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

4.1.17 · D3 · Maths › Calculus I — Limits & Derivatives › Derivatives of sin x, cos x — proofs from first principles

Yeh first-principles proofs ka hands-on companion hai. Wahan humne prove kiya tha ki $\frac{d}{d x} sin x = cos x$ aur $\frac{d}{d x} cos x = - sin x$ . Yahan hum unhe use karte hain har us situation mein jo tumhe mil sakti hai — har sign, har quadrant, do building-block limits akele, ek swing (word problem), aur ek exam twist.

Scenario matrix

Neeche, Quadrant I mein angle ka matlab hai $0 < x < \frac{π}{2}$ , QII ka matlab $\frac{π}{2} < x < π$ , QIII ka matlab $π < x < \frac{3 π}{2}$ , QIV ka matlab $\frac{3 π}{2} < x < 2 π$ . (Ek "quadrant" ek full turn ke chaar hisson mein se ek hota hai — neeche wali figure dekho.)

#	Cell class	Tricky kyon hai	Example
A	Core limit $\frac{s i n h}{h}$ directly use karna	numerator/denominator dono $\to 0$	Ex 1
B	Companion limit $\frac{c o s h - 1}{h}$	galat " $\frac{1}{2}$ " ka trap	Ex 2
C	$sin x$ ka slope, QI vs QII ( $cos$ ka sign flip)	slope pehle positive phir negative	Ex 3
D	$cos x$ ka slope, QIII vs QIV ( $- sin$ ka sign flip)	double sign bookkeeping	Ex 4
E	Degenerate / limiting points: $x = 0, \frac{π}{2}, π, \frac{3 π}{2}$	peaks aur zeros, slope $= 0$ ya $\pm 1$	Ex 5
F	Real-world word problem (ek swing)	units lagao, sirf radians	Ex 6
G	Exam twist: first principles se inner rate $sin (k x)$	chain rule ka preview, factor $k$	Ex 7
H	Degrees-vs-radians degenerate case	$\frac{π}{180}$ ka pitfall	Ex 8

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

Upar ki picture charon quadrants dikhati hai, aur chaar sample angles par sine wave ka slope arrow bhi. Arrow ka jhukav dekho: QI/QIV mein upar, QII/QIII mein neeche — yahi jhukav $cos x$ ka sign hai.

Cell A — $\frac{sin h}{h} \to 1$ ko akele use karna

Worked example Ex 1 ·

$h \to 0 lim \frac{sin 5 h}{3 h}$ evaluate karo Forecast: Aage padhne se pehle ek number guess karo. Kya yeh $1$ hai? $\frac{5}{3}$ ? $\frac{3}{5}$ ? Apna guess likh lo.

Step 1 — mismatch pakdo. Yeh step kyon? Humara jaana-pehchana fact hai $\frac{s i n ( same thing )}{same thing} \to 1$ . Yahan upar $5 h$ hai lekin neeche $3 h$ — yeh same thing nahi hai, toh abhi $1$ nahi padh sakte.

Step 2 — angle ko match karo. Kyon? Multiply aur divide karo taaki denominator $5 h$ ban jaye, jo sine ke argument se match kare: $\frac{s i n 5 h}{3 h} = \frac{s i n 5 h}{5 h} \cdot \frac{5 h}{3 h} = \frac{s i n 5 h}{5 h} \cdot \frac{5}{3} .$ $\frac{5 h}{3 h}$ sirf algebra hai — $h$ cancel hokar constant $\frac{5}{3}$ bach jaata hai.

Step 3 — $h \to 0$ lene do. Kyon? Jab $h \to 0$ , toh $5 h \to 0$ bhi, isliye $\frac{s i n 5 h}{5 h}$ wahi fundamental limit hai aur $\to 1$ . $lim_{h \to 0} \frac{s i n 5 h}{3 h} = 1 \cdot \frac{5}{3} = \frac{5}{3} .$

Verify: $h = 0.001$ try karo: $sin (0.005) = 0.00499998$ , $0.003$ se divide karo toh $1.66666$ milta hai, yaani $\approx \frac{5}{3}$ . ✔ Match karta hai.

Cell B — $\frac{cos h - 1}{h} \to 0$ use karna (aur trap se bachna)

Worked example Ex 2 ·

$h \to 0 lim \frac{1 - cos 4 h}{h}$ evaluate karo Forecast: Parent note ne " $\frac{1}{2}$ " wali galti ki warning di hai. Kya tumhare hisaab se answer $0$ hai, $\frac{1}{2}$ hai, ya $2$ ? Pehle commit karo.

