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

What we are proving

The two ingredients we MUST establish first

Everything hinges on two limits. Without these the proof collapses.

Formula Limit 1 — the fundamental small-angle limit

$\lim_{h\to 0}\frac{\sin h}{h} = 1$

WHY it's true (geometric squeeze): For small $h>0$ draw the unit circle. Compare three areas: $\underbrace{\tfrac12\sin h}_{\text{triangle}} \le \underbrace{\tfrac12 h}_{\text{sector}} \le \underbrace{\tfrac12\tan h}_{\text{big triangle}}$ Divide through by $\tfrac12\sin h$ : $1 \le \frac{h}{\sin h} \le \frac{1}{\cos h}.$ As $h\to 0$ , $\cos h\to 1$ , so both ends $\to 1$ . By the Squeeze Theorem, $\dfrac{h}{\sin h}\to 1$ , hence $\dfrac{\sin h}{h}\to 1$ . (Even function, so it holds for $h<0$ too.)

Proof 1: $\dfrac{d}{dx}\sin x = \cos x$

Worked example Full derivation — each step justified

$\frac{d}{dx}\sin x = \lim_{h\to0}\frac{\sin(x+h)-\sin x}{h}$ Step 1 — expand with the addition formula. Why? We need to separate the " $x$ " part from the " $h$ shrinking to 0" part, and $\sin(x+h)=\sin x\cos h+\cos x\sin h$ does exactly that. $=\lim_{h\to0}\frac{\sin x\cos h + \cos x\sin h-\sin x}{h}$ Step 2 — group the $\sin x$ terms. Why? To build the two known limits. $=\lim_{h\to0}\left[\sin x\cdot\frac{\cos h-1}{h}+\cos x\cdot\frac{\sin h}{h}\right]$ Step 3 — $\sin x,\cos x$ are constants w.r.t. $h$ , so pull them out: $=\sin x\underbrace{\lim_{h\to0}\frac{\cos h-1}{h}}_{=\,0}+\cos x\underbrace{\lim_{h\to0}\frac{\sin h}{h}}_{=\,1}$ Step 4 — substitute the two ingredients: $=\sin x\cdot 0+\cos x\cdot 1=\boxed{\cos x}\ \checkmark$

Proof 2: $\dfrac{d}{dx}\cos x = -\sin x$

Worked example Full derivation — mirror of the above

$\frac{d}{dx}\cos x=\lim_{h\to0}\frac{\cos(x+h)-\cos x}{h}$ Step 1 — addition formula $\cos(x+h)=\cos x\cos h-\sin x\sin h$ . Why? Same reason: split $x$ from $h$ . $=\lim_{h\to0}\frac{\cos x\cos h-\sin x\sin h-\cos x}{h}$ Step 2 — group the $\cos x$ terms: $=\lim_{h\to0}\left[\cos x\cdot\frac{\cos h-1}{h}-\sin x\cdot\frac{\sin h}{h}\right]$ Step 3 — pull out constants & substitute the limits ( $0$ and $1$ ): $=\cos x\cdot 0-\sin x\cdot 1=\boxed{-\sin x}\ \checkmark$

Forecast-then-Verify

Recall Predict before you peek

Q: Without re-deriving, what is $\dfrac{d}{dx}\sin(2x)$ from first principles structure? (Hint: $\sin(2x+2h)$ ...) A: The same proof with $h\to 2h$ effectively scales by the inner rate: you'd get $2\cos(2x)$ . The factor 2 comes from $\frac{\sin 2h}{h}\to 2$ . This previews the chain rule.

Common mistakes (Steel-manned)

Active recall

First-principles definition of

f'(x)

\displaystyle\lim_{h\to0}\frac{f(x+h)-f(x)}{h}

Value of

\lim_{h\to0}\frac{\sin h}{h}

(radians)?

