Differential Calculus for Social Science 201-103 • CEGEP, Montreal

Revision sheet: trigonometric functions and their derivatives (201-103)

The trigonometry chapter of Social Science differential calculus is not a chapter about triangles. Sine serves as the single model of everything that comes back: retail sales over twelve months, emergency room traffic over seven days, average temperature, seasonal migration. All these series are written $f(t)=A\sin(\omega t+\varphi)+k$ , and the derivative answers a question that the data column never shows: at what time is the quantity increasing fastest?

The real obstacle is there, and it is fixed in one sentence: the peak is the moment when $f'=0$ , so the moment when the rise is OVER. Maximum growth falls a quarter of a period earlier, where the curve crosses its mean and where $f''$ vanishes. The rest of the sheet deals with the mechanical errors that cost the most, the calculator's degree mode and the factor $\omega$ left by the chain rule.

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The thread of the chapter

Here sine is not a triangle tool, it is the model of whatever comes back: the peak is the moment when $f'=0$ , maximum growth arrives a quarter of a period earlier, and none of this holds unless the angle is measured in radians.

This chapter is part of 201-103, Differential Calculus for Social Science

What is on this sheet

1. The essentials
2. The rules in table form
3. The mistakes that cost marks
4. Which method to choose
5. How the answer is expected to be written
6. Check before you hand in
7. The typical problem, taken apart
8. Learn by heart
9. Frequently asked questions

Topics covered

Radian
Unit circle
Pythagorean identity
Fundamental limit of sine
Derivatives of trigonometric functions
Chain rule
Sinusoidal model
Amplitude and midline
Period and angular frequency
Phase shift
Inflection point
Angle addition identity

The essentials

The circle and the radian, and what the Pythagorean identity does not say

• $180^{\circ}=\pi$ rad. We convert a degree to a radian by multiplying by $\dfrac{\pi}{180}$ , never by $\dfrac{180}{\pi}$ .
• A radian is a ratio of two lengths, hence a number without a unit: that is what allows us to write $\sin t$ where $t$ counts months.
• Pythagorean identity: $\sin^{2}x+\cos^{2}x=1$ . It gives the SQUARE, so $\cos x=\pm\sqrt{1-\sin^{2}x}$ , and the quadrant chooses the sign.
• Reference angle: memorize only the first quadrant, then attach the sign. So $\cos\dfrac{2\pi}{3}=-\cos\dfrac{\pi}{3}=-0.5$ .
• Related angles: $\sin(\pi-x)=\sin x$ , $\cos(\pi+x)=-\cos x$ , $\sin(-x)=-\sin x$ and $\cos(-x)=\cos x$ .

The two points have the SAME sine,

0.866

, and opposite cosines: the Pythagorean identity could not tell them apart, only the quadrant does.

Writing the sign of the quadrant BEFORE touching the calculator earns the half mark for rigour and removes half the errors on this page.

The periodic model: four letters, four management questions

• $f(t)=A\sin(\omega t+\varphi)+k$ : $k$ is the mean level, $A$ the amplitude, $\omega$ the angular frequency and $\varphi$ the phase shift.
• $k=\dfrac{\max+\min}{2}$ and $A=\dfrac{\max-\min}{2}$ : the mean of the two extremes, then their HALF-difference.
• $T=\dfrac{2\pi}{\omega}$ , so $\omega=\dfrac{2\pi}{T}$ . Over twelve months, $\omega=\dfrac{\pi}{6}\approx 0.5236$ radian per month.
• The maximum is $k+A$ and the minimum $k-A$ : the amplitude is measured from the midline (the axis of oscillation), not from zero.
• The gap between the best and the worst moment is $2A$ , and this is the number a report quotes.

The horizontal line gives

k=110

, the vertical line gives

A=30

from that midline and not from zero, the full cycle gives

T=12

months, and the maximum is indeed

k+A=140

The four parameters are read in a fixed order, $k$ , then $A$ , then $\omega$ , then $\varphi$ : each one needs the previous ones, never the reverse.

