Derivatives of all six trig functions: Visual walkthrough

We will earn one central result from absolute zero: $\frac{d}{d x} sin x = cos x .$ Every symbol below gets a plain-word meaning and a picture before it is used.

Step 1 — What " $sin x$ " even is: a height on a spinning wheel

WHAT. Draw a circle of radius $1$ (a unit circle — "unit" just means the radius equals one). Put a point $P$ on its edge. Let $x$ be the angle, measured from the rightmost point, turning counter-clockwise. Two words:

$cos x$ = the horizontal distance of $P$ from the centre.
$sin x$ = the vertical height of $P$ above the centre.

WHY. We cannot differentiate a thing we cannot picture. Fixing $sin x$ as "the height of a turning point" turns an abstract symbol into a concrete arrow we can watch move.

WHY radians, not degrees. We measure the angle $x$ by arc length — how far the point has actually travelled along the rim. Since the radius is $1$ , a quarter turn is a quarter of the circumference $= \frac{π}{2}$ . Measuring the angle as a length is exactly what makes the pictures below line up; degrees would secretly rescale everything (see the parent's "silent killer" mistake).

PICTURE.

Figure — Derivatives of all six trig functions

Step 2 — What "the derivative of $sin$ " is asking

WHAT. The derivative is a rate: if the angle $x$ increases by a tiny amount, how fast does the height $sin x$ change? We write this question as $\frac{d}{d x} sin x = lim_{h \to 0} \frac{s i n ( x + h ) - s i n x}{h} .$

WHY this exact expression. This is the Limit definition of the derivative applied to $sin$ . Reading it piece by piece, right where each symbol sits:

$h$ — a tiny nudge we add to the angle. Think of it shrinking toward $0$ .
$sin (x + h)$ — the new height after nudging.
$sin (x + h) - sin x$ — the change in height (rise).
dividing by $h$ — change in height per unit of angle-nudge (rise over run, a slope).
$lim_{h \to 0}$ — the question "what does that slope settle to as the nudge vanishes?"

PICTURE. On the sine wave, this is the slope of the tiny chord between $x$ and $x + h$ .

Recall Why a limit and not just a fraction?

Why not just compute the rise-over-run for some small $h$ ? ::: A fixed $h$ gives an average slope over an interval; the instantaneous slope only appears in the limit as $h \to 0$ . That limit is the whole point of a derivative.

Step 3 — Zoom into the wheel: the tiny nudge is a tiny arc

WHAT. Go back to the circle. Nudging the angle from $x$ to $x + h$ moves the point $P$ a short way along the rim to a new point $Q$ . Because the radius is $1$ , that arc from $P$ to $Q$ has length exactly $h$ .

WHY. We want to understand $sin (x + h) - sin x$ — the change in height. That change is the vertical part of the little step from $P$ to $Q$ . So let us look very closely at that step.

PICTURE. Zoomed in, the arc $P Q$ is nearly a straight segment of length $h$ , and it points tangent (perpendicular) to the radius $O P$ .

Step 4 — The little right triangle: rise of the step

WHAT. The tiny step from $P$ to $Q$ (length $h$ ) is the hypotenuse of a little right triangle. Its vertical leg is the change in height, $Δ y = sin (x + h) - sin x$ . Its horizontal leg is the change in width, $Δ x = cos (x + h) - cos x$ (which is negative when we move up-left).

WHY. We need $\frac{Δ y}{h}$ . If we can read the vertical leg off the triangle, we're done. Here comes the punchline: because the step is perpendicular to the radius $O P$ , the little triangle is similar (same shape, same angles) to the big triangle formed by the radius and the axes.

The big radius makes angle $x$ with the horizontal. The step is turned $9 0^{\circ}$ from the radius, so the step's little triangle has its angle $x$ measured against the vertical. Result: $\frac{vertical leg}{hypotenuse} = cos x, so Δ y \approx h cos x .$

Reading term by term:

$Δ y$ — how much the height rose.
$h$ — length of the whole tiny step.
$cos x$ — the fraction of that step that went upward, because the step is tilted at angle $x$ from vertical.

PICTURE. The little triangle, with its vertical leg labelled $h cos x$ .

Step 5 — The same picture, done with algebra (why it's exact)

WHAT. The word " $\approx$ " should make you nervous. Let's confirm the picture using the exact Angle addition formulas: $sin (x + h) = sin x cos h + cos x sin h .$

WHY. Algebra turns the "approximately" into an exact limit, and shows precisely which pieces of the triangle become $cos x$ . Substitute and simplify:

