Single-Variable-Calculus · Unit 3 · Video 2 · Interactive Practice

The Derivative of Cosine | Why A and B Are Really Derivatives at Zero

IKey Formulas

Statement	Name	What it says
$\cos(x + \Delta x) = \cos x \cos \Delta x - \sin x \sin \Delta x$	Cosine addition formula	Each product mixes both angles, and the middle sign is minus
$\dfrac{\cos(x + \Delta x) - \cos x}{\Delta x} = \cos x \cdot \dfrac{\cos \Delta x - 1}{\Delta x} \; - \; \sin x \cdot \dfrac{\sin \Delta x}{\Delta x}$	Regrouped difference quotient	The same two brackets $(A)$ and $(B)$ as for the sine, with the coefficients swapped
$A = \lim\limits_{\Delta x \to 0} \dfrac{\cos \Delta x - 1}{\Delta x} = 0 \qquad B = \lim\limits_{\Delta x \to 0} \dfrac{\sin \Delta x}{\Delta x} = 1$	The two slopes at $x = 0$	$A = \dfrac{d}{dx}\cos x \Big\|_{x = 0}$ and $B = \dfrac{d}{dx}\sin x \Big\|_{x = 0}$
$\dfrac{d}{dx}\sin x = \cos x \qquad \dfrac{d}{dx}\cos x = -\sin x$	The two derivatives	Valid at every $x$ ; only the cosine picks up the minus sign

Key Insight: Both derivative formulas are built from the two numbers $A = 0$ and $B = 1$ : the slope of the cosine at its peak and the slope of the sine at the origin. One slope apiece at the single point $x = 0$ , and the addition formulas supply every other $x$ .

IIWhat $A$ and $B$ Measure

Each bracket is a secant slope at $x = 0$ , and shrinking $\Delta x$ shows where it settles.

Limit Δx 1.50

💡 The two values are only measured here — they are proved in the next video, and both boxed derivative formulas rest on them.

IIIFrom the Difference Quotient to $-\sin x$

Group the terms so that a zero stays over a zero, and the two brackets do the rest.

Step 1

Step 1 — The difference quotient

\frac{\cos(x + \Delta x) - \cos x}{\Delta x}

Step 2 — Expand with the cosine addition formula

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

\Longrightarrow \quad \frac{\cos x \cos \Delta x - \sin x \sin \Delta x - \cos x}{\Delta x}

Step 3 — Pair $\cos x \cos \Delta x$ with the trailing $-\cos x$ , then factor

\cos x \cdot \underbrace{\frac{\cos \Delta x - 1}{\Delta x}}_{(A)} \; - \; \sin x \cdot \underbrace{\frac{\sin \Delta x}{\Delta x}}_{(B)}

Step 4 — Bracket $(A)$ kills the first term

\frac{\cos \Delta x - 1}{\Delta x} \to A = 0 \qquad \Longrightarrow \qquad \cos x \cdot A = 0

Step 5 — Bracket $(B)$ keeps the second term whole

\frac{\sin \Delta x}{\Delta x} \to B = 1 \qquad \Longrightarrow \qquad -\sin x \cdot B = -\sin x

Step 6 — The pair of results

\frac{d}{dx}\sin x = \cos x \qquad \boxed{\frac{d}{dx}\cos x = -\sin x}

IVTwo Numbers, Every $x$

With $A = 0$ and $B = 1$ substituted, the tangent slope at any $x$ is read off a second curve.

Function Base point x₀ 2.00

VQuiz Questions

Problem 1 · Reading the New Formula

Given: the curve $y = \cos x$ — find the slope of its tangent line at $x = \dfrac{\pi}{3}$ .

✅ Correct!

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

, and

\sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}

, so the cosine is falling there with slope

-\frac{\sqrt{3}}{2} \approx -0.87

❌ Close, but check the sign. That is

+\sin\frac{\pi}{3}

— the derivative of the sine. Of the two trig derivatives, the cosine is the one that picks up the minus.

❌ Not quite.

\frac{1}{2}

\cos\frac{\pi}{3}

, the height of the curve there. The slope needs

-\sin

, not

\cos

Show solution

The video's boxed result is

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

Differentiate first, then substitute $x = \frac{\pi}{3}$ :

\left.\frac{d}{dx}\cos x\right|_{x = \pi/3} = -\sin\frac{\pi}{3} = -\frac{\sqrt{3}}{2} \approx -0.866.

The sign is a check on the picture: between $0$ and $\pi$ the cosine falls, so every tangent there must have a negative slope. The value $\frac{1}{2} = \cos\frac{\pi}{3}$ is the height of the point, not the steepness of the curve.

Problem 2 · The Regrouping Step

Given: the expanded quotient $\dfrac{\cos x \cos \Delta x - \sin x \sin \Delta x - \cos x}{\Delta x}$ — find the regrouping that isolates the two brackets $(A)$ and $(B)$ .

✅ Correct! The trailing

-\cos x

belongs with

\cos x \cos \Delta x

, because that is the term whose

\cos \Delta x

tends to

1

; factoring

\cos x

out of the pair leaves bracket

(A)

❌ Check the middle sign. The cosine addition formula subtracts

\sin x \sin \Delta x

, so the second term is carried along with a minus — that minus is the whole difference between this calculation and the sine's.