The table of derivatives, and the factor that follows it everywhere

• $(\sin x)'=\cos x$ , $(\cos x)'=-\sin x$ , $(\tan x)'=\dfrac{1}{\cos^{2}x}=1+\tan^{2}x$ .
• $\dfrac{1}{\cos^{2}x}\ge 1$ everywhere, so the tangent function has NO horizontal tangent line and is never used as a cyclical model.
• On a model, the chain rule gives $f'(t)=A\omega\cos(\omega t+\varphi)$ : the amplitude of the rate of change is $A\omega$ , in units of $f$ per unit of time.
• $f''(t)=-\omega^{2}\big(f(t)-k\big)$ : the acceleration is proportional to the distance from the mean and has the opposite sign.
• The whole table rests on $\displaystyle\lim_{x\to 0}\dfrac{\sin x}{x}=1$ , which is true only in radians and only at ZERO.

The orange curve is the slope of the blue one:

\cos x

is positive when

\sin x

rises, zero at

\frac{\pi}{2}

when

\sin x

peaks, negative when it falls.

The unit is the quickest check on the page: if $f$ is in thousands of dollars and $t$ in months, $f'$ reads in thousands of dollars per month, never in thousands of dollars.

The peak and maximum growth, a quarter of a period apart

• The peak is the moment when $f'=0$ , so the moment when the rise is OVER; $f$ equals $k+A$ there.
• Maximum growth is the moment when $f''=0$ : $f$ equals exactly $k$ there and $f'$ equals $+A\omega$ .
• The two are separated by $\dfrac{T}{4}$ , and maximum growth comes BEFORE the peak.
• Over one period, $f''$ vanishes twice: the time when $f'>0$ is the fastest rise, the one when $f'<0$ is the fastest fall.
• A dashboard that reads only the level sees nothing happening on the day of maximum growth, since $f$ equals its annual mean there.

It is the one thing the derivative brings that no reading of the data column gives: a date when the level is ordinary and the rate of change is extreme.

The rules in table form

Each row reads from left to right: the assumptions, then the result. A red cell is not an answer, it is the finding that the form settles nothing and the instruction to rewrite it. Every case is followed by a worked example.

The derivatives of the chapter, with their numerical check

The first four lines are meant to be copied. The last three are expressions you will see on student papers: a red cell is never an answer, it is an order to rewrite.

What we differentiate	Derivative	Numerical check
$\sin x$	$\cos x$	$f'(0)=1$
Example: $f'(0)=\cos 0=1$ : at zero, sine rises with slope 1, which $\sin(0.05)=0.049979$ confirms.
$\cos x$	$-\sin x$	$f'\left(\dfrac{\pi}{6}\right)=-0.5$
Example: $-\sin\dfrac{\pi}{6}=-\dfrac{1}{2}=-0.5$ : at $\dfrac{\pi}{6}$ cosine is already falling.
$\tan x$	$\dfrac{1}{\cos^{2}x}$	$f'\left(\dfrac{\pi}{4}\right)=2$
Example: $\dfrac{1}{(\sqrt{2}/2)^{2}}=\dfrac{1}{0.5}=2$ , and this derivative is at least 1 everywhere.
$A\sin(\omega t+\varphi)+k$	$A\omega\cos(\omega t+\varphi)$	amplitude $A\omega$
Example: $A=30$ and $\omega=\dfrac{\pi}{6}$ give $A\omega=5\pi\approx 15.71$ thousand dollars per month.
$\sin x$	$-\cos x$	wrong rule that does not exist
Example: At $x=0$ sine RISES, so $f'(0)>0$ : $-\cos 0=-1$ is absurd, $\cos 0=1$ is right. What to do: The minus sign belongs to the other line, $(\cos x)'=-\sin x$ . Check the sign on a graph, not from memory.
$\sin(\omega t)$	$\cos(\omega t)$	wrong forgotten factor
Example: $\left(\sin\dfrac{\pi t}{6}\right)'=\dfrac{\pi}{6}\cos\dfrac{\pi t}{6}$ , whose value at $t=0$ is $0.5236$ and not 1. What to do: The inner layer leaves its coefficient $\omega$ AS A FACTOR in front of the cosine.
$\sin x$ , in degrees	$\cos x$	wrong rule that does not exist
Example: In degrees $\dfrac{\sin(0.1^{\circ})}{0.1}=0.01745$ and not 1, so the derivative is $\dfrac{\pi}{180}\cos x\approx 0.01745\cos x$ . What to do: Set the calculator to radians before the first calculation, and convert every angle given in degrees.