Step 1 — shape pehchano. Kyon? Hum jaante hain $\frac{c o s h - 1}{h} \to 0$ . Note karo ki $1 - cos 4 h = - (cos 4 h - 1)$ , toh sign palat jaata hai lekin limit value phir bhi " $cos - 1$ over first power of $h$ " ki form se bani hai → yeh phir bhi $0$ ki taraf jayegi.

Step 2 — angle ko $4 h$ se match karo. Kyon? Ex 1 wali trick: denominator mein $4 h$ lao. $\frac{1 - c o s 4 h}{h} = - \frac{c o s 4 h - 1}{4 h} \cdot 4.$

Step 3 — limit lo. Kyon? $\frac{c o s 4 h - 1}{4 h} \to 0$ (companion limit jab $4 h \to 0$ ), toh: $lim_{h \to 0} \frac{1 - c o s 4 h}{h} = - 0 \cdot 4 = 0 .$

$\frac{1}{2}$ kyon NAHI? $\frac{1}{2}$ wala result $\frac{1 - c o s h}{h ^{2}} \to \frac{1}{2}$ mein hai — neeche $h$ ki do powers hain. Yahan sirf ek power of $h$ hai, aur ek power itna chhota $cos - 1$ "pakadne" ke liye kaafi nahi, isliye yeh $0$ ho jaata hai.

Verify: $h = 0.001$ : $1 - cos (0.004) = 0.000008$ , $0.001$ se divide karo toh $0.008 \to 0$ milta hai. ✔

Cell C — $sin x$ ka slope ek quadrant boundary cross karte hue ( $cos$ ka sign)

Worked example Ex 3 ·

$y = sin x$ ka slope $x = \frac{π}{3}$ (QI) aur $x = \frac{2 π}{3}$ (QII) par nikalo. Compare karo. Forecast: In mein se ek slope positive hai, ek negative. Kaun sa kaun sa hai — aur kya dono size mein equal hain?

Step 1 — derivative lagao. Kyon? Humne prove kiya hai ki $\frac{d}{d x} sin x = cos x$ ; kisi point par slope bas us point ka $cos$ value hai. $y^{'}_{x = π /3} = cos \frac{π}{3} = \frac{1}{2}, y^{'}_{x = 2 π /3} = cos \frac{2 π}{3} = - \frac{1}{2} .$

Step 2 — geometry padho. Kyon? $\frac{π}{3} = 6 0^{\circ}$ sine wave ke upar jaate hisse par hai (peak $\frac{π}{2}$ se pehle), toh slope $> 0$ . $\frac{2 π}{3} = 12 0^{\circ}$ peak ke baad girte hisse par hai, toh slope $< 0$ .

Step 3 — same magnitude, ulta sign. Kyon? $\frac{π}{3}$ aur $\frac{2 π}{3}$ peak $\frac{π}{2}$ ke baare mein mirror images hain, isliye wave equally steep hai, bas ek upar aur ek neeche. $+ \frac{1}{2} at \frac{π}{3}, - \frac{1}{2} at \frac{2 π}{3} .$

Verify: $cos 6 0^{\circ} = 0.5$ , $cos 12 0^{\circ} = - 0.5$ . Size mein equal, sign opposite. ✔

Cell D — QIII aur QIV mein $cos x$ ka slope (double-sign case)

Worked example Ex 4 ·

$y = cos x$ ka slope $x = \frac{5 π}{4}$ (QIII) aur $x = \frac{7 π}{4}$ (QIV) par nikalo. Forecast: Cosine ke derivative mein already ek minus hai. Ab $sin$ khud QIII/QIV mein negative hai. Do minus... toh kya slopes positive nikalenge?

Step 1 — derivative lagao. Kyon? Humne prove kiya hai ki $\frac{d}{d x} cos x = - sin x$ . $y^{'} = - sin x .$

Step 2 — har angle plug karo, DONO signs track karo. Kyon? Careful bookkeeping is cell ka pura point hai.