1

Value of

\lim_{h\to0}\frac{\cos h-1}{h}

0

Which theorem proves

\frac{\sin h}{h}\to1

The Squeeze (Sandwich) Theorem via unit-circle areas

Three areas compared in the squeeze?

\frac12\sin h \le \frac12 h \le \frac12\tan h

Addition formula used for

\sin

\sin(x+h)=\sin x\cos h+\cos x\sin h

Addition formula used for

\cos

\cos(x+h)=\cos x\cos h-\sin x\sin h

\frac{d}{dx}\sin x

\cos x

\frac{d}{dx}\cos x

-\sin x

Why must angles be in radians?

Because

\frac{\sin h}{h}\to1

only holds in radians (sector area

=\frac12 h

)

Trick to evaluate

\frac{\cos h-1}{h}

Multiply by conjugate

\frac{\cos h+1}{\cos h+1}

Why can't we use L'Hôpital here?

It assumes

\frac{d}{dx}\sin x=\cos x

— circular

Recall Feynman: explain to a 12-year-old

Imagine pushing a swing. When the swing is at the very bottom it's moving fastest; at the top it stops for a moment. The height of the swing is like $\sin x$ , and how fast it's moving is like $\cos x$ . Notice: when height is in the middle (zero), speed is biggest — and when height is at the top, speed is zero. Differentiation is just asking "how fast is it changing right now?" For sine, that answer is always the cosine. To prove it we zoom in super close (that's the limit), use a circle to show a tiny angle and its sine are nearly equal, and the algebra clicks into place.

Connections

Squeeze Theorem — the engine behind $\frac{\sin h}{h}\to1$
Trigonometric addition formulas — the algebraic splitter
Limits — definition and properties
Chain Rule — extends to $\sin(g(x))$
Derivatives of tan, sec, csc, cot — built from these two via quotient rule
Taylor series of sin and cos — alternative viewpoint of the same fact
Radian measure — why the constant comes out clean

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, idea simple hai: agar $\sin x$ ek lehar (wave) hai, to uska rate of change yaani derivative ek aur wave nikalti hai — aur wo wave exactly $\cos x$ hai. Matlab differentiation ne sirf wave ko 90 degree shift kar diya. Aur $\cos x$ ka derivative $-\sin x$ , kyunki cosine apne peak par decrease hona shuru karta hai, isliye minus aata hai.

Proof ka dil do limits hain. Pehla: $\frac{\sin h}{h}\to 1$ jab $h$ chhota hota hai. Yeh hum unit circle mein teen area compare karke (triangle $\le$ sector $\le$ bada triangle) Squeeze Theorem se prove karte hain. Yaad rakho yeh sirf radians mein sach hai — degrees mein galat ho jayega, isliye calculus hamesha radians use karti hai. Dusra limit: $\frac{\cos h-1}{h}\to 0$ , jise hum conjugate $(\cos h+1)$ se multiply karke nikaalte hain.

Ab actual derivative: $\sin(x+h)$ ko addition formula se khol do — $\sin x\cos h + \cos x\sin h$ . Phir $\sin x$ wale terms group karke do known limits ( $0$ aur $1$ ) plug kar do, aur seedha $\cos x$ aa jata hai. Cosine ke liye bilkul same kaam, bas sign flip hone se $-\sin x$ milta hai.

Important warning: L'Hôpital rule mat lagana $\frac{\sin h}{h}$ par — kyunki uske liye pehle se $\frac{d}{dx}\sin x=\cos x$ chahiye, jo hum prove hi kar rahe hain. Yeh circular logic hai. Bas Squeeze + addition formula — yeh 80/20 hai, inhe pakad lo to poora proof 1 minute mein khud bana loge.

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

What we are proving

The two ingredients we MUST establish first

Proof 1: $\dfrac{d}{dx}\sin x = \cos x$

Proof 2: $\dfrac{d}{dx}\cos x = -\sin x$

Forecast-then-Verify

Common mistakes (Steel-manned)

Active recall

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Calculus I — Limits & Derivatives

Connections