Derivatives of sin x, cos x — proofs from first principles: Visual walkthrough

Step 1 — What "derivative" is asking (a slope you shrink)

WHAT. Pick a point on the sine curve. Draw a second point a little to the right, a horizontal gap $h$ away. Connect them with a straight line. That line has a steepness — how much it climbs ( $Δ y$ ) divided by how far it runs ( $h$ ).

WHY. Slope is the only honest way to say "how fast is this curve rising here?" A single point has no slope on its own, so we borrow a nearby friend at distance $h$ , measure the average climb between them, then slide the friend inward ( $h \to 0$ ) until the two points nearly touch. What the slope settles down to is the derivative.

PICTURE. The amber secant line below tilts less and less as the right point creeps left; it is heading toward the cyan tangent line.

Here $f (x) = sin x$ , so the rise is $sin (x + h) - sin x$ . Every step below is about taming that rise.

Step 2 — Why we need one small-angle fact first

WHAT. Before touching $sin (x + h)$ , we isolate the engine of the whole proof: what happens to $\frac{sin h}{h}$ when $h$ is tiny?

WHY. Look ahead: after expanding, the answer will contain $\frac{sin h}{h}$ . If we don't know that number, the proof stalls. So we settle it now, geometrically, using a unit circle (a circle of radius $1$ ).

PICTURE. On the unit circle, sweep out a small angle $h$ (measured in radians — the arc length itself). Three regions nest inside one another: a thin triangle, the pie-slice (sector), and a bigger triangle.

The inner region sits inside the sector which sits inside the outer triangle — the picture is the inequality.

Step 3 — Squeeze the ratio to 1

WHAT. Divide the whole inequality by the positive quantity $\frac{1}{2} sin h$ , then flip it, and let $h \to 0$ .

WHY. We want $\frac{sin h}{h}$ trapped between two things that both march to the same limit. That is exactly what the Squeeze Theorem needs: if a quantity is pinned between two others heading to $1$ , it must also head to $1$ .

PICTURE. As $h$ shrinks, the two straight sides and the curved arc become indistinguishable — the ratio $\frac{sin h}{h}$ is squeezed onto $1$ .

Flipping the trapped ratio gives the star of the show: $h \to 0 lim \frac{sin h}{h} = 1$ It holds for $h < 0$ too, because $\frac{sin h}{h}$ is unchanged when $h \to - h$ (an even ratio).

Step 4 — The cosine companion limit

WHAT. Establish the second number the proof will need: $lim_{h \to 0} \frac{cos h - 1}{h}$ .

WHY. After expanding $sin (x + h)$ a $cos h - 1$ piece will appear. We can't guess its limit — it's a $\frac{0}{0}$ shape. We convert it into the known $\frac{s i n h}{h}$ using the conjugate trick (multiply by $cos h + 1$ top and bottom).

PICTURE. Geometrically $1 - cos h$ is the tiny horizontal sag of the circle point away from the right edge — it shrinks faster than $h$ , so the ratio dies to zero.

$h \to 0 lim \frac{cos h - 1}{h} = 0$

Step 5 — Expand the rise with the addition formula

WHAT. Now attack the actual derivative. Replace $sin (x + h)$ using the angle-addition formula.

WHY. The rise $sin (x + h) - sin x$ mixes $x$ (fixed) and $h$ (vanishing) inside one sine. We must separate them so the $h$ -parts can meet the limits from Steps 3–4. The addition formula is the tool that splits an angle-sum into pieces where $x$ and $h$ stand apart.

PICTURE. Think of rotating a unit vector from angle $x$ by an extra $h$ ; its new height $sin (x + h)$ is built from the old height and width scaled by $cos h, sin h$ .

Step 6 — Group into the two known limits

WHAT. Divide by $h$ , then gather the $sin x$ terms together.

WHY. We want the expression to look like $sin x \cdot (Step 4 limit) + cos x \cdot (Step 3 limit)$ . Grouping is pure bookkeeping so the recognisable pieces surface.

PICTURE. Two labelled "slots" — one waiting for the value $0$ , one waiting for the value $1$ .

Step 7 — Take the limit and read the answer

WHAT. Let $h \to 0$ . Constants $sin x, cos x$ ride outside the limit; the two slots fill with $0$ and $1$ .

WHY. A limit of a sum is the sum of limits (from Limits — definition and properties), and constant factors pass straight through. So each slot just becomes its known value.

PICTURE. The two slots snap to $0$ and $1$ ; only the $cos x$ term survives.

The mirror proof for cosine is identical with $cos (x + h) = cos x cos h - sin x sin h$ , giving $\frac{d}{d x} cos x = cos x \cdot 0 - sin x \cdot 1 = - sin x$ .

Step 8 — The degenerate & edge cases (never skip these)

WHAT. Check the corners the pictures quietly assumed.

WHY. A proof isn't finished until every input is covered.

PICTURE. Where sine crosses zero going up, its slope is maximal ( $cos = + 1$ ); at a peak, slope is zero.

Every $x$ across all four quadrants gives a slope matching the cosine wave, and both signs of $h$ agree — the result is watertight.

The one-picture summary

Below: the sine curve (cyan) with its slope-arrows, and directly beneath the cosine curve (amber) that those slopes trace out. The proof is the statement that these two frames are locked together.

Recall Feynman retelling — say it to a friend

We wanted the slope of the sine wave. Slope means "rise over run", so I put a second dot a little gap $h$ to the right and measured the climb, then slid it in. To find the climb I first needed one magic fact: for a tiny angle, the sine of the angle is basically the angle itself — I proved that by squeezing a pie-slice between two triangles on a circle, which forced $\frac{s i n h}{h} \to 1$ . A cousin trick (multiply by the conjugate) showed the leftover cosine piece $\frac{c o s h - 1}{h}$ just dies to $0$ . Then I broke $sin (x + h)$ into its $x$ -part and $h$ -part with the addition formula, sorted the terms into two slots, dropped in $0$ and $1$ — and out fell $cos x$ . Sanity check with the pictures: where sine crosses zero it climbs fastest (cosine $= 1$ ), at its peak it's flat (cosine $= 0$ ). The slope-wave is the cosine wave, shifted a quarter turn. Cosine works the same way but picks up a minus sign, because at $x = 0$ cosine is at its peak and about to fall.

Connections

Derivatives of sin x, cos x — Proofs from First Principles — the parent proof this page illustrates
Squeeze Theorem — Step 3's engine
Trigonometric addition formulas — Step 5's splitter
Limits — definition and properties — sum/constant rules in Step 7
Radian measure — why the sector area is $\frac{1}{2} h$
Chain Rule — extends this to $sin (g (x))$
Derivatives of tan, sec, csc, cot — built from these two
Taylor series of sin and cos — the same fact from a different angle