Derivatives of sin x, cos x — proofs from first principles: Question bank

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

True or false — justify

lim_{h \to 0} \frac{s i n h}{h} = 1

holds whether

h

is measured in degrees or radians.

False. In degrees the limit is

\frac{π}{180} \approx 0.01745

, because the sector-area bound

\frac{1}{2} h

(valid only on a unit circle, radius

= 1

, where arc length equals the radian angle) requires

h

in radians. The clean "

= 1

" is a radian-only fact — see the unit-circle squeeze in Figure s02.

The derivative

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

shifts the whole sine wave left by

9 0^{\circ}

True.

cos x = sin (x + \frac{π}{2})

, so differentiating advances the wave by a quarter period — the graph of the slope is the sine graph slid left by

\frac{π}{2}

. Figure s03 overlays

sin x

and its slope-curve

cos x

so you can see the phase shift directly.

Since

cos h \to 1

h \to 0

, we may conclude

\frac{c o s h - 1}{h} \to \frac{0}{0}

, which is undefined.

False.

\frac{0}{0}

is not a number but an indeterminate form — a signal to work harder, not a final answer. The conjugate trick resolves it to the definite value

0

\frac{d}{d x} cos x

is negative near

x = 0

because cosine is decreasing there.

True. At

x = 0

cosine sits at its peak and immediately falls, so its slope is negative;

- sin x < 0

for small

x > 0

confirms it. The minus sign is geometrically forced.

The two ingredient limits (

\frac{s i n h}{h} \to 1

and

\frac{c o s h - 1}{h} \to 0

) are logically independent facts.

False. The second is built from the first: the conjugate trick turns

\frac{c o s h - 1}{h}

into

- \frac{s i n h}{h} \cdot \frac{s i n h}{c o s h + 1}

, so it inherits the value

- (1) \cdot \frac{0}{2} = 0

(using

cos h \to 1

by continuity of cosine, so

cos h + 1 \to 2

If we only knew

\frac{s i n h}{h} \to 1

but nothing about addition formulas, we could still finish the proof.

False. The addition formula is what splits

sin (x + h)

into an "

x

part" and an "

h

part"; without it there is nowhere for the limits to plug in. Both ingredients are load-bearing.

\frac{d}{d x} sin (2 x) = cos (2 x)

False. The inner rate matters:

\frac{d}{d x} sin (2 x) = 2 cos (2 x)

, because

\frac{s i n 2 h}{h} \to 2

. To see the

2

: substitute

u = 2 h

, so

\frac{s i n 2 h}{h} = \frac{s i n u}{u /2} = 2 \cdot \frac{s i n u}{u}

, and as

h \to 0

we have

u \to 0

, giving

2 \cdot 1 = 2

. This missing factor of

2

is exactly what the Chain Rule later formalizes.

Spot the error

"Apply L'Hôpital to

\frac{s i n h}{h}

: differentiate top and bottom to get

\frac{c o s h}{1} \to 1

. Done."

The error is circularity. L'Hôpital needs

\frac{d}{d h} sin h = cos h

, which is the very theorem we're proving; using it here assumes the conclusion. The squeeze argument is the honest foundation.

cos h \approx 1 - \frac{h ^{2}}{2}

, so

\frac{c o s h - 1}{h} \approx - \frac{h}{2} \to - \frac{1}{2}

The limit of

- \frac{h}{2}

h \to 0

0

, not

- \frac{1}{2}

. The

\frac{1}{2}

belongs to the different limit

\frac{1 - c o s h}{h ^{2}} \to \frac{1}{2}

— watch the power of

h

in the denominator.

"In the squeeze,

\frac{1}{2} sin h \leq \frac{1}{2} h \leq \frac{1}{2} tan h

holds for all

h

, so the whole argument works for any

h

The area inequality is derived for small positive

h

(a first-quadrant angle on the unit circle, Figure s02). For

h < 0

we invoke evenness —

\frac{s i n h}{h}

is an even function — rather than the raw area picture, which needs a positive angle.

sin (x + h) = sin x + sin h

, so the derivative is

lim_{h \to 0} \frac{s i n h}{h} = 1

Sine is not additive:

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

. The correct expansion is

sin x cos h + cos x sin h

, and it is precisely this that yields

cos x

, not the constant

1

"Since both limits give constants (

0

and

1

), the derivative of

sin x

is the constant

cos (0) \cdot 1 = 1

The factors

sin x

and

cos x

are constants with respect to $h$ , but they still depend on

x

and must stay. The answer is

sin x \cdot 0 + cos x \cdot 1 = cos x

, a function of

x

, not the number

1

tan h

appears in the squeeze, so we secretly assumed we know

\frac{d}{d x} tan x

No —

tan h = \frac{s i n h}{c o s h}

is used purely as a ratio of lengths (the outer triangle's area in Figure s02), a static geometric fact. No derivative of tangent is invoked; Derivatives of tan, sec, csc, cot come later, built on this result.