= \sin x\cdot\underbrace{\frac{\cos h-1}{h}}_{\to\, 0} \;+\; \cos x\cdot\underbrace{\frac{\sin h}{h}}_{\to\, 1}.$$ Reading term by term: - $\dfrac{\cos h - 1}{h}$ — how the **horizontal** part of the step scales; it vanishes because sideways motion is second-order tiny. - $\dfrac{\sin h}{h}$ — how the **vertical** part scales; it heads to $1$, so $\cos x$ survives untouched. **PICTURE.** These are exactly the two foundation limits — the geometric squeeze that gives $\frac{\sin h}{h}\to 1$, and the conjugate trick that gives $\frac{\cos h-1}{h}\to 0$. ![[deepdives/dd-maths-4.1.18-d2-s05.png]] Taking the limit: $$\frac{d}{dx}\sin x = \sin x\cdot 0 + \cos x\cdot 1 = \boxed{\cos x}.$$ > [!intuition] Why one limit is $0$ and the other is $1$ > Near the bottom of a shrinking arc, you move almost **straight up** — full vertical progress ($\frac{\sin h}{h}\to 1$) and essentially **zero sideways** progress ($\frac{\cos h-1}{h}\to 0$). The triangle picture and the algebra are the *same story* told twice. --- ## Step 6 — Every quadrant: does $\cos x$ really predict the slope? **WHAT.** We claimed the slope of the sine wave *is the value of $\cos$*. Let's check this in all four quadrants — no scenario left unshown. **WHY.** A formula you only tested at one point is a guess. The sign of $\cos x$ must match whether the wave is rising or falling everywhere. | angle range | $\cos x$ sign | sine wave is... | matches? | |---|---|---|---| | $0$ to $\tfrac{\pi}{2}$ | $+$ | rising | ✅ | | $\tfrac{\pi}{2}$ to $\pi$ | $-$ | falling | ✅ | | $\pi$ to $\tfrac{3\pi}{2}$ | $-$ | falling | ✅ | | $\tfrac{3\pi}{2}$ to $2\pi$ | $+$ | rising | ✅ | **Degenerate points.** At the very top of the wave, $x=\tfrac{\pi}{2}$: the point $P$ is moving purely **sideways**, so the height isn't changing — slope $=0$, and indeed $\cos\tfrac{\pi}{2}=0$. At $x=0$ the point moves purely **upward** — steepest rise, slope $=1=\cos 0$. Both extreme cases fit. **PICTURE.** ![[deepdives/dd-maths-4.1.18-d2-s06.png]] > [!example] Numeric spot-check at $x=\tfrac{\pi}{3}$ > Slope of $\sin$ at $x=\tfrac{\pi}{3}$ should be $\cos\tfrac{\pi}{3}=\tfrac12$. Nudge with $h=0.001$: > $$\frac{\sin(\tfrac{\pi}{3}+0.001)-\sin\tfrac{\pi}{3}}{0.001}\approx 0.49957\ldots \approx \tfrac12.\ \checkmark$$ --- ## Step 7 — The twin: $\cos' = -\sin$, from the same wheel **WHAT.** By the identical triangle argument, the **horizontal** motion of $P$ is $-\,h\sin x$ (the minus because moving up means moving *left*). So $$\frac{d}{dx}\cos x = -\sin x.$$ **WHY.** The width $\cos x$ *decreases* as we climb the first quadrant, so its rate is negative. This is the origin of the whole "co's are negative" rule the parent note leans on — it is literally the leftward tilt of the step. **PICTURE.** The same little triangle, now reading the horizontal leg. ![[deepdives/dd-maths-4.1.18-d2-s07.png]] Once you own $\sin'=\cos$ and $\cos'=-\sin$, the [[Quotient Rule]] on $\tan=\sin/\cos$, $\sec=1/\cos$, etc. hands you the remaining four (see the parent note). And these two facts are the engine behind [[Simple Harmonic Motion]] and [[Derivatives of inverse trig functions]]. --- ## The one-picture summary ![[deepdives/dd-maths-4.1.18-d2-s08.png]] The single figure stacks it all: the spinning point traces $\sin x$ (blue height wave), and the **slope of that wave at each moment** is the pink $\cos x$ wave beneath it. Where blue is steepest-up, pink is at its peak; where blue crests flat, pink crosses zero. > [!recall]- Feynman retelling of the whole walkthrough > Explain the derivation with no symbols. ::: Spin a dot around a circle of radius one. Its height above the middle is what we call $\sin$. Give the dot a tiny push further round. It moves a short distance, and that short distance is tilted — partly up, partly sideways. The *up* fraction of that push, at any moment, is exactly how wide the dot currently is from the middle, which is $\cos$. So "how fast the height grows" equals "how wide the dot is" — that's why the slope of $\sin$ is $\cos$. The sideways fraction shrinks the width, and that fraction is the height itself — so the slope of $\cos$ is *minus* $\sin$. Two facts, one picture of a tilted little step. > [!mnemonic] The picture in four words > **Up-fraction is cosine.** The rise of the tiny circular step is $h\cos x$; divide by $h$ and the derivative $\cos x$ falls out. --- ## Connections - [[Limit definition of the derivative]] — the rise/run limit of Step 2 - [[The Squeeze Theorem]] — proves $\frac{\sin h}{h}\to 1$ in Step 5 - [[Angle addition formulas]] — the exact expansion in Step 5 - [[Pythagorean identity sin^2 + cos^2 = 1]] — powers the quotient-rule sequels - [[Quotient Rule]] · [[Product Rule]] · [[Chain Rule]] — build the other four functions - [[Simple Harmonic Motion]] — physics where this slope is a velocity - [[Derivatives of inverse trig functions]] — the reverse direction - [[Derivatives of all six trig functions|Back to parent topic]]