Derivatives of sin x, cos x — proofs from first principles: Worked examples

This is the hands-on companion to the first-principles proofs. There we proved $\frac{d}{d x} sin x = cos x$ and $\frac{d}{d x} cos x = - sin x$ . Here we use them in every situation you might meet — every sign, every quadrant, the two building-block limits on their own, a swing (word problem), and an exam twist.

The scenario matrix

Below, an angle in Quadrant I means $0 < x < \frac{π}{2}$ , QII means $\frac{π}{2} < x < π$ , QIII means $π < x < \frac{3 π}{2}$ , QIV means $\frac{3 π}{2} < x < 2 π$ . (A "quadrant" is just one of the four quarters of a full turn — see the figure just below.)

#	Cell class	What makes it tricky	Example that hits it
A	The core limit $\frac{s i n h}{h}$ used directly	numerator/denominator both $\to 0$	Ex 1
B	The companion limit $\frac{c o s h - 1}{h}$	wrong " $\frac{1}{2}$ " trap	Ex 2
C	Slope of $sin x$ , QI vs QII (sign flip of $cos$ )	slope positive then negative	Ex 3
D	Slope of $cos x$ , QIII vs QIV (sign flip of $- sin$ )	double sign bookkeeping	Ex 4
E	Degenerate / limiting points: $x = 0, \frac{π}{2}, π, \frac{3 π}{2}$	peaks & zeros, slope $= 0$ or $\pm 1$	Ex 5
F	Real-world word problem (a swing)	attach units, radians only	Ex 6
G	Exam twist: inner rate $sin (k x)$ from first principles	previews chain rule, factor $k$	Ex 7
H	Degrees-vs-radians degenerate case	the $\frac{π}{180}$ pitfall	Ex 8

Figure — Derivatives of sin x, cos x — proofs from first principles

The picture above shows the four quadrants and, at four sample angles, the slope arrow of the sine wave. Watch the arrow tilt: uphill in QI/QIV, downhill in QII/QIII — that tilt is the sign of $cos x$ .

Cell A — using $\frac{sin h}{h} \to 1$ by itself

Worked example Ex 1 · Evaluate

$h \to 0 lim \frac{sin 5 h}{3 h}$ Forecast: Guess a number before reading on. Is it $1$ ? $\frac{5}{3}$ ? $\frac{3}{5}$ ? Write your guess down.

Step 1 — spot the mismatch. Why this step? Our known fact is $\frac{s i n ( same thing )}{same thing} \to 1$ . Here the top has $5 h$ but the bottom has $3 h$ — they are not the same thing, so we cannot read off $1$ yet.

Step 2 — force the angle to match. Why? Multiply and divide so that the denominator becomes $5 h$ , matching the sine's argument: $\frac{s i n 5 h}{3 h} = \frac{s i n 5 h}{5 h} \cdot \frac{5 h}{3 h} = \frac{s i n 5 h}{5 h} \cdot \frac{5}{3} .$ The $\frac{5 h}{3 h}$ is just algebra — the $h$ 's cancel to leave the constant $\frac{5}{3}$ .

Step 3 — let $h \to 0$ . Why? As $h \to 0$ the piece $5 h \to 0$ too, so $\frac{s i n 5 h}{5 h}$ is the fundamental limit and $\to 1$ . $lim_{h \to 0} \frac{s i n 5 h}{3 h} = 1 \cdot \frac{5}{3} = \frac{5}{3} .$

Verify: Try $h = 0.001$ : $sin (0.005) = 0.00499998$ , divide by $0.003$ gives $1.66666$ , i.e. $\approx \frac{5}{3}$ . ✔ Matches.

Cell B — using $\frac{cos h - 1}{h} \to 0$ (and dodging the trap)

Worked example Ex 2 · Evaluate

$h \to 0 lim \frac{1 - cos 4 h}{h}$ Forecast: The parent warns of a " $\frac{1}{2}$ " mistake here. Do you think the answer is $0$ , $\frac{1}{2}$ , or $2$ ? Commit first.

