2.3.5 Derivatives of sine and cosine

In this lesson

The slopes of sine and cosine at 0, and why they need radians
The slope of $\sin x$ is $\cos x$ , read off the two graphs and checked by central differences
The slope of $\cos x$ is $-\sin x$ , and one unit-circle picture for both rules
Differentiating again: a cycle of four, and a second derivative that pulls the wave back
$\tan x$ by the quotient rule, and where it has no slope
With the chain rule: $\sin(kx)$ , $\cos(\omega t + \varphi)$ and $\sin(x^2)$ , with frequencies and periods
Reading the rules backwards

You need radians, the unit circle and the graphs of sine and cosine from lesson 1.7.1, and the limit of $\frac{\sin x}{x}$ from lesson 2.1.1. The unit-circle picture also uses the quarter turn of lessons 1.7.2 and 1.7.3. You need central differences (lesson 2.2.2) and the product, quotient and chain rules (lessons 2.3.2 and 2.3.3).

Two slopes at 0

Every rule in this lesson grows from two numbers: the slopes of sine and cosine at 0. Both come from chord slopes (lesson 2.2.1), with the gap $h$ in radians.

Sine at 0. Since $\sin 0 = 0$ , the chord slope of $\sin x$ from 0 to $h$ is $\frac{\sin h}{h}$ , the limit of lesson 2.1.1.

gap $h$	$\pm 0.1$	$\pm 0.01$	$\pm 0.001$
chord slope $\frac{\sin h}{h}$	0.998334	0.999983	1.0

Both sides give the same numbers, because $\sin(-h) = -\sin h$ (lesson 1.7.1). So the slope at 0 is 1, and near 0 sine hugs the line $y = x$ : $\sin 0.02 \approx 0.019999$ .

Cosine at 0. Cosine is at the top of its wave at 0, where $\cos 0 = 1$ . So its chord slope is $\frac{\cos h - 1}{h}$ .

gap $h$	0.1	0.01	0.001
from the right, $h$	$-0.049958$	$-0.005$	$-0.0005$
from the left, $-h$	0.049958	0.005	0.0005

The chords slope down to the right and up to the left, and both shrink to 0. The tangent at the top is flat.

Rule: Two limits at 0

\begin{gathered} \lim_{h \to 0} \frac{\sin h}{h} = 1 \\[4pt] \lim_{h \to 0} \frac{\cos h - 1}{h} = 0 \end{gathered}

With $h$ in radians, these are the slopes of sine and cosine at 0. This lesson takes them from the tables; the Proof Gym proves them.

In degrees the slope is not 1

Measure the gap in degrees and the chord slope becomes $\frac{\sin(h°)}{h}$ . One degree is only $\frac{\pi}{180} \approx 0.017453$ radians, so the slope shrinks by that factor.

gap in degrees	1	0.1	0.01
chord slope per degree	0.017452	0.017453	0.017453

In degrees, every rule below would carry that factor. Radians make the slope at 0 exactly 1, as $e$ makes the slope of its exponential 1 at 0 (lesson 2.3.4).

Try it The first loop prints the chord slope of sine at 0, from the right and from the left. The second does the same for cosine, whose two sides have opposite signs. The last loop measures the gap in degrees, converted by math.radians (lesson 1.7.1).

import math for h in [0.1, 0.01, 0.001]: print(h, round(math.sin(h) / h, 6), round(math.sin(-h) / -h, 6)) for h in [0.1, 0.01, 0.001]: print(h, round((math.cos(h) - 1) / h, 6), round((math.cos(-h) - 1) / -h, 6)) for h in [1, 0.1, 0.01]: print(h, round(math.sin(math.radians(h)) / h, 6)) print(round(math.pi / 180, 6))

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The slope of sine is cosine

Read the slope of $\sin x$ off its graph and you trace out $\cos x$ .

At 0, sine crosses zero going up, at its steepest: slope 1.
At $\frac{\pi}{2}$ it is at its top: slope 0.
At $\pi$ it crosses zero going down, at its steepest: slope $-1$ .
At $\frac{3\pi}{2}$ it is at its bottom: slope 0. At $2\pi$ the wave starts again.

These slopes, 1, 0, $-1$ , 0, are the values of $\cos x$ at the same points. Sine is steepest where it crosses zero, which is where cosine peaks.

Tangents to $y = \sin x$ where it crosses zero and at its top and bottom. Plotted at the same $x$ , their slopes 1, 0, $-1$ , 0, 1 sit on $y = \cos x$ .