❌ The coefficients are swapped. That arrangement belongs to the sine calculation. Here

\cos x

is the factor pulled out of the

(A)

pair, and

\sin x

multiplies bracket

(B)

❌ Check the bracket. Factoring

\cos x

from

\cos x \cos \Delta x - \cos x

leaves

\cos \Delta x - 1

, not

1 - \cos \Delta x

; you have flipped the sign inside the bracket without flipping it outside.

Show solution

At $\Delta x = 0$ the numerator is $\cos x - \cos x = 0$ , so the quotient is $0/0$ and the terms must be grouped so that a zero stays over a zero. Since $\cos \Delta x \to 1$ , the term that needs a partner is $\cos x \cos \Delta x$ , and its partner is the trailing $-\cos x$ :

\frac{\overbrace{\cos x \cos \Delta x - \cos x}^{\text{pair}} - \sin x \sin \Delta x}{\Delta x}

Factor $\cos x$ out of the pair and $\sin x$ out of the last term:

\cos x \cdot \frac{\cos \Delta x - 1}{\Delta x} \; - \; \sin x \cdot \frac{\sin \Delta x}{\Delta x}

The same two brackets as in the sine calculation appear, but with the coefficients swapped and a minus sign in front of the $(B)$ term. That minus is exactly why $\frac{d}{dx}\cos x$ comes out negative.

Problem 3 · A Limit in Disguise

Given: $\displaystyle \lim_{h \to 0} \frac{\sin\!\left(\frac{\pi}{2} + h\right) - \sin\frac{\pi}{2}}{h}$ — identify the quantity it computes and evaluate it.

Which derivative is this limit?

What is its value?

✅ Correct! The base point is

\frac{\pi}{2}

, not

0

, so this is the sine's slope at its own peak:

\cos\frac{\pi}{2} = 0

— the same flat-top picture that limit

A

describes for the cosine at

x = 0

❌ Check the base point. A difference quotient

\frac{f(a + h) - f(a)}{h}

names its function by

f

and its point by

a

. Here

f = \sin

and

a = \frac{\pi}{2}

;

h

is the vanishing increment, never the base point.

❌ Not quite. Being

0/0

before the limit is exactly what makes a derivative worth computing — it does not stop the limit existing. Use

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

at the base point

\frac{\pi}{2}

Show solution

Step 1 — Match the pattern. The definition of the derivative at a point $a$ is

\left.\frac{d}{dx}f(x)\right|_{x = a} = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}.

Here $f = \sin$ and $a = \frac{\pi}{2}$ , so the limit is $\frac{d}{dx}\sin x$ evaluated at $x = \frac{\pi}{2}$ .

Step 2 — Evaluate. The video's result $\frac{d}{dx}\sin x = \cos x$ holds at every $x$ , so

\left.\frac{d}{dx}\sin x\right|_{x = \pi/2} = \cos\frac{\pi}{2} = 0.

Step 3 — Check the picture. $x = \frac{\pi}{2}$ is the peak of the sine, and a tangent at a peak is horizontal. With $a = 0$ instead, the same limit would have been $B = 1$ ; the base point, not the letter $h$ , decides the answer.

Problem 4 · Where the Cosine Climbs Fastest

Given: $y = \cos x$ on $[0, 2\pi]$ — find the value of $x$ at which the tangent slope is greatest.

✅ Correct! The slope function is

-\sin x

, and it is largest where

\sin x

is most negative:

-\sin\frac{3\pi}{2} = 1

, the steepest climb on the whole period.

❌ That is where the curve is extreme, not its slope. At

x = 0

and

x = \pi

the cosine is at

1

and

-1

, and

-\sin x = 0

at both — those tangents are horizontal.

❌ Close: that is the steepest point, but the wrong direction.

-\sin\frac{\pi}{2} = -1

is the most negative slope — the fastest fall, not the fastest climb.

Show solution

The tangent slope at $x$ is the value of the derivative there:

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

So the question is: where on $[0, 2\pi]$ is $-\sin x$ largest? Since $\sin x$ has its minimum $-1$ at $x = \frac{3\pi}{2}$ ,

-\sin\frac{3\pi}{2} = -(-1) = 1,

and no larger value is possible because $-\sin x \le 1$ everywhere. The four candidates give

$x = 0$ : slope $-\sin 0 = 0$ (the peak of the cosine — horizontal tangent, which is limit $A$ )
$x = \frac{\pi}{2}$ : slope $-1$ (steepest fall)
$x = \pi$ : slope $-\sin\pi = 0$ (the trough — horizontal again)
$x = \frac{3\pi}{2}$ : slope $+1$ (steepest climb) $\checkmark$

Extremes of $\cos x$ and extremes of its slope never coincide — they are a quarter period apart.

Solved: 0 / 4

The Derivative of Cosine | Why A and B Are Really Derivatives at Zero

IKey Formulas

IIWhat AAA and BBB Measure

IIIFrom the Difference Quotient to −sin⁡x-\sin x−sinx

IVTwo Numbers, Every xxx

VQuiz Questions

IIWhat $A$ and $B$ Measure

IIIFrom the Difference Quotient to $-\sin x$

IVTwo Numbers, Every $x$