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

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1. What we are proving

Notice the sign gymnastics: in cosine the signs flip (plus outside → minus inside), in sine the signs match. That pattern is the whole memory game.

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2. Derivation from scratch — the unit-circle / distance method

We prove ==$\cos(A-B)$== first, because everything else falls out of it.

Step 1 — Place the points

On the unit circle:

• $P = (\cos A, \sin A)$
• $Q = (\cos B, \sin B)$

Why this step? By definition, a point at angle $\theta$ on the unit circle has coordinates $(\cos\theta,\sin\theta)$. This encodes the angles as geometry.

Step 2 — Chord length via coordinates (distance formula)

$PQ^2 = (\cos A-\cos B)^2 + (\sin A - \sin B)^2$ Expand: $= \cos^2 A - 2\cos A\cos B + \cos^2 B + \sin^2 A - 2\sin A\sin B + \sin^2 B$ Group using $\cos^2\theta+\sin^2\theta=1$: $= \underbrace{(\cos^2 A+\sin^2 A)}_{1} + \underbrace{(\cos^2 B+\sin^2 B)}_{1} - 2(\cos A\cos B + \sin A\sin B)$ $\boxed{PQ^2 = 2 - 2(\cos A\cos B + \sin A\sin B)}$

Why this step? The Pythagorean identity collapses four squared terms into a clean constant, leaving only the cross-terms — which is exactly what our formula is about.

Step 3 — Chord length via the angle between them

The angle from $Q$ to $P$ is $(A-B)$. Rotate the whole picture so $Q$ sits at angle $0$, i.e. at $(1,0)$, and $P$ sits at angle $(A-B)$, i.e. at $(\cos(A-B),\sin(A-B))$. Rotation doesn't change distances, so: $PQ^2 = (\cos(A-B)-1)^2 + (\sin(A-B)-0)^2$ $= \cos^2(A-B) - 2\cos(A-B) + 1 + \sin^2(A-B)$ $= 1 - 2\cos(A-B) + 1 = 2 - 2\cos(A-B)$

Why this step? Distance is invariant under rotation, so the same chord can be measured after simplifying the geometry — now only $(A-B)$ appears.

Step 4 — Equate the two expressions

$2 - 2\cos(A-B) = 2 - 2(\cos A\cos B + \sin A\sin B)$ $\therefore \cos(A-B) = \cos A\cos B + \sin A\sin B \quad\checkmark$

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3. Getting the other five for free

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4. Worked examples

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5. Forecast-then-Verify

Recall Predict before expanding: what is

$\cos(A+B)+\cos(A-B)$? Forecast a product... Expand: $(\cos A\cos B - \sin A\sin B)+(\cos A\cos B + \sin A\sin B) = 2\cos A\cos B$ The $\sin\sin$ terms cancel → this is the product-to-sum identity. Sums of these formulas kill half the terms; that's the engine behind all product-to-sum results.

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6. Common mistakes (Steel-manned)

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7. Flashcards

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Recall Feynman: explain to a 12-year-old

Imagine a clock hand spinning. Turning it by angle $A$ and then a bit more by angle $B$ ends in the same place as turning it by $A+B$ all at once. The sine and cosine are just the "height" and "sideways" positions of the tip. These formulas are the recipe that says: to find the new height/sideways after both turns, mix the heights and sideways of each turn together. That's why the answer isn't just adding them — the two turns get stirred together, some parts adding, some subtracting.

Connections

• Unit Circle and Trig Definitions — the coordinates $(\cos\theta,\sin\theta)$ start everything.
• Double Angle Formulas — set $B=A$ in these.
• Product-to-Sum and Sum-to-Product — add/subtract these identities.
• Pythagorean Identity — the collapse trick in Step 2.
• Derivatives of Trig Functions — the limit proof of $\frac{d}{dx}\sin x$ uses $\sin(x+h)$ expansion.
• Cofunction Identities — the bridge from cosine to sine proofs.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, in formulas ka core idea simple hai: agar tumhe do alag angles $A$ aur $B$ ke sine-cosine pata hain, to tum unke combined angle $A+B$ ka sine-cosine nikal sakte ho. Kyun? Kyunki circle pe angle $A$ ghumana phir $B$ ghumana, ekdum $A+B$ ghumane ke barabar hai. Yeh formulas basically "rotation ka algebra" hain.

Proof ka asli jugaad yeh hai: unit circle pe do points rakho angle $A$ aur $B$ par. Unke beech ki chord (seedhi line) ki length sirf $(A-B)$ pe depend karti hai. Ab chord length ko do tareeke se nikalo — ek baar distance formula se (coordinates use karke), doosri baar picture ko ghuma ke jahan ek point $(1,0)$ pe aa jaye. Dono expressions equal karo, aur $\cos(A-B)=\cos A\cos B+\sin A\sin B$ apne aap nikal aata hai. Baaki paanch formulas isi se free milte hain — $B$ ki jagah $-B$ daalo, ya cofunction $\sin\theta=\cos(90°-\theta)$ use karo, ya tangent ke liye upar-neeche $\cos A\cos B$ se divide karo.

Ek baat yaad rakhna jo students galat karte hain: $\sin(A+B)$ kabhi bhi $\sin A+\sin B$ nahi hota. Test karo $A=B=90°$: left side $=\sin180°=0$, right side $=2$. Bilkul galat! Sine linear nahi hai, woh rotation hai. Aur cosine mein signs ulte hote hain: $\cos(A+B)=\cos\cos-\sin\sin$. Formula: "Cosine is contrary".

Yeh matter kyun karta hai? $\sin75°$, $\cos15°$ jaise exact values, double-angle formulas, calculus mein $\sin x$ ka derivative — sab isi se aate hain. Ek baar proof samajh gaye to ratta maarne ki zaroorat hi nahi.