Between those points the match holds too. Central differences (lesson 2.2.2) with $h = 0.001$ give:

$x$	0	$\frac{\pi}{6}$	$\frac{\pi}{3}$	$\frac{\pi}{2}$	$\pi$
slope of $\sin x$	1.0	0.866025	0.5	0.0	$-1.0$
$\cos x$	1	0.866025	0.5	0	$-1$

Rule: Derivative of sine

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

With $x$ in radians. Sine rises where $\cos x > 0$ , falls where $\cos x < 0$ , and is flat where $\cos x = 0$ , at its tops and bottoms.

The slope stays between $-1$ and 1, and sine has no corners or jumps (lesson 2.2.2). Negative inputs work the same way: at $-\frac{\pi}{3}$ the slope is $\cos(-\frac{\pi}{3}) = 0.5$ .

Slide the point along $\sin x$ and watch the slope graph underneath trace the cosine wave. Jump to $\frac{\pi}{2}$ , where the tangent is flat, and to $\pi$ , where it falls most steeply.

Explore

Slide the point along the curve. Its tangent shows the slope there, and the lower graph plots that slope for every x: the derivative.

The slope of y = sin x, which is cos x

Point at x = 0.5

Trace as you go

Point: (0.159π, 0.4794)
Slope there: 0.8776
The curve is: rising: the slope is positive, so the slope graph is above zero

x

import math def central(f, a, h=0.001): return (f(a + h) - f(a - h)) / (2 * h) for a in [0, math.pi / 6, math.pi / 3, math.pi / 2, math.pi, -math.pi / 3]: print(round(a, 6), round(central(math.sin, a), 6) + 0.0, round(math.cos(a), 6) + 0.0)

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The slope of cosine is minus sine

Read the cosine wave the same way: at 0 it is at its top, and flat (the second limit). It falls most steeply as it crosses zero at $\frac{\pi}{2}$ , with slope $-1$ . It is flat again at its bottom, $\pi$ , and climbs most steeply at $\frac{3\pi}{2}$ , with slope 1.

These slopes, 0, $-1$ , 0, 1, are the values of $-\sin x$ . Central differences agree: at $\frac{\pi}{6}$ they give $-0.5$ , which is $-\sin\frac{\pi}{6}$ .

Rule: Derivative of cosine

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

With $x$ in radians. The minus sign matters: cosine falls while sine is positive, from 0 to $\pi$ , and rises while sine is negative.

One picture for both rules

Lesson 1.7.1 put the point $(\cos x, \sin x)$ on the unit circle, an arc $x$ round from $(1, 0)$ . In radians the arc grows as fast as the angle, so the point moves at speed 1.

It moves at right angles to the radius (lesson 1.7.3): in the direction of the radius turned a quarter turn. That turn takes $(a, b)$ to $(-b, a)$ (lesson 1.7.2), giving $(-\sin x, \cos x)$ .

Per unit of angle the point moves 1 along the circle. Its height changes by $\cos x$ : up at $\frac{\pi}{6}$ , down at $\frac{2\pi}{3}$ .

So the height $\sin x$ changes at the rate $\cos x$ , and the across coordinate $\cos x$ at the rate $-\sin x$ . In degrees the point moves only $\frac{\pi}{180}$ of a unit per degree.

x

import math def central(f, a, h=0.001): return (f(a + h) - f(a - h)) / (2 * h) for a in [0, math.pi / 6, math.pi / 2, math.pi, 3 * math.pi / 2]: print(round(a, 6), round(central(math.cos, a), 6) + 0.0, round(-math.sin(a), 6) + 0.0) a = math.pi / 6 for h in [0.1, 0.01, 0.001]: rise = math.sin(a + h) - math.sin(a) across = math.cos(a + h) - math.cos(a) print(h, round(rise / h, 6), round(across / h, 6))

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Differentiating again: a cycle of four

Each rule turns sine into cosine, or cosine into minus sine. So repeated differentiation goes round a cycle of four.

times differentiated	0	1	2	3	4
$\sin x$ becomes	$\sin x$	$\cos x$	$-\sin x$	$-\cos x$	$\sin x$

The fifth derivative is the first again, $\cos x$ . Starting from $\cos x$ , the same cycle begins one step along. Compare $e^x$ (lesson 2.3.4), which comes back after one step.

The second derivative pulls the wave back

Differentiating twice gives the second derivative $f''(x)$ , the slope of the slope; lesson 2.4.2 studies it. For $f(x) = \sin x$ it is $-\sin x$ : the function itself, with its sign flipped.

So above the axis the slope is falling and the wave bends back down; below the axis it bends back up. The bend always pulls the wave towards 0, so it overshoots and swings for ever.

Module 6.2 solves differential equations such as $f'' = -f$ , whose solutions are waves like these.