The three red lines are, in order, the memory error, the chain error and the unit error. All three can be caught by a five-second numerical check.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Answering the month of the peak when the question asks for the month of fastest growth

the whole question, 4 marks out of 10

What not to write

“Sales rise fastest in December, since that is the maximum of the year.”

What to write

“The maximum is in December, but $f'(12)=0$ : the rise is over there. Growth is maximal in September, $t=9$ , where $f''(9)=0$ and $f'(9)=5\pi\approx 15.71$ thousand dollars per month.”

t=9

the distance from the mean is zero while the derivative is at its maximum; at

t=12

it is the reverse. The maximum of one always falls on the zero of the other.

Why: This is the real obstacle of the chapter and the one question only the derivative answers: at the peak, the rate of change is ZERO, and the moment when the rate is extreme is the one where the level is ordinary.

2. Differentiating with a calculator left in degrees

the whole table, so the whole problem

What not to write

“The calculator displays degrees, so I differentiate in degrees, it only changes the display.”

What to write

“I switch to radians before the first calculation: the table $(\sin x)'=\cos x$ is true only in radians, because $\displaystyle\lim_{x\to 0}\dfrac{\sin x}{x}=1$ is true only in radians. In degrees the derivative would be $\dfrac{\pi}{180}\cos x$ .”

Why: The stray factor $\dfrac{\pi}{180}\approx 0.01745$ would contaminate every chain rule and every optimization, and both answers stay dimensionally consistent, so the unit catches nothing.

3. Swapping the derivative of sine and that of cosine

2 marks, and the decision is reversed

What not to write

“The derivative of sine is minus cosine.”

What to write

“ $(\sin x)'=\cos x$ , with no sign at all, and $(\cos x)'=-\sin x$ . At $x=0$ the sine curve rises, so its derivative is positive there, and $\cos 0=1$ fits.”

x=0

, the sine curve RISES: its tangent line (green) has slope

\cos 0=1

. A derivative “

-\cos x

” would give the red line, which goes down.

Why: The memory error is harmless, its effect is not: on a seasonal model it swaps the month of maximum growth and that of maximum decline, so it reverses the management recommendation.

4. Forgetting the factor left by the inner layer

3 marks, and a 91 percent error on the rate of change

What not to write

“ $f(t)=30\sin\left(\dfrac{\pi t}{6}+\dfrac{\pi}{2}\right)+110$ , so $f'(t)=30\cos\left(\dfrac{\pi t}{6}+\dfrac{\pi}{2}\right)$ .”

What to write

“The inner layer $u(t)=\dfrac{\pi t}{6}+\dfrac{\pi}{2}$ has derivative $\dfrac{\pi}{6}$ , so $f'(t)=30\times\dfrac{\pi}{6}\cos\left(\dfrac{\pi t}{6}+\dfrac{\pi}{2}\right)=5\pi\cos\left(\dfrac{\pi t}{6}+\dfrac{\pi}{2}\right)$ .”

The correct rate of change (green) oscillates between

-5\pi

and

5\pi

, about

15.7

thousand dollars per month. Without the factor

\frac{\pi}{6}

from the inner layer, we announce

30

(red): nearly double.

Why: Announcing a rate amplitude of 30 instead of $5\pi\approx 15.708$ overestimates it by a factor of $\dfrac{6}{\pi}\approx 1.91$ : on a cash-flow plan, setting aside nearly double the real need.

5. Multiplying the period by the coefficient instead of dividing

2 marks, and the whole timeline of the problem

What not to write

“The period of $\sin(3t)$ is $6\pi$ .”

What to write

“The period of $\sin(\omega t)$ is $\dfrac{2\pi}{\omega}$ , so that of $\sin(3t)$ is $\dfrac{2\pi}{3}\approx 2.094$ . Check: $\sin\left(3\left(t+\dfrac{2\pi}{3}\right)\right)=\sin(3t+2\pi)=\sin(3t)$ .”

Over

[0, 2\pi]

\sin 3t

makes THREE full cycles: each lasts

\frac{2\pi}{3}\approx 2.09

. The coefficient

3

speeds up the cycle, it divides the period instead of multiplying it.

Why: Multiplying the argument by 3 makes the phenomenon cycle three times FASTER: the coefficient divides the period. The value $6\pi$ would be that of $\sin\left(\dfrac{t}{3}\right)$ .