Step 1 — recognise the shape. Why? We know $\frac{c o s h - 1}{h} \to 0$ . Note $1 - cos 4 h = - (cos 4 h - 1)$ , so the sign flips but the limit value is still built from a " $cos - 1$ over first power of $h$ " form → it will still head to $0$ .

Step 2 — match the angle to $4 h$ . Why? Same trick as Ex 1: make the denominator carry $4 h$ . $\frac{1 - c o s 4 h}{h} = - \frac{c o s 4 h - 1}{4 h} \cdot 4.$

Step 3 — take the limit. Why? $\frac{c o s 4 h - 1}{4 h} \to 0$ (companion limit with $4 h \to 0$ ), so: $lim_{h \to 0} \frac{1 - c o s 4 h}{h} = - 0 \cdot 4 = 0 .$

Why NOT $\frac{1}{2}$ ? The $\frac{1}{2}$ lives in $\frac{1 - c o s h}{h ^{2}} \to \frac{1}{2}$ — two powers of $h$ downstairs. Here we have only one power of $h$ , and one power is not enough to "catch" the tiny $cos - 1$ , so it collapses to $0$ .

Verify: $h = 0.001$ : $1 - cos (0.004) = 0.000008$ , over $0.001$ gives $0.008 \to 0$ . ✔

Cell C — slope of $sin x$ crossing a quadrant boundary (sign of $cos$ )

Worked example Ex 3 · Find the slope of

$y = sin x$ at $x = \frac{π}{3}$ (QI) and at $x = \frac{2 π}{3}$ (QII). Compare. Forecast: One of these slopes is positive, one is negative. Which is which — and are they equal in size?

Step 1 — apply the derivative. Why? We proved $\frac{d}{d x} sin x = cos x$ ; the slope at a point is just $cos$ of that point. $y^{'}_{x = π /3} = cos \frac{π}{3} = \frac{1}{2}, y^{'}_{x = 2 π /3} = cos \frac{2 π}{3} = - \frac{1}{2} .$

Step 2 — read the geometry. Why? $\frac{π}{3} = 6 0^{\circ}$ is on the rising part of the sine wave (before its peak at $\frac{π}{2}$ ), so slope $> 0$ . $\frac{2 π}{3} = 12 0^{\circ}$ is past the peak on the falling side, so slope $< 0$ .

Step 3 — same magnitude, opposite sign. Why? $\frac{π}{3}$ and $\frac{2 π}{3}$ are mirror images about the peak $\frac{π}{2}$ , so the wave is equally steep, just one up and one down. $+ \frac{1}{2} at \frac{π}{3}, - \frac{1}{2} at \frac{2 π}{3} .$

Verify: $cos 6 0^{\circ} = 0.5$ , $cos 12 0^{\circ} = - 0.5$ . Equal in size, opposite sign. ✔

Cell D — slope of $cos x$ in QIII and QIV (the double-sign case)

Worked example Ex 4 · Find the slope of

$y = cos x$ at $x = \frac{5 π}{4}$ (QIII) and $x = \frac{7 π}{4}$ (QIV). Forecast: The derivative of cosine already carries a minus. Now $sin$ itself is negative in QIII/QIV. Two minuses... do the slopes come out positive?

Step 1 — apply the derivative. Why? We proved $\frac{d}{d x} cos x = - sin x$ . $y^{'} = - sin x .$

Step 2 — plug each angle, track BOTH signs. Why? Careful bookkeeping is the whole point of this cell.

$x = \frac{5 π}{4}$ (QIII): $sin \frac{5 π}{4} = - \frac{2}{2}$ , so $y^{'} = - (- \frac{2}{2}) = + \frac{2}{2}$ .
$x = \frac{7 π}{4}$ (QIV): $sin \frac{7 π}{4} = - \frac{2}{2}$ , so $y^{'} = - (- \frac{2}{2}) = + \frac{2}{2}$ .