\sin x

import sympy as sp x = sp.symbols("x") f = sp.sin(x) for k in range(1, 9): f = sp.diff(f, x) print(k, f) print(sp.diff(sp.sin(x), x, 2), sp.diff(sp.cos(x), x, 2)) print(sp.diff(-sp.cos(x), x))

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Tangent, by the quotient rule

$\tan x = \frac{\sin x}{\cos x}$ (lesson 1.7.1), so the quotient rule of lesson 2.3.2, $\frac{f'g - fg'}{g^2}$ , applies with $f = \sin x$ and $g = \cos x$ :

\begin{gathered} \frac{\cos x \cdot \cos x - \sin x \cdot (-\sin x)}{\cos^2 x} \\[4pt] = \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = \frac{1}{\cos^2 x} \end{gathered}

The last step is the identity $\cos^2 x + \sin^2 x = 1$ (lesson 1.7.1). Dividing it by $\cos^2 x$ gives a second form, $1 + \tan^2 x$ .

Rule: Derivative of tangent

\begin{gathered} \frac{d}{dx}\,\tan x = \frac{1}{\cos^2 x} \\[4pt] = 1 + \tan^2 x \\[4pt] \text{where } \cos x \neq 0 \end{gathered}

The slope is never below 1, so every branch of tangent rises. Where tangent has no value, at $\frac{\pi}{2}$ and every half turn from it, it has no slope.

At $\frac{\pi}{4}$ , $\cos^2 x = \frac{1}{2}$ , so the slope is 2. Central differences agree:

gap $h$	0.1	0.01	0.001
central difference at $\frac{\pi}{4}$	2.0271	2.000267	2.000003

The slope is 1 at 0 and 4 at $\frac{\pi}{3}$ . Near $\frac{\pi}{2}$ it blows up: 199.85 at 1.5 and about 1.58 million at 1.57, where tangent itself shoots off (lesson 1.7.1).

\tan x

import math import sympy as sp for h in [0.1, 0.01, 0.001]: print(h, round((math.tan(math.pi / 4 + h) - math.tan(math.pi / 4 - h)) / (2 * h), 6)) for a in [0, math.pi / 4, math.pi / 3, 1.5, 1.57]: print(round(a, 6), round(1 / math.cos(a)**2, 4)) x = sp.symbols("x") print(sp.diff(sp.tan(x), x), "|", sp.simplify(sp.diff(sp.sin(x) / sp.cos(x), x)))

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With the chain rule: rates and periods

With sine or cosine outside, the chain rule of lesson 2.3.3 keeps the inside $u = g(x)$ as it is and multiplies by its slope.

Rule: Sine and cosine of a function

\begin{gathered} \frac{d}{dx}\,\sin u = \cos u \cdot \frac{du}{dx} \\[4pt] \frac{d}{dx}\,\cos u = -\sin u \cdot \frac{du}{dx} \end{gathered}

Here $u = g(x)$ is the inside. The outer slope is taken at $u$ , then multiplied by the slope of the inside.

A linear inside. $\sin(3x)$ has inside $3x$ , with slope 3, so its derivative is $3\cos(3x)$ : 3 at $x = 0$ . Central differences agree:

gap $h$	0.1	0.01	0.001
central difference at 0	2.955202	2.99955	2.999996

$\sin(3x)$ runs through its wave three times as fast, with period $\frac{2\pi}{3} \approx 2.094395$ . Squeezed sideways by 3, the wave is 3 times as steep.

Frequency and phase

A wave in time is written $y = \cos(\omega t + \varphi)$ , with $t$ in seconds. The angular frequency $\omega$ (omega) is in radians per second; the phase $\varphi$ (phi) shifts the wave along. The inside has slope $\omega$ , so

\frac{dy}{dt} = -\omega \sin(\omega t + \varphi)

The period is $\frac{2\pi}{\omega}$ seconds, and the slope swings between $-\omega$ and $\omega$ , in units of $y$ per second. A frequency $\omega$ makes slopes $\omega$ times larger; the phase only moves the wave.

For $y = \cos(2t + \frac{\pi}{6})$ , the slope at $t = 0$ is $-2\sin\frac{\pi}{6} = -1$ , and the period is $\pi$ . One cycle per second is $\omega = 2\pi$ , with slopes up to $2\pi \approx 6.283185$ per second.

Curved insides, powers and degrees

A curved inside. $\sin(x^2)$ has inside $x^2$ , with slope $2x$ , so its derivative is $2x\cos(x^2)$ . At $x = 1$ that is $2\cos 1 \approx 1.080605$ . Central differences give 1.056644, 1.080364 and 1.080602 for gaps 0.1, 0.01 and 0.001.

A power of sine. $\sin^2 x$ means $(\sin x)^2$ : the outer function is $u^2$ and the inside is $\sin x$ . Its slope is $2\sin x\cos x$ , which is 1 at $\frac{\pi}{4}$ .