6. Confusing the amplitude with the maximum value

2 marks, and the derivative follows

What not to write

“The amplitude of the sales model is 140, since that is the top of the curve.”

What to write

“The amplitude is $A=\dfrac{\max-\min}{2}=\dfrac{140-80}{2}=30$ , measured from the midline $y=k=110$ . The maximum value is $k+A=140$ .”

Why: The error is double: taking 140 for $A$ multiplies the amplitude of the rate $A\omega$ by more than four, so it also spoils the whole derivative part of the problem.

7. Moving the fundamental limit from zero to infinity

1.5 marks, and the question on asymptotic behaviour

What not to write

“ $\dfrac{\sin x}{x}$ tends to 1 as $x$ tends to infinity.”

What to write

“The limit 1 is the one at ZERO, and it is a $\dfrac{0}{0}$ form. At infinity, $\left|\dfrac{\sin x}{x}\right|\le\dfrac{1}{x}\to 0$ , so $\displaystyle\lim_{x\to+\infty}\dfrac{\sin x}{x}=0$ .”

Near

0

, the curve of

\frac{\sin x}{x}

arrives at height

1

(green open dot). Toward infinity, it stays squeezed between

-\frac{1}{x}

and

\frac{1}{x}

(dashed) and flattens onto

0

Why: The two situations are unrelated: at zero the numerator and the denominator vanish together, at infinity the numerator stays bounded while the denominator grows. At $x=100$ the quotient is about $-0.0051$ .

8. Answering 1 to a sine limit whose coefficients differ

2 marks, and the same reflex repeats on every limit of the page

What not to write

“ $\displaystyle\lim_{x\to 0}\dfrac{\sin 5x}{3x}=1$ , it is the fundamental limit.”

What to write

“Force the same argument on top and bottom: $\dfrac{\sin 5x}{3x}=\dfrac{5}{3}\times\dfrac{\sin 5x}{5x}$ , and the second factor tends to 1, so the limit is $\dfrac{5}{3}\approx 1.6667$ .”

The curve of

\frac{\sin 5x}{3x}

arrives at

\left(0, \frac{5}{3}\right)

(green), not at height

1

(red): the reference limit asks for the SAME argument on top and bottom.

Why: The reference limit requires the SAME argument in the numerator and the denominator. When they differ, the limit is the ratio of the coefficients, never 1.

9. Adding the amplitudes of two out-of-phase cycles

3 marks, and a seasonality overestimated by 31 percent

What not to write

“ $E(t)=5\sin u+12\cos u$ , so the amplitude is $5+12=17$ .”

What to write

“Two cycles of the same period add up to a single one: $E(t)=R\sin(u+\varphi)$ with $R=\sqrt{5^{2}+12^{2}}=13$ . The amplitude is 13, and the gap between the best and the worst month is $2R=26$ .”

5\sin u

and

12\cos u

do not peak at the same moment: their sum (green) peaks at

\sqrt{5^{2}+12^{2}}=13

, never at

5+12=17

(red).

Why: The two effects are a quarter turn out of phase, so they never fully reinforce each other: we add the SQUARES of the amplitudes, not the amplitudes.

Which method to choose

Level or rate of change? The verb of the question decides

Read the verb of the question before touching the model. Only three families, and they do not open the same line of calculation.

If “what will the value be”, “which month is the busiest”, “what is the maximum” → we stay on $f$ : evaluate $f(t)$ , or solve $f'(t)=0$ then study the SIGN of $f'$

Example: $f'(t)=0$ at $t=0$ , 6 and 12; the sign gives a minimum in June, $f(6)=80$

the derivative vanishing is not enough to conclude, it is the change of sign that names the maximum
If “fastest”, “most rapidly”, “strongest acceleration” → we move to $f'$ and look for ITS maximum, so we solve $f''(t)=0$

Example: $f''(t)=0$ at $t=3$ and $t=9$ ; since $f'(9)>0$ , the fastest rise is in September

of the two roots, the one where $f'>0$ is the rise, the one where $f'<0$ is the fall
If “by how much per month”, “at what rate”, “what monthly change” → it is $f'$ EVALUATED at one time, given with its full unit

Example: $f'(3)=-5\pi\approx -15.71$ thousand dollars per month, so a decrease

an answer without a unit is not worth half the marks of the question
If “how many times a year”, “how often”, “what is the period” → $T=\dfrac{2\pi}{\omega}$ , and we check that the result is consistent with the spacing of the data

Example: $\omega=\dfrac{\pi}{6}$ gives $T=12$ months, which is indeed a yearly cycle

a $T$ that is not a natural multiple of the data spacing signals a reading error

If no branch applies, the question is about the MODEL itself, so about reading the four parameters: see the model write-up.