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

Step 3 — wave se sanity check. Kyon? Cosine ka trough (minimum) $x = π$ par hai; trough ke thoda baad yeh wapas upar chadhta hai, toh QIII aur QIV mein slope positive hona chahiye — dono $+ \frac{2}{2}$ se match karta hai. $y^{'} = + \frac{2}{2} \approx 0.707 dono points par.$

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

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

Worked example Ex 5 · Chaar special points

$x = 0, \frac{π}{2}, π, \frac{3 π}{2}$ par $sin x$ AND $cos x$ dono ka slope do. Forecast: Kisi peak ya trough par tangent line flat hoti hai. Toh in aath numbers mein se kuch exactly $0$ honge. Guess karo kitne.

Step 1 — do derivatives se table banao. Kyon? $\frac{d}{d x} sin x = cos x$ aur $\frac{d}{d x} cos x = - sin x$ ; bas evaluate karo.

$x$	$sin x$ (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 — pattern padho. Kyon? Jahan height peak/trough ( $\pm 1$ ) ho, wahan slope $0$ hai (flat tangent). Jahan height $0$ ho (ek crossing), wahan slope $\pm 1$ hai (steepest). Aath mein se chaar slopes zero hain — forecast se match karta hai.

Step 3 — $9 0^{\circ}$ shift, visible ho gaya. Kyon? "slope of $sin$ " column ko " $cos x$ height" column se compare karo — yeh identical hain. Yahi wave-shift claim hai, ab ek concrete table ki shakal mein.

Verify: Jaise, $x = \frac{π}{2}$ par $sin$ ka slope $cos \frac{π}{2} = 0$ hai; $x = \frac{π}{2}$ par $cos$ ka slope $- sin \frac{π}{2} = - 1$ hai. ✔

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

Worked example Ex 6 · Ek bacha swing par hai jiska lowest point se upar height

$y (t) = 0.8 sin (1.5 t)$ metres hai, jahan $t$ seconds mein hai aur argument $1.5 t$ radians mein hai. $t = 0$ par woh kitni tezi se upar ja rahi hai? Forecast: Swing ke sabse neeche tum sabse tezi se move karte ho. Toh $t = 0$ par speed sabse zyada honi chahiye, zero nahi. m/s mein koi number guess karo.

Step 1 — time ke saath differentiate karo. Kyon? "Height kitni tezi se badal rahi hai" = $y$ ka $t$ ke saath derivative. Outer function $sin$ hai jiska derivative $cos$ hai; inner angle $1.5 t$ har second $1.5$ ki rate se badalta hai, jo aage multiply ho jaata hai (yeh factor bilkul wahi hai jo parent note ka Chain Rule preview hai). $y^{'} (t) = 0.8 \cdot cos (1.5 t) \cdot 1.5 = 1.2 cos (1.5 t) m/s .$

Step 2 — $t = 0$ par evaluate karo. Kyon? $t = 0$ tab hai jab woh sabse neeche se guzarti hai. $y^{'} (0) = 1.2 cos (0) = 1.2 \cdot 1 = 1.2 m/s .$

Step 3 — units aur sanity. Kyon? Metres divided by seconds → m/s, speed ke liye sahi. Aur $1.2$ , $1.2 cos (1.5 t)$ ka maximum hai, toh yeh sach mein sabse tez wala moment hai — forecast se match karta hai. $y^{'} (0) = 1.2 m/s (upar ki taraf, aur yeh maximum speed hai).$

Verify: derivative $1.2 cos (1.5 \cdot 0) = 1.2$ . Aur $1.2 cos (\cdot)$ ka max $1.2$ hai. ✔

Cell G — exam twist: first principles se inner rate

Worked example Ex 7 · Limit definition se seedha dikhao ki

$\frac{d}{d x} sin (k x) = k cos (k x)$ ek constant $k$ ke liye. Forecast: Parent ne hint diya tha ki factor $k$ , $\frac{s i n k h}{h} \to k$ ke through aata hai. Predict karo ki $k$ kahan se nikalta hai.

Step 1 — definition likho. Kyon? Humse derive karne ko kaha gaya hai, koi rule quote nahi karna. $\frac{d}{d x} sin (k x) = lim_{h \to 0} \frac{s i n ( k ( x + h )) - s i n ( k x )}{h} = lim_{h \to 0} \frac{s i n ( k x + k h ) - s i n ( k x )}{h} .$

Step 2 — $sin (k x + k h)$ par addition formula lagao. Kyon? Trigonometric addition formulas fixed part $k x$ ko shrinking part $k h$ se alag kar deta hai: $sin (k x + k h) = sin (k x) cos (k h) + cos (k x) sin (k h) .$ Substituting karke aur $sin (k x)$ terms group karke: $= lim_{h \to 0} [sin (k x) \cdot \frac{c o s ( k h ) - 1}{h} + cos (k x) \cdot \frac{s i n ( k h )}{h}] .$