Degrees. Sine in degrees is $\sin\left(\frac{\pi x}{180}\right)$ . Its slope, $\frac{\pi}{180}\cos\left(\frac{\pi x}{180}\right)$ , carries the factor $\frac{\pi}{180}$ at every point.

\sin(3x)

import math import sympy as sp def central(f, a, h): return (f(a + h) - f(a - h)) / (2 * h) for h in [0.1, 0.01, 0.001]: print(h, round(central(lambda v: math.sin(3 * v), 0, h), 6), round(central(lambda v: math.sin(v**2), 1, h), 6)) x, t = sp.symbols("x t") for g in [sp.sin(3 * x), sp.sin(x**2), sp.sin(x)**2, sp.sin(sp.pi * x / 180)]: print(g, "->", sp.diff(g, x)) dy = sp.diff(sp.cos(2 * t + sp.pi / 6), t) print(dy, "| at 0:", dy.subs(t, 0))

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Reading the rules backwards

The rules also answer the reverse question: which function has a given slope? $\sin x$ has slope $\cos x$ . Since $\cos x$ has slope $-\sin x$ , the function with slope $\sin x$ is $-\cos x$ , not $\cos x$ .

Lesson 2.5.2 calls such a function an antiderivative, and lists both rules backwards. Lesson 2.6.2 uses both rules to integrate $e^x \sin x$ .

Common mistakes

Use radians. In degrees every slope has an extra factor $\frac{\pi}{180}$ . The slope of sine at 0 is then 0.017453 per degree.
The minus sign belongs to cosine. $\frac{d}{dx}\cos x = -\sin x$ , but $\frac{d}{dx}\sin x = \cos x$ .
Keep the chain factor, and leave the inside alone. $\sin(3x)$ gives $3\cos(3x)$ , not $\cos(3x)$ or $3\cos x$ . $\cos(\omega t + \varphi)$ gives $-\omega\sin(\omega t + \varphi)$ .
$\sin^2 x$ is not $\sin(x^2)$ . Their slopes are $2\sin x\cos x$ and $2x\cos(x^2)$ .
Tangent has no slope at its gaps. At $\frac{\pi}{2}$ there is no $\tan x$ and no slope. Close by, the slope is huge.
Backwards, mind the sign. The function with slope $\sin x$ is $-\cos x$ .

Derivatives of sine and cosine

0 of 12 correct

1.
Near 0, $\sin x$ is close to its tangent $y = x$ . Use it to estimate $\sin 0.03$
2.
With $x$ in degrees, what is the slope of $\sin x$ at $x = 0$ , per degree? (type pi for $\pi$ )
3.
Differentiate $\cos(5x)$ . Type sine and cosine with brackets, like sin(5x)
4.
Differentiate $x \sin x$
5.
Find the slope of $3\sin x - 2\cos x$ at $x = \frac{\pi}{2}$
6.
Find the third derivative of $\sin x$ at $x = 0$
7.
Find the second derivative of $\sin(2x)$ at $x = \frac{\pi}{4}$
8.
Find the slope of $\tan x$ at $x = \frac{\pi}{3}$
9.
Find the slope of $\sin(3x)$ at $x = \frac{\pi}{9}$
10.
$y = \cos(3t + \frac{\pi}{6})$ . Find $\frac{dy}{dt}$ at $t = 0$
11.
Find the slope of $\sin(x^2)$ at $x = \sqrt{\pi}$ (type sqrt and pi)
12.
Find the slope of $\sin^2 x$ at $x = \frac{\pi}{4}$

How to type answers

Powers with ^: x^7, 2^(1/3), a^-2.
Multiply with * or by writing side by side: 3x^2, 2(x+1).
ln(x) is the natural log, log(x, 2) is log base 2, e is Euler's number, sqrt(x) is a square root.
sin(x), cos(x) and tan(x) take radians; write pi for π and sin(x)^2 for sin²x.
Your first answer to each question is saved: it counts towards your score and brings the question back for review.
Exact answers like ln(20)/ln(3) are best. Decimals need about 4 significant figures.

\frac{\sin h}{h} \to 1

Derivatives of sine and cosine

Two slopes at 0

Rule: Two limits at 0

In degrees the slope is not 1

The slope of sine is cosine

Rule: Derivative of sine

The slope of cosine is minus sine

Rule: Derivative of cosine

One picture for both rules

Differentiating again: a cycle of four

The second derivative pulls the wave back

Tangent, by the quotient rule

Rule: Derivative of tangent

With the chain rule: rates and periods

Rule: Sine and cosine of a function

Frequency and phase

Curved insides, powers and degrees

Reading the rules backwards

Common mistakes

Derivatives of sine and cosine

Summary card

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