Which technique for which form of expression

Look at the written form, not the theme of the chapter. Five forms cover the whole page.

If $\dfrac{\sin(ax)}{bx}$ with $x\to 0$ → $\dfrac{0}{0}$ form: force the same argument, the limit is $\dfrac{a}{b}$

Example: $\dfrac{\sin 5x}{3x}=\dfrac{5}{3}\times\dfrac{\sin 5x}{5x}\to\dfrac{5}{3}$

same move for $\dfrac{\sin ax}{\sin bx}$ , by building two reference quotients
If $\dfrac{1-\cos x}{x}$ with $x\to 0$ → multiply top and bottom by the conjugate $1+\cos x$ , then use Pythagoras

Example: $\dfrac{\sin x}{x}\times\dfrac{\sin x}{1+\cos x}\to 1\times 0=0$

it is the conjugate technique from the chapter on limits, only the identity changes
If $\dfrac{\sin x}{x}$ with $x\to+\infty$ → this is NOT an indeterminate form: squeeze between $\pm\dfrac{1}{x}$ , the limit is 0

Example: at $x=100$ , the quotient is $-0.0051$

the numerator stays bounded, the denominator grows: the squeeze theorem is enough
If $a\sin u+b\cos u$ , same argument $u$ in both terms → identify $R=\sqrt{a^{2}+b^{2}}$ and $\tan\varphi=\dfrac{b}{a}$ , then work with $R\sin(u+\varphi)$

Example: $5\sin u+12\cos u=13\sin(u+1.176)$

the compact form gives the extreme values with no sign study at all
If $A\sin(\omega t+\varphi)+k$ to differentiate → chain rule: $A\omega\cos(\omega t+\varphi)$ , the $k$ disappears

Example: $30\sin\left(\dfrac{\pi t}{6}+\dfrac{\pi}{2}\right)+110$ gives $5\pi\cos\left(\dfrac{\pi t}{6}+\dfrac{\pi}{2}\right)$

the mean level has no effect on the rate of change, two series with very different levels can have the same derivative

L'Hospital's rule is not on the outline of this course at this stage: a trigonometric $\dfrac{0}{0}$ form is handled with the two reference limits, never otherwise.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Fitting a sinusoidal model to data

When to use it: The statement gives a table of values, a graph, or simply the maximum and the minimum of a phenomenon that comes back.

1 Read off the observed maximum and minimum, with the time of each, and write them with their unit.
2 Compute the midline $k=\dfrac{\max+\min}{2}$ : it comes first because it depends on no other parameter.
3 Compute the amplitude $A=\dfrac{\max-\min}{2}$ , the HALF-difference, and state that it is measured from the midline.
4 Read the duration of a full cycle off the data, call it $T$ , then set $\omega=\dfrac{2\pi}{T}$ in radians per unit of time.
5 Write the argument at the time of the first maximum and set it equal to $\dfrac{\pi}{2}$ : this gives $\varphi$ .
6 Check the model on a third data point, other than the maximum and the minimum.

Concluding sentence

“On $[0, 12]$ , monthly sales are modelled by $f(t)=30\sin\left(\dfrac{\pi t}{6}+\dfrac{\pi}{2}\right)+110$ , in thousands of dollars, where $t$ is the month number.”

The trap: Skipping step 4 and writing $\omega=T$ : the angular frequency is not the period, it is its inverse up to a factor of $2\pi$ .

Marking: 1 mark per parameter, 1 mark for the units, 1 mark for the check on a third point

Answering “at what time is growth fastest?”

When to use it: The statement uses a comparative of speed: fastest, most rapidly, strongest increase, maximum acceleration.