Step 3 — sanity via the wave. Why? Cosine has its trough (minimum) at $x = π$ ; just after the trough it climbs back up, so slope should be positive through QIII and QIV — matching both $+ \frac{2}{2}$ . $y^{'} = + \frac{2}{2} \approx 0.707 at both points.$

Verify: $- sin (5 π /4) = 0.7071$ , $- sin (7 π /4) = 0.7071$ . Both positive. ✔

Cell E — degenerate / limiting points (peaks and zeros)

Worked example Ex 5 · Give the slope of

$sin x$ AND of $cos x$ at the four special points $x = 0, \frac{π}{2}, π, \frac{3 π}{2}$ . Forecast: At a peak or trough the tangent line is flat. So some of these eight numbers must be exactly $0$ . Guess how many.

Step 1 — build a table from the two derivatives. Why? $\frac{d}{d x} sin x = cos x$ and $\frac{d}{d x} cos x = - sin x$ ; just evaluate.

$x$	$sin x$ (height)	slope of $sin$ $= cos x$	$cos x$ (height)	slope of $cos$ $= - sin x$
$0$	$0$	$1$	$1$ (peak)	$0$
$\frac{π}{2}$	$1$ (peak)	$0$	$0$	$- 1$
$π$	$0$	$- 1$	$- 1$ (trough)	$0$
$\frac{3 π}{2}$	$- 1$ (trough)	$0$	$0$	$1$

Step 2 — read the pattern. Why? Wherever the height is a peak/trough ( $\pm 1$ ), the slope is $0$ (flat tangent). Wherever the height is $0$ (a crossing), the slope is $\pm 1$ (steepest). Four of the eight slopes are zero — matching the forecast.

Step 3 — the $9 0^{\circ}$ shift, made visible. Why? Compare the "slope of $sin$ " column to the " $cos x$ height" column — they are identical. That is the wave-shift claim, now a concrete table.

Verify: e.g. slope of $sin$ at $x = \frac{π}{2}$ is $cos \frac{π}{2} = 0$ ; slope of $cos$ at $x = \frac{π}{2}$ is $- sin \frac{π}{2} = - 1$ . ✔

Cell F — real-world word problem (units matter)

Worked example Ex 6 · A child on a swing has height above the lowest point

$y (t) = 0.8 sin (1.5 t)$ metres, where $t$ is in seconds and the argument $1.5 t$ is in radians. How fast is she rising at $t = 0$ ? Forecast: At the bottom of a swing you move fastest. So the speed at $t = 0$ should be the biggest speed, not zero. Guess a number in m/s.

Step 1 — differentiate w.r.t. time. Why? "How fast is height changing" = derivative of $y$ with respect to $t$ . The outer function is $sin$ , whose derivative is $cos$ ; the inner angle $1.5 t$ changes at rate $1.5$ per second, which multiplies out front (this factor is exactly the Chain Rule preview from the parent note). $y^{'} (t) = 0.8 \cdot cos (1.5 t) \cdot 1.5 = 1.2 cos (1.5 t) m/s .$

Step 2 — evaluate at $t = 0$ . Why? $t = 0$ is when she passes the bottom. $y^{'} (0) = 1.2 cos (0) = 1.2 \cdot 1 = 1.2 m/s .$

Step 3 — units and sanity. Why? Metres divided by seconds → m/s, correct for a speed. And $1.2$ is the maximum of $1.2 cos (1.5 t)$ , so this is indeed the fastest moment — matching the forecast. $y^{'} (0) = 1.2 m/s (upward, and it’s the maximum speed).$

Verify: derivative $1.2 cos (1.5 \cdot 0) = 1.2$ . And the max of $1.2 cos (\cdot)$ is $1.2$ . ✔

Cell G — exam twist: inner rate from first principles

Worked example Ex 7 · Straight from the limit definition, show

$\frac{d}{d x} sin (k x) = k cos (k x)$ for a constant $k$ . Forecast: The parent hinted the factor $k$ appears through $\frac{s i n k h}{h} \to k$ . Predict where the $k$ pops out.