Why questions

Why do we expand with the addition formula instead of just plugging

h = 0

directly?

Plugging

h = 0

gives

\frac{0}{0}

— the average slope over a zero-width interval is undefined. The addition formula reshapes the quotient into pieces whose limits do exist.

Why must the angle be in radians for calculus, geometrically?

The squeeze's middle term, the sector area

\frac{1}{2} h

, equals the actual circular-sector area only on a unit circle (radius $1$ ) with $h$ in radians, where arc length equals the angle. Any other unit inserts a conversion constant into every derivative.

Why does the limit-flow

cos x cos h - sin x sin h

collapse neatly to

cos x \cdot 1 - sin x \cdot 0

Because as

h \to 0

, cosine is continuous so

cos h \to cos 0 = 1

and sine is continuous so

sin h \to sin 0 = 0

; the

x

-factors just ride along unchanged. Figure s03's companion idea: the "

h

-part" fades to a clean

1

and

0

, leaving the "

x

-part" standing.

Why does differentiating rotate the sine/cosine family by exactly

9 0^{\circ}

each time?

Because the slope of a wave peaks where the wave crosses zero and vanishes where the wave peaks — a quarter-cycle offset (visible in Figure s03). Iterating gives the cycle

sin \to cos \to - sin \to - cos \to sin

Why is the conjugate multiplier

cos h + 1

chosen and not something else?

Because

(cos h - 1) (cos h + 1) = cos^{2} h - 1 = - sin^{2} h

converts the stubborn

cos h - 1

into a

sin

we already control, letting us reuse

\frac{s i n h}{h} \to 1

Why does the sanity check (peak of cosine ⇒ negative slope) count as evidence but not proof?

It confirms the sign is consistent, ruling out a sign slip, but it says nothing about the magnitude (

sin x

vs. anything else). Only the full limit argument pins the exact function.

Why do we prove

cos h - 1

over

h

(linear power) rather than over

h^{2}

Because the derivative's definition divides the difference by the interval width

h

to the first power. That single power forces the limit to

0

; the

h^{2}

version is a separate, higher-order fact not needed here.

Edge cases

x = 0

, is the first-principles limit for

sin

still

cos 0 = 1

, and does the graph confirm it?

Yes. At

x = 0

the sine graph crosses zero with its steepest upward slope (Figure s03), and the derivation gives

cos 0 = 1

— the maximum possible slope of

sin

, matching the picture exactly.

What happens to

\frac{s i n h}{h}

from the left, i.e. as

h \to 0^{-}

It still

\to 1

, because

\frac{s i n h}{h}

is an even function (

sin (- h) / (- h) = sin h / h

). The two-sided limit exists and equals

1

, so the derivative is well-defined at every

x

sin x

differentiable everywhere, including at its peaks

x = \frac{π}{2}

Yes. At

x = \frac{π}{2}

the derivation gives

cos \frac{π}{2} = 0

— a smooth horizontal tangent, not a corner. Sine and cosine are differentiable on all of

R

If we replaced

h

by a sequence like

h = \frac{1}{n} \to 0

(where

n

is a positive integer growing without bound,

n \to \infty

), would we get the same limit?

Yes, since the limit exists as a genuine (two-sided) limit, every sequence

h_{k} \to 0

with

h_{k} \neq = 0

yields

1

. Path-independence is guaranteed once the limit is established.

Does the proof break if

x

itself is huge, say

x = 1000

No. The limit is taken over

h \to 0

with

x

held fixed as a constant; the size of

x

never enters the two ingredient limits. The result

cos x

holds for every real

x

What if someone defines the derivative with

\frac{f ( x ) - f ( x - h )}{h}

(backward difference) instead?

You get the same derivative

cos x

; the backward, forward, and symmetric quotients all converge to the identical limit for a differentiable function. The choice is cosmetic here.

Recall One-line summary of every trap

Almost all traps reduce to three sins: wrong units (degrees), wrong power of $h$ (mixing the two cosine limits), or circular logic (L'Hôpital / assuming additivity). Guard those three and the topic is airtight.

Connections

Squeeze Theorem — the engine the "spot the error" items probe
Trigonometric addition formulas — the non-additivity trap
Radian measure — the degrees-vs-radians trap
Chain Rule — the $sin (2 x)$ factor-of-2 trap
Derivatives of tan, sec, csc, cot — where $tan h$ later reappears
Taylor series of sin and cos — the $1 - \frac{h ^{2}}{2}$ approximation trap
Limits — definition and properties — indeterminate-form reasoning