1 Differentiate the model by the chain rule, $f'(t)=A\omega\cos(\omega t+\varphi)$ , keeping the factor $\omega$ and writing the unit of $f'$ .
2 State explicitly that the question is about the maximum of $f'$ , so about $f''$ vanishing, and NOT about $f'$ vanishing.
3 Differentiate a second time and solve $f''(t)=0$ on the required interval, writing the condition in the form argument $=\dfrac{\pi}{2}+n\pi$ .
4 Among the roots, choose the one where $f'>0$ , since the other is the moment of fastest fall, and justify the choice by the sign.
5 Evaluate $f'$ at that time and give the rate with its full unit.
6 Check with the quarter-period rule: the time found falls $\dfrac{T}{4}$ before a peak.

Concluding sentence

“Growth is maximal at $t=9$ , that is, in September: $f''(9)=0$ and $f'(9)=5\pi\approx 15.71$ , so sales then increase by $15.71$ thousand dollars per month, the highest rate of the year.”

The trap: Concluding at step 3 without choosing between the two roots: both make $f''$ vanish, but one is the fastest fall.

Marking: 2 marks for the derivative with its factor, 2 marks for $f''=0$ , 1 mark for the choice of root, 1 mark for the unit

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

The calculator mode, before the first calculation

Type $\sin(1)$ and read the screen. In radians the display gives $0.8415$ ; in degrees it gives $0.0175$ . On a TI, the setting is under MODE, the RADIAN row; on a Casio, under SET UP, the Angle row.

A screen that shows $0.0175$ says that every derivative on the paper is off by a factor of $\dfrac{\pi}{180}$ , including the ones already written.
The unit of the derivative

$f'$ reads in units of $f$ DIVIDED by units of $t$ . If $f$ is in thousands of dollars and $t$ in months, then $A\omega$ is in thousands of dollars per month. A rate amplitude identical to the level amplitude signals a forgotten factor $\omega$ .

Reading $A\omega=30$ when $A=30$ is impossible, unless $\omega=1$ : here $\omega=\dfrac{\pi}{6}$ , so $A\omega=5\pi\approx 15.71$ .
The quarter period

Subtract the time of maximum growth from that of the next peak: the gap must be exactly $\dfrac{T}{4}$ . The check is independent of all the calculation that precedes it.

$12-9=3$ and $\dfrac{12}{4}=3$ : the check passes. A gap of 6 would say that we solved $f'=0$ instead of $f''=0$ .
The maximum is indeed $k+A$

Plug the time of the peak into the written model and compare with the observed maximum. Then check that the mean of the two extreme values gives back $k$ and that their half-gap gives back $A$ .

$f(12)=140$ recovers the observed maximum; $\dfrac{140+80}{2}=110$ and $\dfrac{140-80}{2}=30$ close the loop.
The table of values against the data

Enter the model in Y1, display the TABLE with a step of 1, and compare three rows with the data of the statement. A gap of a few percent is normal for a smoothing; a gap that keeps the same sign over all the months signals an underlying trend that was not captured.

The model predicts 110 in September against 112 observed, which is $1.8$ percent: acceptable. Three positive gaps in a row would not be.

The typical problem, taken apart

Average monthly temperature: from the scatter plot to the month that warms fastest

A city records its average monthly temperatures. The warmest month is July, $t=7$ , with $22.0$ degrees Celsius; the coldest month is January, $t=1$ , with $-8.0$ degrees Celsius. We model with $\theta(t)=A\sin(\omega t+\varphi)+k$ , where $t$ is the month number.

a) Give the four parameters and write $\theta(t)$ . b) Give $\theta'(t)$ with its unit. c) Find the month when the temperature rises fastest and give this rate. d) Check with the quarter-period rule.

Step 1

$k=\dfrac{22+(-8)}{2}=7$ degrees Celsius, and $A=\dfrac{22-(-8)}{2}=15$ degrees Celsius.

Why

The midline is written first because it depends on nothing else. Writing $A=22$ here would confuse the amplitude with the maximum, and the error would carry through to the rate in question c).

Step 2

The cycle is yearly, so $T=12$ months and $\omega=\dfrac{2\pi}{12}=\dfrac{\pi}{6}\approx 0.5236$ radian per month.

Why

The angular frequency is not the period: it is what we put INSIDE the sine. Naming its unit, radian per month, is what will make the unit of $\theta'$ obvious two steps later.