Step 1 — write the definition. Why? We are asked to derive, not quote a rule. $\frac{d}{d x} sin (k x) = lim_{h \to 0} \frac{s i n ( k ( x + h )) - s i n ( k x )}{h} = lim_{h \to 0} \frac{s i n ( k x + k h ) - s i n ( k x )}{h} .$

Step 2 — addition formula on $sin (k x + k h)$ . Why? Trigonometric addition formulas separate the fixed part $k x$ from the shrinking part $k h$ : $sin (k x + k h) = sin (k x) cos (k h) + cos (k x) sin (k h) .$ Substituting and grouping the $sin (k x)$ terms: $= lim_{h \to 0} [sin (k x) \cdot \frac{c o s ( k h ) - 1}{h} + cos (k x) \cdot \frac{s i n ( k h )}{h}] .$

Step 3 — evaluate the two pieces with a matched angle. Why? Force the angle $k h$ into each denominator: $\frac{c o s ( k h ) - 1}{h} = k \cdot \frac{c o s ( k h ) - 1}{k h} \to k \cdot 0 = 0, \frac{s i n ( k h )}{h} = k \cdot \frac{s i n ( k h )}{k h} \to k \cdot 1 = k .$ There's the $k$ : it comes from converting $\frac{( \cdot )}{h}$ into $\frac{( \cdot )}{k h}$ , which costs a factor $k$ .

Step 4 — assemble. Why? Put the limit values back: $= sin (k x) \cdot 0 + cos (k x) \cdot k = k cos (k x) .$

Verify: For $k = 2$ , $x = \frac{π}{6}$ : predicted $2 cos \frac{π}{3} = 2 \cdot \frac{1}{2} = 1$ . Numerical slope of $sin (2 x)$ there $\approx 1$ . ✔

Cell H — the degrees degenerate case (why radians are non-negotiable)

Worked example Ex 8 · A student computes

$\frac{d}{d x} sin x$ but keeps $x$ in degrees. What extra constant appears, and what is the "slope" of $sin$ at $x = 0^{\circ}$ they'd get? Forecast: The parent says degrees break $\frac{s i n h}{h} \to 1$ . Guess the constant.

Step 1 — convert degrees to radians inside the sine. Why? Our proof only works when the sine's input is in radians. If $x$ is measured in degrees, the actual angle in radians is $\frac{π}{180} x$ , so the function is really $sin (\frac{π}{180} x)$ .

Step 2 — differentiate that. Why? This is exactly Ex 7 with $k = \frac{π}{180}$ : $\frac{d}{d x} sin (\frac{π}{180} x) = \frac{π}{180} cos (\frac{π}{180} x) .$ So the "extra constant" is $\frac{π}{180} \approx 0.01745$ .

Step 3 — evaluate the slope at $x = 0^{\circ}$ . Why? At $0$ , $cos 0 = 1$ : $slope = \frac{π}{180} \cdot 1 = \frac{π}{180} \approx 0.01745.$ In radians the same slope is $cos 0 = 1$ . The mismatch (a factor $57.3$ !) is exactly why calculus insists on radians. $Degrees give \frac{π}{180} cos x, not cos x .$

Verify: $\frac{π}{180} = 0.0174533$ . ✔ (Compare the honest radian answer $cos 0 = 1$ .)

Recall Checkpoint — cover the answers

Which cell forces you to convert the sine's argument before differentiating? ::: Cell H (degrees) — turn $x^{\circ}$ into $\frac{π}{180} x$ radians first Why does $lim \frac{1 - c o s 4 h}{h} = 0$ and not $\frac{1}{2}$ ? ::: Only ONE power of $h$ downstairs; the $\frac{1}{2}$ needs $h^{2}$ Where does the factor $k$ in $\frac{d}{d x} sin (k x)$ come from? ::: From rewriting $\frac{s i n k h}{h} = k \cdot \frac{s i n k h}{k h} \to k$ At a peak of a wave, the slope is always what? ::: $0$ (flat tangent)

Connections

Derivatives of sin x, cos x — proofs from first principles — the parent this page drills
Squeeze Theorem — behind every $\frac{s i n h}{h}$ step (Ex 1, 7)
Trigonometric addition formulas — the splitter in Ex 7
Limits — definition and properties — how we manipulate Ex 1, 2
Chain Rule — the inner-rate factor $k$ (Ex 6, 7, 8)
Radian measure — the whole reason Ex 8 exists
Derivatives of tan, sec, csc, cot — next stop once these are automatic