Step 3 — matched angle ke saath do pieces evaluate karo. Kyon? Angle $k h$ ko har denominator mein force karo: $\frac{c o s ( k h ) - 1}{h} = k \cdot \frac{c o s ( k h ) - 1}{k h} \to k \cdot 0 = 0, \frac{s i n ( k h )}{h} = k \cdot \frac{s i n ( k h )}{k h} \to k \cdot 1 = k .$ Yahan se $k$ aata hai: $\frac{( \cdot )}{h}$ ko $\frac{( \cdot )}{k h}$ mein convert karne se, jisme $k$ ka ek factor lagta hai.

Step 4 — assemble karo. Kyon? Limit values wapas daalo: $= sin (k x) \cdot 0 + cos (k x) \cdot k = k cos (k x) .$

Verify: $k = 2$ , $x = \frac{π}{6}$ ke liye: predicted $2 cos \frac{π}{3} = 2 \cdot \frac{1}{2} = 1$ . Wahan $sin (2 x)$ ka numerical slope $\approx 1$ . ✔

Cell H — degrees wala degenerate case (radians kyon zaroori hain)

Worked example Ex 8 · Ek student

$\frac{d}{d x} sin x$ compute karta hai lekin $x$ degrees mein rakhta hai. Kaun sa extra constant aata hai, aur $x = 0^{\circ}$ par unhe $sin$ ka "slope" kya milega? Forecast: Parent kehta hai degrees $\frac{s i n h}{h} \to 1$ ko tod dete hain. Constant guess karo.

Step 1 — degrees ko radians mein convert karo sine ke andar. Kyon? Hamara proof tabhi kaam karta hai jab sine ka input radians mein ho. Agar $x$ degrees mein measure kiya gaya hai, toh actual angle radians mein $\frac{π}{180} x$ hai, toh function aslaan $sin (\frac{π}{180} x)$ hai.

Step 2 — use differentiate karo. Kyon? Yeh exactly Ex 7 hai jahan $k = \frac{π}{180}$ : $\frac{d}{d x} sin (\frac{π}{180} x) = \frac{π}{180} cos (\frac{π}{180} x) .$ Toh "extra constant" $\frac{π}{180} \approx 0.01745$ hai.

Step 3 — $x = 0^{\circ}$ par slope evaluate karo. Kyon? $0$ par, $cos 0 = 1$ : $slope = \frac{π}{180} \cdot 1 = \frac{π}{180} \approx 0.01745.$ Radians mein same slope $cos 0 = 1$ hota hai. Yeh mismatch ( $57.3$ ka factor!) exactly wahi reason hai kyon calculus radians par insist karta hai. $Degrees se \frac{π}{180} cos x milta hai, na ki cos x .$

Verify: $\frac{π}{180} = 0.0174533$ . ✔ (Honest radian answer $cos 0 = 1$ se compare karo.)

Recall Checkpoint — answers cover karo

Kaun sa cell tumhe differentiate karne se pehle sine ka argument convert karne par majboor karta hai? ::: Cell H (degrees) — pehle $x^{\circ}$ ko $\frac{π}{180} x$ radians mein badlo $lim \frac{1 - c o s 4 h}{h} = 0$ kyon hai aur $\frac{1}{2}$ kyon nahi? ::: Neeche sirf EK power of $h$ hai; $\frac{1}{2}$ ke liye $h^{2}$ chahiye $\frac{d}{d x} sin (k x)$ mein factor $k$ kahan se aata hai? ::: $\frac{s i n k h}{h} = k \cdot \frac{s i n k h}{k h} \to k$ likhne se Kisi wave ke peak par, slope hamesha kya hota hai? ::: $0$ (flat tangent)

Connections

Derivatives of sin x, cos x — proofs from first principles — woh parent jise yeh page drill karta hai
Squeeze Theorem — har $\frac{s i n h}{h}$ step ke peeche (Ex 1, 7)
Trigonometric addition formulas — Ex 7 mein splitter
Limits — definition and properties — Ex 1, 2 mein manipulation ka basis
Chain Rule — inner-rate factor $k$ (Ex 6, 7, 8)
Radian measure — woh poora reason jiske liye Ex 8 exist karta hai
Derivatives of tan, sec, csc, cot — agla padaav jab yeh automatic ho jayein