Step 3

The maximum is reached when the argument equals $\dfrac{\pi}{2}$ : $\dfrac{7\pi}{6}+\varphi=\dfrac{\pi}{2}$ , so $\varphi=\dfrac{\pi}{2}-\dfrac{7\pi}{6}=-\dfrac{2\pi}{3}$ . Thus $\theta(t)=15\sin\left(\dfrac{\pi t}{6}-\dfrac{2\pi}{3}\right)+7$ .

Why

The phase shift is read from the position of the FIRST maximum, never from the first value in the table. It is the only parameter that requires solving an equation, which is why it comes last.

Step 4

Check of the model in January: $\theta(1)=15\sin\left(\dfrac{\pi}{6}-\dfrac{2\pi}{3}\right)+7=15\sin\left(-\dfrac{\pi}{2}\right)+7=-15+7=-8$ .

Why

A model is checked on a data point that was not used to build it, or at the very least on the second extreme value. This line earns the check mark of the marking scheme and catches a sign error on $\varphi$ .

Step 5

Chain rule: $\theta'(t)=15\times\dfrac{\pi}{6}\cos\left(\dfrac{\pi t}{6}-\dfrac{2\pi}{3}\right)=\dfrac{5\pi}{2}\cos\left(\dfrac{\pi t}{6}-\dfrac{2\pi}{3}\right)$ , in degrees Celsius per month, with amplitude $\dfrac{5\pi}{2}\approx 7.854$ .

Why

The factor $\dfrac{\pi}{6}$ comes from the inner layer and it is the number one oversight of the chapter: without it, we would announce a maximum rate of 15 degrees per month instead of $7.85$ , nearly double.

Step 6

The question is about the maximum of $\theta'$ , so about $\theta''=0$ . Now $\theta''(t)=-\dfrac{5\pi^{2}}{12}\sin\left(\dfrac{\pi t}{6}-\dfrac{2\pi}{3}\right)$ , which vanishes when $\dfrac{\pi t}{6}-\dfrac{2\pi}{3}=n\pi$ , that is $t=4+6n$ : over one year, $t=4$ and $t=10$ .

Why

This is the step everyone skips: solving $\theta'=0$ would give July and January, that is, the months when the temperature has FINISHED rising or falling. Saying so out loud is worth a mark for rigour.

Step 7

$\theta'(4)=\dfrac{5\pi}{2}\cos 0=+\dfrac{5\pi}{2}\approx +7.854$ and $\theta'(10)=\dfrac{5\pi}{2}\cos\pi\approx -7.854$ . We keep $t=4$ , April, where the derivative is positive. Incidentally, $\theta(4)=7$ , the annual mean.

Why

Of the two roots, only the one where $\theta'>0$ answers the question asked: the other is the month that cools fastest. The fact that $\theta(4)$ equals exactly $k$ is no coincidence, it is the signature of the inflection point.

Step 8

Check: the peak is at $t=7$ and $\dfrac{T}{4}=\dfrac{12}{4}=3$ , and $7-3=4$ . The time found does fall a quarter of a period before the peak.

Why

This check is independent of everything that precedes it: it catches an error on $\varphi$ just as well as a mix-up between $\theta'=0$ and $\theta''=0$ , in five seconds and without redoing the problem.

The conclusion, written out

“The model is $\theta(t)=15\sin\left(\dfrac{\pi t}{6}-\dfrac{2\pi}{3}\right)+7$ , in degrees Celsius. The temperature rises fastest in April, $t=4$ : $\theta''(4)=0$ and $\theta'(4)=\dfrac{5\pi}{2}\approx 7.85$ degree Celsius per month, the fastest monthly warming of the year, while the temperature there equals exactly the annual mean of 7 degrees.”

The classic mistake on this problem: Answering July, because it is the warmest month. In July $\theta'(7)=0$ : the warming is over there. The question asked for the maximum of the DERIVATIVE, not that of the function.

