theta ,\sin \theta )$ are defined for every angle, in every quadrant, so the proof holds for all real $A,B$ — no case is special.

---

The one-picture summary

Left half: chord measured from raw coordinates → $2-2(\cos A\cos B+\sin A\sin B)$. Right half: same chord after a rigid rotation → $2-2\cos (A-B)$. The equals sign in the middle is the entire theorem. Everything else in the parent note — $\cos (A+B)$ (send $B\to -B$), the sines (via Cofunction Identities), the tangents (divide), the Double Angle Formulas (set $B=A$) — is downstream of this single equality.

Recall Feynman retelling — say it to a friend with no notation

Two ants sit on a round track of radius one. One is parked at "$A$ o'clock", the other at "$B$ o'clock". Stretch a straight string between them — that's the chord. Now I measure the string two ways. First way: I read off each ant's east–west and north–south position and use Pythagoras — this spits out $\cos A\cos B+\sin A\sin B$ (the circle-parts ${\cos }^2+{\sin }^2$ all quietly add to one and vanish). Second way: I spin the whole track until one ant sits exactly at "3 o'clock". The string never changed length — spinning doesn't stretch anything — but now the only thing that matters is the gap between the ants, which is $A-B$, and Pythagoras this time spits out $\cos (A-B)$. Same string, two answers, so they're equal. Cancel the boring $2$'s and you're left with $\cos (A-B)=\cos A\cos B+\sin A\sin B$. That's it — a string measured twice.

Connections

• Unit Circle and Trig Definitions — supplies $(\cos \theta ,\sin \theta )$, the only fact we start from.
• Pythagorean Identity — the collapse in Steps 3 and 5.
• Cofunction Identities — turns this cosine result into the sine formulas.
• Double Angle Formulas — the $A=B$ special case.
• Product-to-Sum and Sum-to-Product — add/subtract sibling identities.
• Derivatives of Trig Functions — these formulas power the limit definitions downstream.