Learn by heart

• $180^{\circ}=\pi$ rad, and we ALWAYS differentiate in radians: in degrees, $(\sin x)'=\dfrac{\pi}{180}\cos x$ .
• $(\sin x)'=\cos x$ , $(\cos x)'=-\sin x$ , $(\tan x)'=\dfrac{1}{\cos^{2}x}=1+\tan^{2}x\ge 1$ .
• $\displaystyle\lim_{x\to 0}\dfrac{\sin x}{x}=1$ at ZERO, and $\displaystyle\lim_{x\to+\infty}\dfrac{\sin x}{x}=0$ : the two are unrelated.
• For $f(t)=A\sin(\omega t+\varphi)+k$ : $k=\dfrac{\max+\min}{2}$ , $A=\dfrac{\max-\min}{2}$ , $T=\dfrac{2\pi}{\omega}$ , and $\max=k+A$ .
• $f'(t)=A\omega\cos(\omega t+\varphi)$ : the amplitude of the RATE OF CHANGE is $A\omega$ , in units of $f$ per unit of time.
• The peak is where $f'=0$ ; maximum growth is a QUARTER OF A PERIOD earlier, where $f''=0$ and where $f$ equals $k$ .
• $a\sin u+b\cos u=R\sin(u+\varphi)$ with $R=\sqrt{a^{2}+b^{2}}$ : we add the SQUARES, never the amplitudes.
• The Pythagorean identity gives the square, so two possible signs: the quadrant decides, never the calculator.

Frequently asked questions

Why must we differentiate in radians and never in degrees?

Because the table of derivatives rests on the limit of sine of x over x as x tends to zero, which equals 1 in radians only. In degrees this limit is pi over 180, about 0.01745, and the derivative of sine would become 0.01745 times cosine. This stray factor would spread into every chain rule. So we set the calculator to radians before the first calculation.

How do we find the month when a periodic quantity increases fastest?

We do not look for the maximum of the function but for that of its derivative. Differentiate once to get the rate of change, a second time, then solve the second derivative equal to zero. Of the two times found over one period, keep the one where the first derivative is positive: it is the fastest rise, the other is the fastest fall.

What is the difference between the amplitude and the maximum value of a sinusoidal model?

The amplitude is the half-difference between the maximum and the minimum, and it is measured from the midline, not from zero. The maximum value is the midline plus the amplitude. On a sales model with midline 110 and amplitude 30, the maximum is 140 thousand dollars, not 30. Confusing the two also throws off the derivative, whose amplitude is the amplitude times the angular frequency.

How do we read the four parameters of a sinusoid off a graph of data?

In a fixed order. The mean of the maximum and the minimum gives the midline. Their half-difference gives the amplitude. The duration of a full cycle gives the period, from which we get the angular frequency by dividing two pi by this duration. Finally, the position of the first maximum gives the phase shift, by writing that the argument equals pi over two there.

What is trigonometry used for in a Social Science program?

To model everything that comes back: retail sales over twelve months, emergency room traffic over seven days, average temperature, seasonal migration, the circadian rhythm. Sine is not a triangle tool here, it is the one simple model of a cyclical phenomenon, and its derivative tells us when the quantity is changing fastest.

Practise it

Corrected exercises: Trigonometric functions and their derivatives, 201-103

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

11 corrected exercises
110 points
150 minutes

Do the exercises

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© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/cegep-103-trigonometric-derivatives. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: trigonometric functions and their derivatives (201-103)

The essentials

The circle and the radian, and what the Pythagorean identity does not say

The periodic model: four letters, four management questions

The table of derivatives, and the factor that follows it everywhere

The peak and maximum growth, a quarter of a period apart

The rules in table form

The derivatives of the chapter, with their numerical check

The mistakes that cost marks

1. Answering the month of the peak when the question asks for the month of fastest growth

2. Differentiating with a calculator left in degrees

3. Swapping the derivative of sine and that of cosine

4. Forgetting the factor left by the inner layer

5. Multiplying the period by the coefficient instead of dividing

6. Confusing the amplitude with the maximum value

7. Moving the fundamental limit from zero to infinity

8. Answering 1 to a sine limit whose coefficients differ

9. Adding the amplitudes of two out-of-phase cycles

Which method to choose

Level or rate of change? The verb of the question decides

Which technique for which form of expression

How the answer is expected to be written

Fitting a sinusoidal model to data

Answering “at what time is growth fastest?”

Check before you hand in

The typical problem, taken apart

Average monthly temperature: from the scatter plot to the month that warms fastest

Learn by heart

Frequently asked questions

Why must we differentiate in radians and never in degrees?

How do we find the month when a periodic quantity increases fastest?

What is the difference between the amplitude and the maximum value of a sinusoidal model?

How do we read the four parameters of a sinusoid off a graph of data?

What is trigonometry used for in a Social Science program?

Corrected exercises: Trigonometric functions and their derivatives, 201-103

See also

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